1.1 Introduction
What is Integers :- It can be Positive,negative, or zero but do not include any fractional (5/6 ,3/4,7/2) or decimal parts ( 2.13,-1.22,0.08 etc..) The set of integers denoted by "Z"
Integers Example :- Z= { .....-2 -1 0 1 2....}
What is Rational Numbers :- Numbers which can be written in the form of p/q where both p and q are integers and Q =/=0 (Q should not be Zero ). All rational numbers can be written either in the form of terminating decimals or non-terminating repeating decimals.
The set of rational numbers denoted by" q "
Example :-
q= { 3 can be written in p/q form as 3/1 }
{ 0.01 can be written in p/q form as 1/100 }
What is Irrational Numbers :- Numbers which cannot be expressed in the form of p/q are irrational. These include numbers like _/2, (root 2), _/3, _/5 and mathematical quantities like "pi". When these are written as decimals they are non-terminating,non-recurring
Example :- _/3 (root 3) = 1.7320508.........so on (non-terminating).
What is Real Numbers :- The set of rational and irrational numbers together are called real numbers.
Exercise -1.1
What is Terminating :- It is a decimal That finally ends after a finite numbers of digits.
Example :- 1/4 = 0.25 (It end after few decimal places)
What is Non-terminating :- It is a decimal that continues endlessly
Example :- "pi" or 22/7= 3.1415.....so on (if You keep on dividing it wont end)
1. which of the following rational numbers are terminating and which are non-terminating repeating in their decimal form?
(1) 2/5
Ans:- Terminating , because if we divide we get 2/5 = 0.4
2) 17/18
Ans :- Non-terminating, because if we divide it keep repeating 17/18 = 0.94444.....
3) 15/16
Ans :- Terminating, because 15/16 = 0.9375
4) 7/40
Ans :- Terminating ,because 7/40 = 0.175
5) 9/11
Ans :- Non-Terminating, because 9/11= 0.1818....so on
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3. Classify the numbers given below as rational or irrational.
1) 2* 1
2
Ans :- Rational , converting mixed fraction 2*1/2 (2*1/2 = 4+1/2 = 5/2) into improper fraction we get 5/2 (p/q)
2) √24
Ans :- Irrational, Because √24 ( 4.89897948.....) doesn't have perfect square it never terminates or repeats .
3) √16
Ans :- Rational, have perfect square 4*4=16 ......_/16=4 or 4/1 (p/q)
4) 7.7777..
Ans :- Clearly irrational.
5) √ 4/9
Ans:- Rational.. 2*2/3*3 = 2/3 (p/q)
6) - √30
Ans:- Irrational, - _/30 = 5.47722.....dont have perfect square ...cant have p/q form
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4.) Represent the following real numbers on the number line.
(1) 3/4
Ans :-
(2) -9/10
Ans ;-
(3) 27/3
Ans :- 27/3 = 9
<----0---1---2---3---4---5---6---7---8---9----->
(4) find Root 5 on number line.
Ans :- In order to find square root of any numbers always imagine any root number on hypotenuse side first
Pythagoras Theorem :- In a right angled triangle the square of the hypotenuse is equal to the sum of the squares of its other two sides.
That is , AC^2 =AB^2 + BC^2
Now Lets solve _/5
Step 1 :- Take 5 in form of sum of perfect squares ,
that is 2^2 (4) + 1^2 (1) = _/ 5
Step1 :- Express "5" in such a manner that is written as sum of perfect squares.
Step2 :- 5 = 4 + 1 , as we know perfect sq of 4 =2^2 and for 1 = 1^2
Step3:- 5 = 2^2 (AB) + 1^2 (BC). The distance between AB = 2 units and BC =1unit
Step4 :- According to Pytho.Theorem
Ac^2 = AB^2 + BC^2
= 2^2 + 1^2
= 4 + 1
= 5
If you transfer square of AC on right side than 5 become _/5 (root 5)
AC = _/5
Step5 :- Take a compass and take distance of AC as Radius and draw an arc passing through C till it reach the point "S" on the number line ..click and see the figure above.
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(5) find √ -16 on number line.
Ans :- As we know 16 have perfect square
- √16 =4*4 = -4 .
<-------|------|------|--------|-------|------>
-4 -3 -2 -1 0
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Q? Are all integers also in Real numbers?why
A) Yes,Because Set of Integer ( Z) is a subset of Real numbers ( R ) .
Also we can say All Rational,whole,natural too is a subset of Real numbers See fig.
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Fundamental theorem of arithmetic :- All natural numbers, except 1, can be written as a product of their prime factors.(2,3,5,7,11,13.....)
NOTE : 1 is neither a composite nor a prime.
Example :- 3 =3, ...6 =2*3....45= 3*3*5.....253 = 11*23.....52 = 2*2*13 so on....
*) What is prime numbers
A) A number that is divisible only by itself and 1 (2,3,5,7,11,13.... will only divide by itself or
with 1 not with any other numbers)
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Example-1. Find the HCF and LCM of 12 and 18 by the prime factorization method.
sol) We have
12 = 2 * 2 * 3 = 2^2 * 3^1
18 = 2 * 3 * 3 = 2^1 * 3^2
Note that
HCF (12, 18) = 2^1 * 3^1 = 6 = product of the smallest power of each common prime factors in the numbers.
LCM (12, 18) = 2^2 * 3^2 = 36 = Product of the greatest power of each prime factors,, in the numbers.
From the example above, you might have noticed that HCF(12,18) * LCM( 12, 18) = 12 * 18.
In fact, we can verify that for any two positive integers "a" and "b",
HCF(a,b) * LCM(a,b) = a*b.
We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers.
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Example 2 . Consider the number 4^n, where "n" is a natural number. Check whether there is any value of "n" for which 4^n ends with the digit zero?
sol) For the number 4^n to end with digit zero for any natural number "n" it should be divisible by "5"
This means that the prime factorization of 4^n should contain the prime number "5"
But it is not possible because
4^n = (2)^2n
so "2" is the only prime in the factorization of 4^n.
Since "5" is not present in the prime factorization, so there is no natural number "n" for which "4^n" ends with the digit zero.
Exercise- 1.2
Q) What is Prime factors
A) It is a process of finding which prime numbers( 2,3,5,7 ....) can be multiplied together to make the original number
1. Express 140 as a product of its prime factors.
A)
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2) Express 156 number as a product of prime factors.
A) 156 = { 2*2*3*13 } = 2^2 * 3 * 13
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3) Express 3825 number as a product of prime factors.
A) 3825 = { 3*3*5*5*17} = 3^2 *5^2 *17
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4) Express 5005 number as a product of prime factors.
A) 5005 = 5*7*11*13
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5) Express 7429 number as a product of prime factors.
A) 7429 = 17*19*23
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2) Find the LCM and HCF of the following integers by the prime factorization method.
(1) Find the LCM and HCF of 12, 15 and 21
Sol) 12 = 2*2*3
15 = 3*5
21 = 3*7
HCF : Product of smallest power of each common prime factor
= 3^1 = 3
LCM :Product of greatest power of each prime factor
=> 2*2*3*5*7=420
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(2) Find the LCM and HCF of 17,23, and 29
Sol) 17 = 1*17
23 = 1*23
29 = 1*29
HCF :- 1
LCM :- 1*17*29*23 = 11,339
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(3) Find the LCM and HCF of 8,9,25
sol) 8 = 2*2*2
9 = 3*3
25 = 5*5
HCF :- Since there are no common prime factors we can take HCF =1
LCM :- 2*2*2*3*3*5*5=1800
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(4) Find the LCM and HCF of 72 and 108
sol) 72 = 2*2*2*3*3
108 = 2*2*3*3*3
HCF :- 2*2*3*3 = 36
LCM :- 2*2*2*3*3*3 = 216
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(5) Find LCM and HCF 306 and 657
sol) 306 = 2*3*3*17
657 = 3*3*73
HCF : 3*3 =9
LCM : 2*3*3*17*73 = 22338
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(3) check whether 6^n can end with the digit 0 for any natural number n.
Sol) Take an example of a numbers which ends with the digit "zero"
Ex: 20 = 2*2*5 or 100 = 2*2*5*5
Here we can clearly see that numbers ending with "0" has both 2 & 5 as their prime factors
As we know the prime factors for 6= 2*3.
Clearly it does not end with "0" because it does not have prime factor "5"
Therefore, 6^n cannot end with zero for any natural number "n" ( n=1,2,3,...)
ex : (6)^1= 6, (6) ^2= 36 , (6)^3 =216 so on....as u can see it never ends with zero.
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What is composite number :- If a number can be divided by any other numbers other than 1 and itself
Ex : all prime numbers (2,3,5,7...) cannot be divided by any other number except 1 or itself .so we can say all prime numbers are not composite numbers
similarly if u take 8 it can divided exactly by 2 (beside 1 and itself(8)) hence 8 can be said as composite number
Also product of primes will be a composite number.
4) Explain why 7*11*13+13 and 7*6*5*4*3*2*1+5 are composite numbers.
Sol) 7*11*13+13 = 1014 is a composite number as it can be divided by [ 1014/2=507}
other number "2" beside 1 and itself
Second method :
7*11*13+13
Take 13 as common
13[7*11+1] or 13 {odd +1 }
13 [77+1] or odd {odd+odd}
13 [78]=1014 or odd { even) => odd *even => even
!!) 7*6*5*4*3*2*1+5 = 5045 composite number as its divisible by "5" 5045/5 = 1009 (beside 1 and itself 5045 )
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5) how will you show that ( 17*11*2)+( 17*11*5) is a composite number? Explain.
sol) 17*11 (2+5)
= 17*11 (7)
= 17*11 (odd)
= odd (odd) = 17*11*7 = product of primes will be a composite number .
Also 17*11*7=1309 which is divisible by 1309/7=187 (beside 1 and itself(1309)
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(*) Rational Numbers and their Decimal expansions
Let us consider the following terminating decimal forms of some rational numbers:
Now let us express them in the form of p/q
(!) 0.375
sol) 375 = 375
1000 10^3
(!!) 1.04
sol) 104 = 104
100 10^2
(!!!) 0.0875
sol) 875 = 875
10000 10^4
(!V) 12.5
sol) 125 = 125
10 10^1
(V) 0.00025
sol) 25 = 25
100000 10^5
We see that all terminating decimals taken by us can be expressed as rational numbers whose denominators are powers of "10".
Let us now prime factorize the numerator and denominator and then express in the simplest rational form:
Now
(!) 0.375
sol) 375 = 3 * 5^3 = 3 = 3
10^3 2^3 * 5^3 2^3 8
(!!) 1.04
sol) 104 = 2^3 * 13 = 26 = 26
100 2^2 * 5^2 5^2 25
(!!!) 0.0875
sol) 875 = 5^3 * 7 = 7
10^4 2^4 * 5^4 2^4 * 5
(!V) 12.5
sol) 125 = 5^3 = 25
10 2*5 2
(V) 0.00025
sol) 25 = 5^2 = 1 = 1
10^5 2^5 * 5^5 2^5 * 5^3 4000
Do you see a pattern in the denominators?
It appears that when the decimal expression is expressed in its simplest rational form then "p" and "q" are co-prime and the denominator (i.e, q) has only
powers of 2, or powers of 5, or both.
This is because the powers of 10 can only have powers of "2" and "5" as factors.
Let us conclude
Any rational number which has a decimal expansion that terminates can be expressed as a rational number whose denominator is a power of "10".
The only prime factors of "10" are "2" and "5".
So, when we simplify the rational number, we find that the number if of the form p/q,
where the prime factorization of "q" is of the form
2^n * 5^m, and n,m are some non-negative integers.
Theorem-1.2 : Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form
p , where p and q are co-prime, and the prime
q
factorization of q is of the form 2^n*5^m, where n,m are non-negative integers.
If we have a rational number in the form p/q, and the prime factorization of q is of the form 2^n*5^m, where n,m are non-negative integers, then does p/q have a terminating decimal expansion?
So, it seems to make sense to convert a rational number of the form p/q, where q is of the form 2^n*5^m, to an equivalent rational number of the form a/b, where b is a power of 10. let us go back to our examples above and work backwards.
(!) 25
2
sol) 5^2 multiply and divide by "5"
2
5^3 = 125 = 12.5
2 * 5 10
(!!) 26
25
sol) 13 * 2 multiply and divide by "2^2"
5^2
13 * 2^3 = 104 = 1.04
2^2 * 5^2 10^2
similarly we can show we can convert a rational number of the form p/q, where "q" is of the form 2^n * 5^m, to an equivalent rational number of the form a/b, where "b" is a power of 10.
.^. , the decimal expansion of such a rational number terminates. We find that a rational number of the form p/q, where "q" is a power of 10, will have a terminating decimal expansion.
So, we find that the converse of theorem 1.2 is also true and can be formally stated as :
Theorem 1.3 : Let x = p/q be a rational number, such that the prime factorization of "q" is of the form 2^n * 5^m, where n,m are non-negative integers. Then x has a decimal expansion which terminates.
(*) Non-terminating, recurring decimals in rational numbers
Let us look at the decimal conversion of 1/7
1/7 = 0.1428571428571........which is a non-terminating and recurring decimal.
Notice , the block of digits '1428571' is repeating in the quotient.
Notice that the denominator here, i.e., 7 is not of the form 2^n*5^m
Theorem-1.4 : Let x = p/q be a rational number, such that the prime factorization of q is not of the form 2^n * 5^m , where n,m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating(recurring).
we can conclude that the decimal form of every rational number is either terminating or non-terminating repeating.
Example-3. Using the above theorems, without actual division, state whether the following rational numbers are terminating or non-terminating decimals.
(!) 16
125
sol) 16 = 16 is terminating decimal.
5*5*5 5^3
(!!) 25
32
sol) 25
2*2*2*2*2
= 25 is terminating decimal
2^5
(!!!) 100
81
sol) 100
3*3*3*3
= 100 is non-terminating, repeating decimal
3^4
(!V) 41
75
sol) 41
3* 5 * 5
= 41 is non-terminating, repeating decimal
3* 5^2
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Example-4 : Write the decimal expansion of the following rational numbers without actual division.
(!) 35
50
sol) 7 * 5
2*5*5
= 7
2* 5
= 7
10^1
= 0.7
(!!) 21
25
sol) 21 multiply and divide by "2^2"
5*5
= 21 * 2^2
5*5*2^2
= 21 * 4
5^2 * 2^2
= 84 = 0.84
10^2
(!!!) 7
8
sol) 7
2* 2* 2
= 7
2^3
= 7 * 5^3
2^3 * 5^3
= 7 * 125
(2 * 5)^3
= 875
(10)^3
= 0.875
Exercise 1.3
1) Write the following rational numbers in their decimal form and also state which are terminating and which have non-terminating , repeating decimal.
1) 3/8
Sol) 3/8 = 0.375 is a terminating decimal
2) 229/400
Sol ) 229/400 = 0.5725 is a terminating decimal.
3) 4 1/5
Sol) convert this mixed fraction into improper fraction we get
What is Proper Fraction :- if the numerator is lesser than denominator ( ex : 2/3)
What is improper fraction :- if the numerator is greater than denominator ( ex : 4/3)
=> 4*1/5
=> 5*4+1
5
=> 20+1
5
=> 21 (improper fraction)
5
=> 21/5 is a terminating decimal
4) 2/11
Sol ) 2/11 = 0.181818.... is a non-terminating decimal.
5) 8/125
sol) 8/125 =0.064.... is a terminating decimal
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Coprime:- Any two numbers "x" and "y" is said to be co-prime if they have no common factor except "1"
Example for co-prime :- x => 14 = 1 *2*7 and 15 = 1 *3*5 ( only '1' common)
Example for relatively or mutually prime :- x => 14 = 1*2*7 and 21 = 1*3*7 ( 1 and 7 common)
Theorem 1.2 :- Let "x" be a rational number whose decimal expansion "terminates". Then
"x" can be expressed in the form p/q, where p and q are "co-prime" , and the prime
factorization of "q" is of the form " 2^n * 5^m ", where n, m are non-negative integers.
---------------------------------------------------------------------------------------------------2) Without actually performing division, state whether the following rational numbers have a terminating decimal form or a non-terminating, repeating decimal form.
In order to prove whether rational numbers terminating or not two conditions should be met.
1) p/q should be co-prime
2) "q" should be in the form 2^n * 5^m
1) 13/3125 = p/q
sol) 1st condition :- Is the fraction is co-prime?
p= 13= 1*13
q= 3125
= 1*5*5*5*5*5 = 1 * 5^5
= 2^0 *5^5 ( 2^n *5 ^m form)
Note :- 2^0 or 3^0 or N ^0 ......= is always 1
Both 13 and 3125 is co-prime because only "1" is common factor
2nd condition :- is the denominator q= 2^n * 5^m form?
q= 1* 5^2 = 2^0 * 5^2 == 2^n * 5^m
As the two condition met .
Thus , 13/3125 is a terminating decimal.
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2) 11 = p/q
12
Sol) 1st condition :- is the fraction is co-prime?
11 = 1*11
12 = 1*2*2*3 yes its co-prime because only "1" is common factor
2nd condition :- is q= 2^n* 5^m form?
q= 2^2 * 3^1 =/= (not equal to) 2^n * 5^m
As the 2nd condition not met 11/12 is non-terminating, repeating decimal.
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3) 64 = p/q
225
Sol ) 64 = 1*2*2*2*2*2*2 =(co-prime)
225 1*5*91
q = 5*91 =/= 2^n * 5^m
Thus, 64/ 225 is non-terminating, repeating decimal.
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4) 15 = 3 = 3 = ( Co-prime)
1600 320 2*2*2*2*2*2*5
q= 2^6 * 5^2 = 2^n * 5^m
Two condition met,, Thus its is 15/1600 is terminating.
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5) 29 = 29 = (co-prime )
343 7*7*7
q= 1 *7^3 = 2^0 * 7^3 =/= 2^n * 5^m
2nd condition not met Thus , 29/343 is non-terminating, repeating decimal.
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6) 23 = (co-prime)
2^3 5^2
q= 2^3 * 5^2 = 2^n *5^m form
Thus it is terminating decimal.
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7) 129
2^2 * 5^7 * 7^5
Sol) Clearly It is non-terminating, repeating decimal.
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8) 9 = 3 * 3 = 3
15 3 * 5 5
Sol) 3/5 (co-prime)
q= 1*5 = 2^0 * 5^1 = 2^n * 5^m
Clearly 9/15 is terminating decimal.
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9) 36 = 9
100 25
Sol ) p= 9 = 3*3
q = 25 = 5*5
1) is it co-prime?
A) yes as 9 and 25 both have different factors
2) is q= 2^n * 5^m form?
A) Yes we can write 25 as 2^0 * 5^2 = 2^n * 5^m (where 2^0=1)
Thus 36/100 is terminating decimal.
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10) 77/210
Sol) p =77 = 1*77
q = 210 = 3 *2*7*5
1) is it co-prime
A) yes
2) Is q= 2^n * 5^m form?
A) No.
Thus, 77/210 is non-terminating, repeating decimal.
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Theorem 1.1 :- Every composite number can be expressed (factorized) as a product or primes, and this factorization is unique, apart from the order in which the prime factors occur.
*) Write the following rationals in decimal form using Theorem 1.1
1) 13
25
Sol) 13/25 = p/q , We know that any number we can be convert into product (multiply) of
primes (2,3,5,7,11.....)
=13
25
= 13 * 2^2 (Multiply and divide by 2^2)
5^2 * 2^2
= 52
(10)^2
transfer q= (10)^2 in numerator it become 0.52
( 1 / 10^2 = 1/100 = 0.01 = 0.01*52 =0.52)
Thus, Decimal form of 13/25 = 0.52
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2) 15
16
Sol) 15 * 5^4 (Multiply and divide by 5^4)
2^4 * 5^4
= 9375
(10)^4
= 0.9375
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3) 23
2^3 * 5^2
Sol) 23 * 5^1 (multiply and divide by 5^1)
2^3 *5 ^2 *5^1
= 23 * 5^1
2^3 * (5)^3
= 115
(10)^3
= 0.115
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4) 7218
3^2 *5^2
Sol) 7218 * 2^2 (multiply and divide by 2^2)
3^2 * 5^2 * 2^2
= 7218 * 4
3^2 * (10) ^2
= 7218 *4 * 1
9 100
= 28,872 * 1
9 100
= 3208 *0.01
= 32.08
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5) -143
110
Sol) 143
11* 10
= 143 *10 (Multiply and divide by 10)
11*10*10
= 143 *10 * 1
11 (10)^2
= 1430 * 1
11 100
= 130 *0.01
= 1.3
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4) The decimal form of some real numbers are given below. In each case,decide whether the number is rational or not. If it is rational, and expressed in form p/q, what can you say about the prime factors of q?
(1) 43.123456789
Sol) Express in p = 43123456789 (Rational)
q 1000000000
=> 43123456789
(10) ^9
=> 43123456789
(2*5) ^9
=> 43123456789 ( co-prime)
2^9 * 5^9
= > q = 2^9 * 5^9 ( 2^n * 5^m form)
As it is satisfied the two condition ( 1) co-prime and (2) q = 2^n and 5^m for rational number to be terminating
Thus ,prime factors of "q" are in terms of 2 and 5 only.
Thus, 43.123456789 is terminating.
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2) 0.120120012000120000....
Sol) 1) It is not rational (p/q form)
2) It is non-terminating and non-repeating
3) No prime factors
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3) 43.123456789
Sol 1) It is rational but in repeating form.
2) It is non-terminating.
3) As it's non-terminating "q" will have factors other than prime (2,and 5) factors
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Irrational numbers:-
A Real numbers is called irrational if it is cannot be written in the form p/q, where p and q are integers and q=/= 0.
Some examples of irrational number, with which you are already familiar, are :
√2, √3, √15, 𝛑 , - √2 / √3, 0.10110111011110...etc
With the help of the fundamental theorem of arithmetic. We will prove that √2, √3, √5 and in general, √p is irrational, where p is a prime
Statement-1 : Let p be a prime number. If p divided a^2, (where a is a positive integer), then p divides a.
Proof: Let a be any positive integer. Then the prime factorization of a is as follows :
a = p1 p2......pn,
where p1,p2.....pn are primes, not necessarily distinct.
.^. a^2 = (p1 p2....pn) (p1 p2...pn)
= p1^2 p2^2......pn^2
Now, here we have been given that p divides a^2.
.^., from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a^2.
Also, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realize that the only prime factors of a^2 are
p1 p2,,,,pn.
So p is one of p1, p2, .....pn
Now, since p is one of p1 p2....pn, it divides a.
Example-5. prove that √2 is irrational.
Proof : Using proof of contradiction, let us assume the contrary,
i.e., √2 is rational.
If it is rational, then there must exist two integers r and s(s=/=0) such that √2 = r/s
Suppose r and s have a common factor other than 1.
Then, we divide by the common factor to get √2 = a/b, where a and b are co-prime.
So, b√2 = a.
On squaring both sides and rearranging , we get
2b^2 = a^2.
.^., 2 divides a^2.
Now, by statement 1, it follows that 2 divides a^2 it also divides a.
So, we can write a = 2c for some integer c.
substituting for a, we get 2b^2 = 4c^2 ,
that is, b^2 = 2c^2.
This means that 2 divides b^2, and so 2 divides b (again using statement 1 with p=2).
.^., both a and b have 2 as a common factor.
But this contradicts the facts that a and b are co-prime and have no common factors other than 1.
This contradiction has arisen because of our assumption that
√2 is rational. So we conclude that √2 is irrational.
In general, it can be shown that √d is irrational whenever d is a positive integer which is not the square of an integer.
As such, it follows that √6, √8, √15, √24 etc, are all irrational number.
Example-6, Show that 5 - √3 is irrational.
sol) Let us assume, to the contrary, that 5 - √3 is rational.
that is, we can find co-primes a and b (b=/=0) such that
5 - √3 = a/b
.^., 5 - a/b = √3
Rearranging this equation, we get
√3 = 5 - a/b
√3 = 5b - a
b
Since a and b are integers , we get
5 - a/b is rational so √3 is rational.
But this contradicts the fact that √3 is irrational.
This contradiction has arisen because of our incorrect assumption that 5 - √3 is rational.
So, we conclude that 5 -√3 is irrational.
*******************************************************
Example-7. Show that 3√2 is irrational.
sol) let us assume, the contrary, that 3√2 is rational.
i.e, we can find co-primes a and b (b=/=0) such that
3√2 = a/b.
Rearranging, we get √2 = a/3b
Since 3, a and b are integers, a/3b is rational, and so √2 is rational.
But this contradicts the fact that √2 is irrational.
So, we conclude that 3√2 is irrational.
******************************************************
Example-8. prove that √2 + √3 is irrational.
sol) Let us suppose that √2 + √3 is rational.
Let √2 + √3 = a/b,
where a,b are integers and b=/=0
.^., √2 = a/b - √3
squaring on both sides, we get
2 = a^2 + 3 - 2a √3
b^2 b
Rearranging
2a√3 = a^2 + 3 - 2
b b^2
= a^2 + 1
b^2
√3 = a^2 + b^2
2ab
Since a,b are integers,
a^2 + b^2 is rational, and so √3 is rational.
2ab
This contradicts the fact that √3 is irrational.
Hence, √2 + √3 is irrational.
NOTE:
1) The sum of the two irrational numbers need not be irrational.
example : if a = √2 and b = -√2, then both a and b are irrational, but a +b = 0 which is rational.
2) The product of two irrational numbers need not be irrational.
For example, a = √2 and b = √8, then both a and b are irrational, but ab = √16 = 4 which is rational.
Exercise -1.4
1) Prove that 1 are irrational.
_/2
Sol) R.T.P ;- 1 is irrational.
_/2
We will use a technique called proof by contradiction
Let us assume the contrary(opposite)
,i.e., 1 is rational.
_/2
As we know if a number is rational it should be in the form of p/q (q=/=0) and "p" and 'q" should be co-prime (no common factors other than 1)
Thus we can write
1 = p
_/2 q
q (rational) = _/2 (irrational)
p
Since , Rational =/= Irrational. This is a Contradiction
Thus __1__ is irrational.
_/2
Hence proved.
-----------------------------------------------------------------------------------------------
Note 1) The sum of the two irrational numbers need not be irrational.
Example :- if a = _/2 and b = -- _/2 , then both 'a" and "b" are irrational, but a+b=0 which is rational.
Note 2). The product of two irrational numbers need not be irrational.
Example :- if a = _/2 and b= _/8 , then both 'a" and "b" are irrational, but "ab= _/16 = 4 which is rational.
2) Prove that _/3 + _/5 is irrational.
Sol) Let us suppose that _/3 + _/5 is rational/
Let
_/3+_/5 = a
b
where a,b are integers and b =/= 0
Transfer _/5 on right side we get
_/3 = a - _/5
b
squaring on both sides,
we get
=>( _/3 )^2 = ( a - _/5 ) ^2
b
{it is in {a-b}^2 = a^2 +b ^2 - 2 a b form}
Where a = a/b and b = _/5
=> 3 = a ^2 + 5 - 2 * a *_/5
b^2 b
=> Rearranging we get
2a _/5 = a^2 + 5 - 3
b b^2
" = a^2 + 2
b^2
" = a^2 + 2b^2
b^2
Transfer 2a/b to right side it becomes b/2a
_/5 =b { a^2 + 2b^2 }
2ab^2
_/5 = 1 * ( a^2 + 2b^2 )
2a b
_/5 = a^2 + 2b^2
2ab
Since a,b are integers,
a^2 + b^2 is rational, and so_/5 is rational.
2ab
This contradicts the fact that _/5 is irrational.
Hence _/3 + _/5 is irrational.
----------------------------------------------------------------------------------------------------
3) prove that 6 +_/2 is irrational.
Sol) Let us suppose that 6 + _/2 is rational.
That is 6 + _/2 = a
b
where a,b are integers and b=/=0
Transfer '+6" on right side it become " - 6 "
Therefore _/2 = a - 6 --------- (1)
b
Rearranging this equation, (1) we get
_/2 = a-6b
b
Since a,b are integer, a-6b is rational ,
b
and so, _/2 is rational
This contradicts the fact that _/2 is irrational.
Hence 6+ _/2 is irrational.
----------------------------------------------------------------------------------
statement 1:- Let "p" be a prime number. If "p" divides a^2, ( where a is a positive integer), then "p" divides "a".
*) In general , it can be shown that _/d is irrational whenever 'd" is a positive integer which is not the square of an integer. As such , it follows that _/6, _/8, _/15, _/24 etc., are all irrational numbers.
4) Prove that_/5 irrational.
Sol) Since we are using proof by contradiction,
let us assume the contrary i.e, _/5 is rational.
If it is rational, then there must exist two integers "a' and "b' (b=/=0) such that _/5 = a/b
Suppose "a' and "b" have a common factor other than 1. Then, we divide by the common
factor to get _/5 = a ,
b
where "a' and "b' are co-prime.
So, b_/5 = a.
On squaring both sides and rearranging, we get
( b _/5) ^2 = (a)^2
5b^2 = a^2
.Therefore , 5 divides a^2
Now, by statement 1, it follows that if "5" divides "a^2" it also divides "a'.
So, we can write a= 5c for some integer "c".
Substituting for "a",
we get 5b^2 = 25c^2
=> b^2 = 5c^2
This means that "5" divides "b^2" , and so "5" divides "b"
( again using statement 1 with p=5).
Therefore, both "a" and "b" have "5" as a common factor.
But this contradicts the fact that " a" and "b" are co-prime and have no-common factor other than "1'.
This contradiction has arisen because of our assumption that _/5 is rational.
So, we conclude that _/5 is irrational.
------------------------------------------------------------------------------------------------------
5) Prove that 3+ 2_/5 is irrational.
Sol) Let us suppose that 3+2_/5 is rational.
therefore 3 + 2_/5 = a/b, where a,b are integers and b=/=0.
Thus, _/5 = a -- 3
2b
Rearranging the equation we get,
_/5 = a-6b
2b
Since a,b are integers, a-6b is rational.
2b
so, _/5 is rational.
This contradicts the fact that _/5 is irrational.
This contradiction has arisen because of our assumption that 3+ 2_/5 is rational.
So, we conclude that 3+2_/5 is irrational.
-----------------------------------------------------------------------------------------------
2) Prove that _/p + _/q is irrational, where p,q are primes.
Sol). Let us suppose that _/p + _/q is rational.
. ^. _/p + _/q = a/b, where a,b are integers and b=/= 0.
Squaring on both sides, we get.
p^2 +q^2 + 2 _/pq = a^2
b^2
2 _/pq = a^2 --p^2 -- q^2
b^2
__/pq = 1 ( a^2 -- p^2 --q^2 )
2 b^2
Since a,b are integer, a^2 -- p^2 --q^2
b^2
is rational
So _/pq is rational.
This contradicts the fact that __/pq is irrational.
Hence __/p + __/q is irrational.
---------------------------------------------------------------------------------------------------------
We know that when 81 is written as 3^4 it is said to be written in its exponential form. that is, in 81 = 3^4, the number 4 is the exponent or index and 3 is the base. We say that
81 is the 4th power of the base 3 or
81 is the 4th power of 3.
Similarly, 27 = 3^3
Now, suppose we want to multiply 27 and 81; one way of doing this is by directly multiplying. But multiplication could get long and tedious if the numbers were much larger than 81 and 27. Can we use powers to makes our work easier?
We know that
81 = 3^4. We also know that 27 = 3^3
Using the law of exponents
a^m * a^n = a^m+n
we can write
27 * 81 = 3^3 * 3^4 = 3^7
Now, if we had a table containing the values for the powers of 3, it would be straight forward task to find the value of 3^7 and obtain the result of 81 *27 = 2187.
Similarly, if we want to divide 81 by 27 we can use the law of exponents
a^m ➗ a^n = a^m-n
where m>n. Then,
81 ➗ 27 = 3^4 ➗ 3^3 = 3^1 or simply 3
Notice that by using powers, we have changed a multiplication problem into one involving addition and a division problem into one of subtraction i.e, the addition of powers, 4 and 3 and the subtraction of the powers 4 and 3.
(*) Writing exponents as Logarithms
We know that 10000 = 10^4.
Here, 10 is the base and 4 is the exponent. Writing a number in the form of a base raised to a power is known as exponentiation. We can also write this in another way called logarithms as
log10 10000 = 4.
This is stated as "log of 10000 to the base 10 is equal to 4",
We observe that the base in the original expression becomes the base of the logarithmic form. Thus,
10000 = 10^4 is the same as
log10 10000 = 4.
In general, if a^n =x ; we write it as loga x = n where a and x are positive numbers and a =/= 1.
Example-9. Write
(!) 64 = 8^2
sol) The logarithmic form of 64 = 8^2 is
log8 64 = 2
(!!) 64 = 4^3
sol) The logarithmic form of 64 = 4^3 is
log4 64 = 3.
In the above example, we find that
log base 8 of 64 is 2 and
log base 4 of 64 is 3.
So, the logarithms of the same number to different bases are different.
Example-10. Write the exponential form of the following.
(!) Log10 100 = 2
sol) 10^2 = 100
(!!) Log5 25 = 2
sol) 5^2 = 25
(!!!) Log2 2 = 1
sol) 2^1 = 2
(!V) Log10 10 = 1
sol) 10^1 = 10
in cases (!!!) and (!V), we notice that
Log10 10 = 1
Log2 2 = 1.
In general, for any base a
a^1 = a
so
Loga a = 1
******************************************************
Example-11. determine the value of the following logarithms.
(!) Log 3 9
sol) Let Log 3 9 = x,
then the exponential form is
3^x = 9
3^x = 3^2
x = 2
(!!) Log8 2
sol) Let Log8 2 = y
Exponential form => 8^y = 2
(2^3)^y = 2^1
3y = 1
y = 1
3
****************************************
(!!!) Logc √c
sol) let Log c √c = z
exponential form => c^z = √c
c^z = c^1/2
z = 1/2
********************************************
Example-12. Expand Log15
sol) Log 15 = Log (3*5)
*******************
we know
Log xy = log x + log y
********************
=> Log3 + Log5
Example-13. Expand Log 343
125
sol) We know
Log x = Log x - Log y
y
Now,
=> Log 343 - Log 125
=> Log 7^3 - Log 5^3
******************
we have
Log x^m = m Log x
*******************
3 Log 7 - 3Log 5
So
Log 343 = 3(Log7- Log5)
125
****************************************************
Example-14. Write 2log3 + 3log5 - 5log2 as a single logarithm.
sol) log 3^2 + log 5^3 - log 3^5
***********************
since
m log x = log x^m
***********************
= log 9 + log 125 - log 32
***********************
log x + log y = log xy
************************
= log (9 * 125) - log 32
= log (1125) - log 32
**********************
log x - log y = log x
y
**********************
=> log 1125
32
Exercise -1.5
1) Write the following in logarithmic form.
1) 3^5 =243
Sol) Apply "log" on both side ,We get
log 3^5 = log 243
-------------------------------------------------------------------------------------------------------------------
2). Write the following in exponential form.
3) Determine the value of the following :-
1) and 2)
Note :- As we know 2 * 1/2 = 1 only
For "5" we can write 5^2 *1/2 = 25 ^1/2
similarly for "3" => 3^4 * 1/4 => 81^1/4
------------------------------------------------------------------------------------------------
3) and 4)
5) and 6)
7) and 8)
----------------------------------------------------------------------------------------------------
Formulas :-
1) log a + log b = log ( a * b )
2) log a - log b = log (a/b)
3) a log b = log b^a
4) log (a *b* c) = log a +log b +log c
4) Write each of the following expressions as log N. determine the value of N (You can assume the base is 10, but the results are identical which ever base used).
(1) log 2 + log 5
Sol) log a +log b = log (a*b)
=> log ( 2 * 5)
=> log (10) = log N
(2) log 16 - log 2
Sol) log a - log b = log (a/b)
=> log (16/2)
=> log 8 = log N
(3) 3log4
sol) => log 4^3
=> log 64 = logN
(4) 2log3 - 3log2
sol) log3^2 - log2^3
=> log9 - log8
=> log (9/8) = log N
(5) log 243 + log 1
sol) => log ( 243 * 1)
=> log 243 = log N
(6) log10 + 2 log 3 - log 2
Sol) => log 10 + log 3^2 - log 2
=> log 10 + log 9 - log 2
=> log (10 * 9) - log2
=> log (90) - log2
=> log 90/2
=> log 45 = log N
-------------------------------------------------------------------------------------------
5. Expand the following.
1) log 1000
Sol) = log 10^3
= 3log10
= 3log( 2*5)
= 3 {log2 + log5}
2) log 128/625
sol) => log 128 - log 625
=> log 2^7 - log 5^4
=> 7log2 - 4log5
3) log x^2 . y^3 .z^4
sol) log ( a *b* c) = log a +log b + log c
=> log x ^2 + log y^3 +log z^4
= 2logx +3logy +4logz
4) log p^2 q^3
r
Sol) log (a/b) = log a - log b
=> log (p^2 . q^3) - logr
= log p^2 + log q^3 - log r
= 2logp + 3logq -logr
5) log _/x^3
_/y^3
sol)
(*) Optional Exercise
1) Can the number 6^n, n being a natural number, end with digit 5? Give reason.
sol) Given: n is a natural number.
Natural Number :- it includes 1,2,3,4.....(also called the positive numbers) or 0,1,2,3.....(also called the non-negative integers or whole numbers)
The number 6^n, never end with "5"
Reason:- lets take "n= 1,2,3,4......"
6^1 = 6
6^2= 6 * 6 =36
6^3= 6 * 6 * 6 = 216 etc...
all ending with "6"
6 is an even number
6^n will also be even, it cannot end with "5"
or
if a number is ending with "5" it should contain "5" in its factors
example :- for 25 = 5*5, 35 = 7*5
but for 6^n = (2*3)^n
it does not contain "5" as factor so for any natural number "n" it cannot end with "5"
*****************************************
2) Is 7*5*3*2+3 a composite number? Justify your answer.
sol) Composite number : A whole number that can be formed by multiplying other whole numbers.
example : 6 = 2*3, 8 =2*4 etc...
If we can express 7*5*3*2+3 as a product of two numbers then we can call it as composite numbers.
we can write the above numbers as
taking "3" as common
3 *(7*5*2+1)
as "7*5*3*2+1" is expressed as product of 2 numbers we can call it as composite numbers.
********************************************************
3) Check whether 12^n can end with digit "0" for any natural number "n" ?
sol) we know that
2 12
2 6
3 3
1
12^n = (2*2*3)^n
=> 2^n * 2^n * 3^n
If the number 12^n ends with the digit "0", then it is divisible by "5"
example:- 10 = 2*5, 100= 2*2*5*5
.^. , the prime factorization of 12^n should contain "5".
This is not possible because the prime factorization of "12" are "2" and "3".
So, the uniqueness of Fundamental Theory of Arithmetic guarantees there are no other primes in the factorization of "12"
Hence, there is no natural number "n" for which "12^n" ends with the digit zero.
********************************************************
4) Show that one and only one out of n, n+2 or n+4 is divisible by 3, where n is any positive integer.
sol) We apply Euclid Division algorithm i.e
a = bq + r where , 0<r<3
put a= n and b = 3 we get
1) n = 3q + 0 = 3q
2) n = 3q+1
3) n = 3q + 2
Now,
condition (1) when n = 3q
n= 3q is clearly divisible by "3"
n+2 = 3q + 2 is not divisible by "3" as remainder is "2"
n+4 = 3q + 4 = 3q + 3 + 1
is not divisible by "3" as remainder is "1"
.^. , in this case one out of n, n+2, n+4 is divisible by "3"
similarly we can prove
condition (2) : when n = 3q +1
(*) n+2 = 3q+1+2 = 3q + 3 is divisible by "3"
condition(3) : when n = 3q+ 2
(*) n+4 = 3q+2+4 = 3q + 6 is divisible by "3"
********************************************************
5) prove that ( 2 √3 + √5 ) is an irrational number. Also check whether ( 2 √3 + √5) ( 2 √3 - √5 ) is rational or irrational..
sol) R.T.P:- ( 2 √3 + √5 ) is an irrational number
By using proof of contradiction,let us assume the contrary.i.e,
( 2 √3 + √5 ) is rational
If it is rational then, there must exists two integers "p" and "q"(q=/=0) such that
p = ( 2 √3 + √5 )
q
squaring on both sides,
p^2 = ( 2 √3 + √5 )^2
q^2
(a)^2 + (b)^2 + 2 (a * b )
p^2 = (2√3)^2 + (√5)^2 + 2 (2√3 *√5)√√√√√
q^2
p^2 = 12 + 5 + 4√15
q^2
p^2 = 17 + 4√15
q^2
p^2 -17q^2 = √15
4q^2
p^2 -17q^2 =Rational
4q^2
√15 = Irrational
Since , rational =/= Irrational
This is a contradiction
.^. Our assumption is incorrect
Hence, ( 2 √3 + √5 ) is an irrational number
Hence proved.
**************************************************
!!) ( 2 √3 + √5) ( 2 √3 - √5 )
sol) Its in (a+b) (a-b) = a^2 - b^2 form
=> (2 √3)^2 - (√5)^2
=> 4 * 3 - 5
=> 12 - 5 = 7
we can write "7" as
7 = p
1 q
both 'P' and 'q' are co-prime numbers.
Hence, ( 2 √3 + √5) ( 2 √3 - √5 ) is rational Number
*********************************************************
6) Without actual division, find after how many places of decimals in the decimal expansion of the following rational numbers terminates. verify by actual division. What do you infer?
Solution :- We know that if the denominator of a rational number had no prime factors other than 2 or 5, then it is expressible as a terminating,
otherwise it has non-terminating repeating decimal representation. Thus, we will have to check the prime factors of the denominators of each of the given rational numbers.
or
p is terminating if
q
(*) p & q are co-prime
and
(*) q or denominator is of the form 2^n * 5^m
Where "n" and "m" are non-negative integers
(!) 5
16
sol) Denominator : 16 = 2*2*2*2 = 2^4
So we can write denominator in the form = 2^4 * 5^0(5^0 =1)
where n= 4 and m = 0
Thus, it is a terminating decimal
Division:-
5 = 0.3125 terminating
16
*****************************************************
(!!) 13
2^2
sol) Denominator : 2^2
we can write it in the form : 2^2 * 5^0
where n= 3, m = 0
Thus its a terminating decimal.
(*) Division :
13 = 3.25
2^2
*******************************************************
(!!!) 17
125
sol) Denominator : 125 = 5*5*5 = 5^3
Its in the form : 2^0 * 5^3
n = 0 m = 3
Thus, its terminates
(*) Division :-
17 = 0.136
125
(!v) 13
80
sol) Denominator : 80 = 2*2*2*2*5 = 2^4 * 5^1
Here n = 4 , m = 1
Thus, it terminates
(*) Division :-
13 = 0.1625
80
******************************************************
(v) 15
32
sol) denominator : 32 = 2*2*2*2*2 = 2^5 * 5^0
Here, n = 5, m = 0
Thus, its terminates
(*) Division :-
15 = 0.46875
32
******************************************************
(V!) 33
2^2 * 5
sol) Denominator : 2^2 * 5 = 2^n * 5^m
Here, n = 2 m = 1
Thus, its terminates
(*) Division :-
33 = 1.65
2^2 * 5
*******************************************************
7) If x^2 +y^2 = 6xy, prove that 2 log(x+y) = logx + logy + 3log2
sol) Given :- x^2 + y^2 = 6xy -----(1)
we know (a+b)^2 = a^2 + b ^2 + 2ab
**********************
(x+y)^2 = x^2 + y ^2 + 2xy
(x+y)^2 - 2xy = x^2 + y^2 ------(2)
***********************
substitute (2) in (1) we get,
(x+y)^2 - 2xy = 6xy
(x+y)^2 = 8xy
(x+y)^2 = 2^3 * x * y
Applying log on both sides
log (x+y)^2 = log (2^3 * x *y)
*******************************
Log (x+y)^2 : Log (a)^b = b log a
Log (2^3 * x * y) :Log (a*b*c) = Log a + Log b + Log c
*****************************
2log(x+y) = log 2^3 + log x + log y
2log(x+y) = 3log2 + log x + log y
Hence proved.
****************************************************
8) Find the number of digits in 4^2013, if log10 2 = 0.3010
sol) Let x = 4^2013
=> x = (2^2)^ 2013
=> x = 2^ 2*2013
=> x = 2^ 4026
Apply log on both sides we get
log x = log 2^4026
log x = 4026 log 2
log x = 4026 * 0.3010
log x = 1211.826
*****************************
1211 : characteristic
826 : mantisa
***********************
No. of digits in 4^2013 : characteristic +1
1211 + 1 = 1212
******************************************************
Application of Logarithm
Example-15. The magnitude of an earthquake was defined in 1935 by Charles richer with the expression M = log I
S
where I is the intensity of the earthquake tremor and S is the intensity of a "threshold earthquake".
(a) if the intensity of an earthquake is 10 times the intensity of a threshold earthquake, then what is its magnitude?
sol) Let the intensity of the earthquake be I, then we are given
I = 10 S
The magnitude of an earthquake is given by-
M = log 10S
S
.^. The magnitude of the earthquake will be
M = log I
S
= log 10
= 1
**************************************************
(b) Let x be the number of times the intensity of the earthquake to that of a threshold earthquake. So the intensity of earthquake is-
I = xS
We know that
M = log I
S
So, the magnitude of the earthquake is
M = log xs
s
M = log x
We know that M = 10
So log x = 10 and
.^. x = 10^10
What is Integers :- It can be Positive,negative, or zero but do not include any fractional (5/6 ,3/4,7/2) or decimal parts ( 2.13,-1.22,0.08 etc..) The set of integers denoted by "Z"
Integers Example :- Z= { .....-2 -1 0 1 2....}
What is Rational Numbers :- Numbers which can be written in the form of p/q where both p and q are integers and Q =/=0 (Q should not be Zero ). All rational numbers can be written either in the form of terminating decimals or non-terminating repeating decimals.
The set of rational numbers denoted by" q "
Example :-
q= { 3 can be written in p/q form as 3/1 }
{ 0.01 can be written in p/q form as 1/100 }
What is Irrational Numbers :- Numbers which cannot be expressed in the form of p/q are irrational. These include numbers like _/2, (root 2), _/3, _/5 and mathematical quantities like "pi". When these are written as decimals they are non-terminating,non-recurring
Example :- _/3 (root 3) = 1.7320508.........so on (non-terminating).
What is Real Numbers :- The set of rational and irrational numbers together are called real numbers.
What is Terminating :- It is a decimal That finally ends after a finite numbers of digits.
Example :- 1/4 = 0.25 (It end after few decimal places)
What is Non-terminating :- It is a decimal that continues endlessly
Example :- "pi" or 22/7= 3.1415.....so on (if You keep on dividing it wont end)
1. which of the following rational numbers are terminating and which are non-terminating repeating in their decimal form?
(1) 2/5
Ans:- Terminating , because if we divide we get 2/5 = 0.4
2) 17/18
Ans :- Non-terminating, because if we divide it keep repeating 17/18 = 0.94444.....
3) 15/16
Ans :- Terminating, because 15/16 = 0.9375
4) 7/40
Ans :- Terminating ,because 7/40 = 0.175
5) 9/11
Ans :- Non-Terminating, because 9/11= 0.1818....so on
=======================================================================
3. Classify the numbers given below as rational or irrational.
1) 2* 1
2
Ans :- Rational , converting mixed fraction 2*1/2 (2*1/2 = 4+1/2 = 5/2) into improper fraction we get 5/2 (p/q)
2) √24
Ans :- Irrational, Because √24 ( 4.89897948.....) doesn't have perfect square it never terminates or repeats .
3) √16
Ans :- Rational, have perfect square 4*4=16 ......_/16=4 or 4/1 (p/q)
4) 7.7777..
Ans :- Clearly irrational.
5) √ 4/9
Ans:- Rational.. 2*2/3*3 = 2/3 (p/q)
6) - √30
Ans:- Irrational, - _/30 = 5.47722.....dont have perfect square ...cant have p/q form
============================================================
4.) Represent the following real numbers on the number line.
(1) 3/4
Ans :-
(2) -9/10
Ans ;-
(3) 27/3
Ans :- 27/3 = 9
<----0---1---2---3---4---5---6---7---8---9----->
(4) find Root 5 on number line.
Ans :- In order to find square root of any numbers always imagine any root number on hypotenuse side first
Pythagoras Theorem :- In a right angled triangle the square of the hypotenuse is equal to the sum of the squares of its other two sides.
That is , AC^2 =AB^2 + BC^2
Now Lets solve _/5
Step 1 :- Take 5 in form of sum of perfect squares ,
that is 2^2 (4) + 1^2 (1) = _/ 5
Step1 :- Express "5" in such a manner that is written as sum of perfect squares.
Step2 :- 5 = 4 + 1 , as we know perfect sq of 4 =2^2 and for 1 = 1^2
Step3:- 5 = 2^2 (AB) + 1^2 (BC). The distance between AB = 2 units and BC =1unit
Step4 :- According to Pytho.Theorem
Ac^2 = AB^2 + BC^2
= 2^2 + 1^2
= 4 + 1
= 5
If you transfer square of AC on right side than 5 become _/5 (root 5)
AC = _/5
Step5 :- Take a compass and take distance of AC as Radius and draw an arc passing through C till it reach the point "S" on the number line ..click and see the figure above.
(5) find √ -16 on number line.
Ans :- As we know 16 have perfect square
- √16 =4*4 = -4 .
-4 -3 -2 -1 0
=============================================================
Q? Are all integers also in Real numbers?why
A) Yes,Because Set of Integer ( Z) is a subset of Real numbers ( R ) .
Also we can say All Rational,whole,natural too is a subset of Real numbers See fig.
------------------------------------------------------------------------------------------------------------
Fundamental theorem of arithmetic :- All natural numbers, except 1, can be written as a product of their prime factors.(2,3,5,7,11,13.....)
NOTE : 1 is neither a composite nor a prime.
Example :- 3 =3, ...6 =2*3....45= 3*3*5.....253 = 11*23.....52 = 2*2*13 so on....
*) What is prime numbers
A) A number that is divisible only by itself and 1 (2,3,5,7,11,13.... will only divide by itself or
with 1 not with any other numbers)
-----------------------------------------------------------------------------------------------------------
Example-1. Find the HCF and LCM of 12 and 18 by the prime factorization method.
sol) We have
12 = 2 * 2 * 3 = 2^2 * 3^1
18 = 2 * 3 * 3 = 2^1 * 3^2
Note that
HCF (12, 18) = 2^1 * 3^1 = 6 = product of the smallest power of each common prime factors in the numbers.
LCM (12, 18) = 2^2 * 3^2 = 36 = Product of the greatest power of each prime factors,, in the numbers.
From the example above, you might have noticed that HCF(12,18) * LCM( 12, 18) = 12 * 18.
In fact, we can verify that for any two positive integers "a" and "b",
HCF(a,b) * LCM(a,b) = a*b.
We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers.
*******************************************************
Example 2 . Consider the number 4^n, where "n" is a natural number. Check whether there is any value of "n" for which 4^n ends with the digit zero?
sol) For the number 4^n to end with digit zero for any natural number "n" it should be divisible by "5"
This means that the prime factorization of 4^n should contain the prime number "5"
But it is not possible because
4^n = (2)^2n
so "2" is the only prime in the factorization of 4^n.
Since "5" is not present in the prime factorization, so there is no natural number "n" for which "4^n" ends with the digit zero.
Exercise- 1.2
Q) What is Prime factors
A) It is a process of finding which prime numbers( 2,3,5,7 ....) can be multiplied together to make the original number
1. Express 140 as a product of its prime factors.
A)
----------------------------------------------------------------------------------------
2) Express 156 number as a product of prime factors.
A) 156 = { 2*2*3*13 } = 2^2 * 3 * 13
----------------------------------------------------------------------------------------
3) Express 3825 number as a product of prime factors.
A) 3825 = { 3*3*5*5*17} = 3^2 *5^2 *17
----------------------------------------------------------------------------------------
4) Express 5005 number as a product of prime factors.
A) 5005 = 5*7*11*13
---------------------------------------------------------------------------------------
5) Express 7429 number as a product of prime factors.
A) 7429 = 17*19*23
====================================================
2) Find the LCM and HCF of the following integers by the prime factorization method.
(1) Find the LCM and HCF of 12, 15 and 21
Sol) 12 = 2*2*3
15 = 3*5
21 = 3*7
HCF : Product of smallest power of each common prime factor
= 3^1 = 3
LCM :Product of greatest power of each prime factor
=> 2*2*3*5*7=420
------------------------------------------------------------------------------------------------
(2) Find the LCM and HCF of 17,23, and 29
Sol) 17 = 1*17
23 = 1*23
29 = 1*29
HCF :- 1
LCM :- 1*17*29*23 = 11,339
-------------------------------------------------------------------------------------
(3) Find the LCM and HCF of 8,9,25
sol) 8 = 2*2*2
9 = 3*3
25 = 5*5
HCF :- Since there are no common prime factors we can take HCF =1
LCM :- 2*2*2*3*3*5*5=1800
-----------------------------------------------------------------------------------
(4) Find the LCM and HCF of 72 and 108
sol) 72 = 2*2*2*3*3
108 = 2*2*3*3*3
HCF :- 2*2*3*3 = 36
LCM :- 2*2*2*3*3*3 = 216
----------------------------------------------------------------------------------------------------------
(5) Find LCM and HCF 306 and 657
sol) 306 = 2*3*3*17
657 = 3*3*73
HCF : 3*3 =9
LCM : 2*3*3*17*73 = 22338
---------------------------------------------------------------------------------------------------
(3) check whether 6^n can end with the digit 0 for any natural number n.
Sol) Take an example of a numbers which ends with the digit "zero"
Ex: 20 = 2*2*5 or 100 = 2*2*5*5
Here we can clearly see that numbers ending with "0" has both 2 & 5 as their prime factors
As we know the prime factors for 6= 2*3.
Clearly it does not end with "0" because it does not have prime factor "5"
Therefore, 6^n cannot end with zero for any natural number "n" ( n=1,2,3,...)
ex : (6)^1= 6, (6) ^2= 36 , (6)^3 =216 so on....as u can see it never ends with zero.
----------------------------------------------------------------------------------------------------------------
What is composite number :- If a number can be divided by any other numbers other than 1 and itself
Ex : all prime numbers (2,3,5,7...) cannot be divided by any other number except 1 or itself .so we can say all prime numbers are not composite numbers
similarly if u take 8 it can divided exactly by 2 (beside 1 and itself(8)) hence 8 can be said as composite number
Also product of primes will be a composite number.
4) Explain why 7*11*13+13 and 7*6*5*4*3*2*1+5 are composite numbers.
Sol) 7*11*13+13 = 1014 is a composite number as it can be divided by [ 1014/2=507}
other number "2" beside 1 and itself
Second method :
7*11*13+13
Take 13 as common
13[7*11+1] or 13 {odd +1 }
13 [77+1] or odd {odd+odd}
13 [78]=1014 or odd { even) => odd *even => even
!!) 7*6*5*4*3*2*1+5 = 5045 composite number as its divisible by "5" 5045/5 = 1009 (beside 1 and itself 5045 )
-----------------------------------------------------------------------------------------------
5) how will you show that ( 17*11*2)+( 17*11*5) is a composite number? Explain.
sol) 17*11 (2+5)
= 17*11 (7)
= 17*11 (odd)
= odd (odd) = 17*11*7 = product of primes will be a composite number .
Also 17*11*7=1309 which is divisible by 1309/7=187 (beside 1 and itself(1309)
---------------------------------------------------------------------------------------------------------
(*) Rational Numbers and their Decimal expansions
Let us consider the following terminating decimal forms of some rational numbers:
Now let us express them in the form of p/q
(!) 0.375
sol) 375 = 375
1000 10^3
(!!) 1.04
sol) 104 = 104
100 10^2
(!!!) 0.0875
sol) 875 = 875
10000 10^4
(!V) 12.5
sol) 125 = 125
10 10^1
(V) 0.00025
sol) 25 = 25
100000 10^5
We see that all terminating decimals taken by us can be expressed as rational numbers whose denominators are powers of "10".
Let us now prime factorize the numerator and denominator and then express in the simplest rational form:
Now
(!) 0.375
sol) 375 = 3 * 5^3 = 3 = 3
10^3 2^3 * 5^3 2^3 8
(!!) 1.04
sol) 104 = 2^3 * 13 = 26 = 26
100 2^2 * 5^2 5^2 25
(!!!) 0.0875
sol) 875 = 5^3 * 7 = 7
10^4 2^4 * 5^4 2^4 * 5
(!V) 12.5
sol) 125 = 5^3 = 25
10 2*5 2
(V) 0.00025
sol) 25 = 5^2 = 1 = 1
10^5 2^5 * 5^5 2^5 * 5^3 4000
Do you see a pattern in the denominators?
It appears that when the decimal expression is expressed in its simplest rational form then "p" and "q" are co-prime and the denominator (i.e, q) has only
powers of 2, or powers of 5, or both.
This is because the powers of 10 can only have powers of "2" and "5" as factors.
Let us conclude
Any rational number which has a decimal expansion that terminates can be expressed as a rational number whose denominator is a power of "10".
The only prime factors of "10" are "2" and "5".
So, when we simplify the rational number, we find that the number if of the form p/q,
where the prime factorization of "q" is of the form
2^n * 5^m, and n,m are some non-negative integers.
Theorem-1.2 : Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form
p , where p and q are co-prime, and the prime
q
factorization of q is of the form 2^n*5^m, where n,m are non-negative integers.
If we have a rational number in the form p/q, and the prime factorization of q is of the form 2^n*5^m, where n,m are non-negative integers, then does p/q have a terminating decimal expansion?
So, it seems to make sense to convert a rational number of the form p/q, where q is of the form 2^n*5^m, to an equivalent rational number of the form a/b, where b is a power of 10. let us go back to our examples above and work backwards.
(!) 25
2
sol) 5^2 multiply and divide by "5"
2
5^3 = 125 = 12.5
2 * 5 10
(!!) 26
25
sol) 13 * 2 multiply and divide by "2^2"
5^2
13 * 2^3 = 104 = 1.04
2^2 * 5^2 10^2
similarly we can show we can convert a rational number of the form p/q, where "q" is of the form 2^n * 5^m, to an equivalent rational number of the form a/b, where "b" is a power of 10.
.^. , the decimal expansion of such a rational number terminates. We find that a rational number of the form p/q, where "q" is a power of 10, will have a terminating decimal expansion.
So, we find that the converse of theorem 1.2 is also true and can be formally stated as :
Theorem 1.3 : Let x = p/q be a rational number, such that the prime factorization of "q" is of the form 2^n * 5^m, where n,m are non-negative integers. Then x has a decimal expansion which terminates.
(*) Non-terminating, recurring decimals in rational numbers
Let us look at the decimal conversion of 1/7
1/7 = 0.1428571428571........which is a non-terminating and recurring decimal.
Notice , the block of digits '1428571' is repeating in the quotient.
Notice that the denominator here, i.e., 7 is not of the form 2^n*5^m
Theorem-1.4 : Let x = p/q be a rational number, such that the prime factorization of q is not of the form 2^n * 5^m , where n,m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating(recurring).
we can conclude that the decimal form of every rational number is either terminating or non-terminating repeating.
Example-3. Using the above theorems, without actual division, state whether the following rational numbers are terminating or non-terminating decimals.
(!) 16
125
sol) 16 = 16 is terminating decimal.
5*5*5 5^3
(!!) 25
32
sol) 25
2*2*2*2*2
= 25 is terminating decimal
2^5
(!!!) 100
81
sol) 100
3*3*3*3
= 100 is non-terminating, repeating decimal
3^4
(!V) 41
75
sol) 41
3* 5 * 5
= 41 is non-terminating, repeating decimal
3* 5^2
*******************************************************
Example-4 : Write the decimal expansion of the following rational numbers without actual division.
(!) 35
50
sol) 7 * 5
2*5*5
= 7
2* 5
= 7
10^1
= 0.7
(!!) 21
25
sol) 21 multiply and divide by "2^2"
5*5
= 21 * 2^2
5*5*2^2
= 21 * 4
5^2 * 2^2
= 84 = 0.84
10^2
(!!!) 7
8
sol) 7
2* 2* 2
= 7
2^3
= 7 * 5^3
2^3 * 5^3
= 7 * 125
(2 * 5)^3
= 875
(10)^3
= 0.875
Exercise 1.3
1) Write the following rational numbers in their decimal form and also state which are terminating and which have non-terminating , repeating decimal.
1) 3/8
Sol) 3/8 = 0.375 is a terminating decimal
2) 229/400
Sol ) 229/400 = 0.5725 is a terminating decimal.
3) 4 1/5
Sol) convert this mixed fraction into improper fraction we get
What is Proper Fraction :- if the numerator is lesser than denominator ( ex : 2/3)
What is improper fraction :- if the numerator is greater than denominator ( ex : 4/3)
=> 4*1/5
=> 5*4+1
5
=> 20+1
5
=> 21 (improper fraction)
5
=> 21/5 is a terminating decimal
4) 2/11
Sol ) 2/11 = 0.181818.... is a non-terminating decimal.
5) 8/125
sol) 8/125 =0.064.... is a terminating decimal
Coprime:- Any two numbers "x" and "y" is said to be co-prime if they have no common factor except "1"
Example for co-prime :- x => 14 = 1 *2*7 and 15 = 1 *3*5 ( only '1' common)
Example for relatively or mutually prime :- x => 14 = 1*2*7 and 21 = 1*3*7 ( 1 and 7 common)
Theorem 1.2 :- Let "x" be a rational number whose decimal expansion "terminates". Then
"x" can be expressed in the form p/q, where p and q are "co-prime" , and the prime
factorization of "q" is of the form " 2^n * 5^m ", where n, m are non-negative integers.
---------------------------------------------------------------------------------------------------2) Without actually performing division, state whether the following rational numbers have a terminating decimal form or a non-terminating, repeating decimal form.
In order to prove whether rational numbers terminating or not two conditions should be met.
1) p/q should be co-prime
2) "q" should be in the form 2^n * 5^m
1) 13/3125 = p/q
sol) 1st condition :- Is the fraction is co-prime?
p= 13= 1*13
q= 3125
= 1*5*5*5*5*5 = 1 * 5^5
= 2^0 *5^5 ( 2^n *5 ^m form)
Note :- 2^0 or 3^0 or N ^0 ......= is always 1
Both 13 and 3125 is co-prime because only "1" is common factor
2nd condition :- is the denominator q= 2^n * 5^m form?
q= 1* 5^2 = 2^0 * 5^2 == 2^n * 5^m
As the two condition met .
Thus , 13/3125 is a terminating decimal.
----------------------------------------------------------------------------------------------------------
2) 11 = p/q
12
Sol) 1st condition :- is the fraction is co-prime?
11 = 1*11
12 = 1*2*2*3 yes its co-prime because only "1" is common factor
2nd condition :- is q= 2^n* 5^m form?
q= 2^2 * 3^1 =/= (not equal to) 2^n * 5^m
As the 2nd condition not met 11/12 is non-terminating, repeating decimal.
------------------------------------------------------------------------------------------------------------
3) 64 = p/q
225
Sol ) 64 = 1*2*2*2*2*2*2 =(co-prime)
225 1*5*91
q = 5*91 =/= 2^n * 5^m
Thus, 64/ 225 is non-terminating, repeating decimal.
--------------------------------------------------------------------------
4) 15 = 3 = 3 = ( Co-prime)
1600 320 2*2*2*2*2*2*5
q= 2^6 * 5^2 = 2^n * 5^m
Two condition met,, Thus its is 15/1600 is terminating.
----------------------------------------------------------------------------
5) 29 = 29 = (co-prime )
343 7*7*7
q= 1 *7^3 = 2^0 * 7^3 =/= 2^n * 5^m
2nd condition not met Thus , 29/343 is non-terminating, repeating decimal.
------------------------------------------------------------------------------------------------
6) 23 = (co-prime)
2^3 5^2
q= 2^3 * 5^2 = 2^n *5^m form
Thus it is terminating decimal.
---------------------------------------------------------------------------------------
7) 129
2^2 * 5^7 * 7^5
Sol) Clearly It is non-terminating, repeating decimal.
-------------------------------------------------------------------------
8) 9 = 3 * 3 = 3
15 3 * 5 5
Sol) 3/5 (co-prime)
q= 1*5 = 2^0 * 5^1 = 2^n * 5^m
Clearly 9/15 is terminating decimal.
----------------------------------------------------------------------------
9) 36 = 9
100 25
Sol ) p= 9 = 3*3
q = 25 = 5*5
1) is it co-prime?
A) yes as 9 and 25 both have different factors
2) is q= 2^n * 5^m form?
A) Yes we can write 25 as 2^0 * 5^2 = 2^n * 5^m (where 2^0=1)
Thus 36/100 is terminating decimal.
-------------------------------------------------------------------------------------
10) 77/210
Sol) p =77 = 1*77
q = 210 = 3 *2*7*5
1) is it co-prime
A) yes
2) Is q= 2^n * 5^m form?
A) No.
Thus, 77/210 is non-terminating, repeating decimal.
-----------------------------------------------------------------------------------------------------
Theorem 1.1 :- Every composite number can be expressed (factorized) as a product or primes, and this factorization is unique, apart from the order in which the prime factors occur.
*) Write the following rationals in decimal form using Theorem 1.1
1) 13
25
Sol) 13/25 = p/q , We know that any number we can be convert into product (multiply) of
primes (2,3,5,7,11.....)
=13
25
= 13 * 2^2 (Multiply and divide by 2^2)
5^2 * 2^2
= 52
(10)^2
transfer q= (10)^2 in numerator it become 0.52
( 1 / 10^2 = 1/100 = 0.01 = 0.01*52 =0.52)
Thus, Decimal form of 13/25 = 0.52
----------------------------------------------------------------------------
2) 15
16
Sol) 15 * 5^4 (Multiply and divide by 5^4)
2^4 * 5^4
= 9375
(10)^4
= 0.9375
--------------------------------------------------------------------------------
3) 23
2^3 * 5^2
Sol) 23 * 5^1 (multiply and divide by 5^1)
2^3 *5 ^2 *5^1
= 23 * 5^1
2^3 * (5)^3
= 115
(10)^3
= 0.115
---------------------------------------------------------------------------------------------------
4) 7218
3^2 *5^2
Sol) 7218 * 2^2 (multiply and divide by 2^2)
3^2 * 5^2 * 2^2
= 7218 * 4
3^2 * (10) ^2
= 7218 *4 * 1
9 100
= 28,872 * 1
9 100
= 3208 *0.01
= 32.08
-----------------------------------------------------------------------------------------------------
5) -143
110
Sol) 143
11* 10
= 143 *10 (Multiply and divide by 10)
11*10*10
= 143 *10 * 1
11 (10)^2
= 1430 * 1
11 100
= 130 *0.01
= 1.3
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4) The decimal form of some real numbers are given below. In each case,decide whether the number is rational or not. If it is rational, and expressed in form p/q, what can you say about the prime factors of q?
(1) 43.123456789
Sol) Express in p = 43123456789 (Rational)
q 1000000000
=> 43123456789
(10) ^9
=> 43123456789
(2*5) ^9
=> 43123456789 ( co-prime)
2^9 * 5^9
= > q = 2^9 * 5^9 ( 2^n * 5^m form)
As it is satisfied the two condition ( 1) co-prime and (2) q = 2^n and 5^m for rational number to be terminating
Thus ,prime factors of "q" are in terms of 2 and 5 only.
Thus, 43.123456789 is terminating.
----------------------------------------------------------------------------------------
2) 0.120120012000120000....
Sol) 1) It is not rational (p/q form)
2) It is non-terminating and non-repeating
3) No prime factors
----------------------------------------------------------------------------------------
3) 43.123456789
Sol 1) It is rational but in repeating form.
2) It is non-terminating.
3) As it's non-terminating "q" will have factors other than prime (2,and 5) factors
--------------------------------------------------------------------------------------------------
Irrational numbers:-
A Real numbers is called irrational if it is cannot be written in the form p/q, where p and q are integers and q=/= 0.
Some examples of irrational number, with which you are already familiar, are :
√2, √3, √15, 𝛑 , - √2 / √3, 0.10110111011110...etc
With the help of the fundamental theorem of arithmetic. We will prove that √2, √3, √5 and in general, √p is irrational, where p is a prime
Statement-1 : Let p be a prime number. If p divided a^2, (where a is a positive integer), then p divides a.
Proof: Let a be any positive integer. Then the prime factorization of a is as follows :
a = p1 p2......pn,
where p1,p2.....pn are primes, not necessarily distinct.
.^. a^2 = (p1 p2....pn) (p1 p2...pn)
= p1^2 p2^2......pn^2
Now, here we have been given that p divides a^2.
.^., from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a^2.
Also, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realize that the only prime factors of a^2 are
p1 p2,,,,pn.
So p is one of p1, p2, .....pn
Now, since p is one of p1 p2....pn, it divides a.
Example-5. prove that √2 is irrational.
Proof : Using proof of contradiction, let us assume the contrary,
i.e., √2 is rational.
If it is rational, then there must exist two integers r and s(s=/=0) such that √2 = r/s
Suppose r and s have a common factor other than 1.
Then, we divide by the common factor to get √2 = a/b, where a and b are co-prime.
So, b√2 = a.
On squaring both sides and rearranging , we get
2b^2 = a^2.
.^., 2 divides a^2.
Now, by statement 1, it follows that 2 divides a^2 it also divides a.
So, we can write a = 2c for some integer c.
substituting for a, we get 2b^2 = 4c^2 ,
that is, b^2 = 2c^2.
This means that 2 divides b^2, and so 2 divides b (again using statement 1 with p=2).
.^., both a and b have 2 as a common factor.
But this contradicts the facts that a and b are co-prime and have no common factors other than 1.
This contradiction has arisen because of our assumption that
√2 is rational. So we conclude that √2 is irrational.
In general, it can be shown that √d is irrational whenever d is a positive integer which is not the square of an integer.
As such, it follows that √6, √8, √15, √24 etc, are all irrational number.
Example-6, Show that 5 - √3 is irrational.
sol) Let us assume, to the contrary, that 5 - √3 is rational.
that is, we can find co-primes a and b (b=/=0) such that
5 - √3 = a/b
.^., 5 - a/b = √3
Rearranging this equation, we get
√3 = 5 - a/b
√3 = 5b - a
b
Since a and b are integers , we get
5 - a/b is rational so √3 is rational.
But this contradicts the fact that √3 is irrational.
This contradiction has arisen because of our incorrect assumption that 5 - √3 is rational.
So, we conclude that 5 -√3 is irrational.
*******************************************************
Example-7. Show that 3√2 is irrational.
sol) let us assume, the contrary, that 3√2 is rational.
i.e, we can find co-primes a and b (b=/=0) such that
3√2 = a/b.
Rearranging, we get √2 = a/3b
Since 3, a and b are integers, a/3b is rational, and so √2 is rational.
But this contradicts the fact that √2 is irrational.
So, we conclude that 3√2 is irrational.
******************************************************
Example-8. prove that √2 + √3 is irrational.
sol) Let us suppose that √2 + √3 is rational.
Let √2 + √3 = a/b,
where a,b are integers and b=/=0
.^., √2 = a/b - √3
squaring on both sides, we get
2 = a^2 + 3 - 2a √3
b^2 b
Rearranging
2a√3 = a^2 + 3 - 2
b b^2
= a^2 + 1
b^2
√3 = a^2 + b^2
2ab
Since a,b are integers,
a^2 + b^2 is rational, and so √3 is rational.
2ab
This contradicts the fact that √3 is irrational.
Hence, √2 + √3 is irrational.
NOTE:
1) The sum of the two irrational numbers need not be irrational.
example : if a = √2 and b = -√2, then both a and b are irrational, but a +b = 0 which is rational.
2) The product of two irrational numbers need not be irrational.
For example, a = √2 and b = √8, then both a and b are irrational, but ab = √16 = 4 which is rational.
Exercise -1.4
1) Prove that 1 are irrational.
_/2
Sol) R.T.P ;- 1 is irrational.
_/2
We will use a technique called proof by contradiction
Let us assume the contrary(opposite)
,i.e., 1 is rational.
_/2
As we know if a number is rational it should be in the form of p/q (q=/=0) and "p" and 'q" should be co-prime (no common factors other than 1)
Thus we can write
1 = p
_/2 q
q (rational) = _/2 (irrational)
p
Since , Rational =/= Irrational. This is a Contradiction
Thus __1__ is irrational.
_/2
Hence proved.
-----------------------------------------------------------------------------------------------
Note 1) The sum of the two irrational numbers need not be irrational.
Example :- if a = _/2 and b = -- _/2 , then both 'a" and "b" are irrational, but a+b=0 which is rational.
Note 2). The product of two irrational numbers need not be irrational.
Example :- if a = _/2 and b= _/8 , then both 'a" and "b" are irrational, but "ab= _/16 = 4 which is rational.
2) Prove that _/3 + _/5 is irrational.
Sol) Let us suppose that _/3 + _/5 is rational/
Let
_/3+_/5 = a
b
where a,b are integers and b =/= 0
Transfer _/5 on right side we get
_/3 = a - _/5
b
squaring on both sides,
we get
=>( _/3 )^2 = ( a - _/5 ) ^2
b
{it is in {a-b}^2 = a^2 +b ^2 - 2 a b form}
Where a = a/b and b = _/5
=> 3 = a ^2 + 5 - 2 * a *_/5
b^2 b
=> Rearranging we get
2a _/5 = a^2 + 5 - 3
b b^2
" = a^2 + 2
b^2
" = a^2 + 2b^2
b^2
Transfer 2a/b to right side it becomes b/2a
_/5 =
2a
2a b
_/5 = a^2 + 2b^2
2ab
Since a,b are integers,
a^2 + b^2 is rational, and so_/5 is rational.
2ab
This contradicts the fact that _/5 is irrational.
Hence _/3 + _/5 is irrational.
----------------------------------------------------------------------------------------------------
3) prove that 6 +_/2 is irrational.
Sol) Let us suppose that 6 + _/2 is rational.
That is 6 + _/2 = a
b
where a,b are integers and b=/=0
Transfer '+6" on right side it become " - 6 "
Therefore _/2 = a - 6 --------- (1)
b
Rearranging this equation, (1) we get
_/2 = a-6b
b
Since a,b are integer, a-6b is rational ,
b
and so, _/2 is rational
This contradicts the fact that _/2 is irrational.
Hence 6+ _/2 is irrational.
----------------------------------------------------------------------------------
statement 1:- Let "p" be a prime number. If "p" divides a^2, ( where a is a positive integer), then "p" divides "a".
*) In general , it can be shown that _/d is irrational whenever 'd" is a positive integer which is not the square of an integer. As such , it follows that _/6, _/8, _/15, _/24 etc., are all irrational numbers.
4) Prove that_/5 irrational.
Sol) Since we are using proof by contradiction,
let us assume the contrary i.e, _/5 is rational.
If it is rational, then there must exist two integers "a' and "b' (b=/=0) such that _/5 = a/b
Suppose "a' and "b" have a common factor other than 1. Then, we divide by the common
factor to get _/5 = a ,
b
where "a' and "b' are co-prime.
So, b_/5 = a.
On squaring both sides and rearranging, we get
( b _/5) ^2 = (a)^2
5b^2 = a^2
.Therefore , 5 divides a^2
Now, by statement 1, it follows that if "5" divides "a^2" it also divides "a'.
So, we can write a= 5c for some integer "c".
Substituting for "a",
we get 5b^2 = 25c^2
=> b^2 = 5c^2
This means that "5" divides "b^2" , and so "5" divides "b"
( again using statement 1 with p=5).
Therefore, both "a" and "b" have "5" as a common factor.
But this contradicts the fact that " a" and "b" are co-prime and have no-common factor other than "1'.
This contradiction has arisen because of our assumption that _/5 is rational.
So, we conclude that _/5 is irrational.
------------------------------------------------------------------------------------------------------
5) Prove that 3+ 2_/5 is irrational.
Sol) Let us suppose that 3+2_/5 is rational.
therefore 3 + 2_/5 = a/b, where a,b are integers and b=/=0.
Thus, _/5 = a -- 3
2b
Rearranging the equation we get,
_/5 = a-6b
2b
Since a,b are integers, a-6b is rational.
2b
so, _/5 is rational.
This contradicts the fact that _/5 is irrational.
This contradiction has arisen because of our assumption that 3+ 2_/5 is rational.
So, we conclude that 3+2_/5 is irrational.
-----------------------------------------------------------------------------------------------
2) Prove that _/p + _/q is irrational, where p,q are primes.
Sol). Let us suppose that _/p + _/q is rational.
. ^. _/p + _/q = a/b, where a,b are integers and b=/= 0.
Squaring on both sides, we get.
p^2 +q^2 + 2 _/pq = a^2
b^2
2 _/pq = a^2 --p^2 -- q^2
b^2
__/pq = 1 ( a^2 -- p^2 --q^2 )
2 b^2
Since a,b are integer, a^2 -- p^2 --q^2
b^2
is rational
So _/pq is rational.
This contradicts the fact that __/pq is irrational.
Hence __/p + __/q is irrational.
---------------------------------------------------------------------------------------------------------
We know that when 81 is written as 3^4 it is said to be written in its exponential form. that is, in 81 = 3^4, the number 4 is the exponent or index and 3 is the base. We say that
81 is the 4th power of the base 3 or
81 is the 4th power of 3.
Similarly, 27 = 3^3
Now, suppose we want to multiply 27 and 81; one way of doing this is by directly multiplying. But multiplication could get long and tedious if the numbers were much larger than 81 and 27. Can we use powers to makes our work easier?
We know that
81 = 3^4. We also know that 27 = 3^3
Using the law of exponents
a^m * a^n = a^m+n
we can write
27 * 81 = 3^3 * 3^4 = 3^7
Now, if we had a table containing the values for the powers of 3, it would be straight forward task to find the value of 3^7 and obtain the result of 81 *27 = 2187.
Similarly, if we want to divide 81 by 27 we can use the law of exponents
a^m ➗ a^n = a^m-n
where m>n. Then,
81 ➗ 27 = 3^4 ➗ 3^3 = 3^1 or simply 3
Notice that by using powers, we have changed a multiplication problem into one involving addition and a division problem into one of subtraction i.e, the addition of powers, 4 and 3 and the subtraction of the powers 4 and 3.
(*) Writing exponents as Logarithms
We know that 10000 = 10^4.
Here, 10 is the base and 4 is the exponent. Writing a number in the form of a base raised to a power is known as exponentiation. We can also write this in another way called logarithms as
log10 10000 = 4.
This is stated as "log of 10000 to the base 10 is equal to 4",
We observe that the base in the original expression becomes the base of the logarithmic form. Thus,
10000 = 10^4 is the same as
log10 10000 = 4.
In general, if a^n =x ; we write it as loga x = n where a and x are positive numbers and a =/= 1.
Example-9. Write
(!) 64 = 8^2
sol) The logarithmic form of 64 = 8^2 is
log8 64 = 2
(!!) 64 = 4^3
sol) The logarithmic form of 64 = 4^3 is
log4 64 = 3.
In the above example, we find that
log base 8 of 64 is 2 and
log base 4 of 64 is 3.
So, the logarithms of the same number to different bases are different.
Example-10. Write the exponential form of the following.
(!) Log10 100 = 2
sol) 10^2 = 100
(!!) Log5 25 = 2
sol) 5^2 = 25
(!!!) Log2 2 = 1
sol) 2^1 = 2
(!V) Log10 10 = 1
sol) 10^1 = 10
in cases (!!!) and (!V), we notice that
Log10 10 = 1
Log2 2 = 1.
In general, for any base a
a^1 = a
so
Loga a = 1
******************************************************
Example-11. determine the value of the following logarithms.
(!) Log 3 9
sol) Let Log 3 9 = x,
then the exponential form is
3^x = 9
3^x = 3^2
x = 2
(!!) Log8 2
sol) Let Log8 2 = y
Exponential form => 8^y = 2
(2^3)^y = 2^1
3y = 1
y = 1
3
****************************************
(!!!) Logc √c
sol) let Log c √c = z
exponential form => c^z = √c
c^z = c^1/2
z = 1/2
********************************************
Example-12. Expand Log15
sol) Log 15 = Log (3*5)
*******************
we know
Log xy = log x + log y
********************
=> Log3 + Log5
Example-13. Expand Log 343
125
sol) We know
Log x = Log x - Log y
y
Now,
=> Log 343 - Log 125
=> Log 7^3 - Log 5^3
******************
we have
Log x^m = m Log x
*******************
3 Log 7 - 3Log 5
So
Log 343 = 3(Log7- Log5)
125
****************************************************
Example-14. Write 2log3 + 3log5 - 5log2 as a single logarithm.
sol) log 3^2 + log 5^3 - log 3^5
***********************
since
m log x = log x^m
***********************
= log 9 + log 125 - log 32
***********************
log x + log y = log xy
************************
= log (9 * 125) - log 32
= log (1125) - log 32
**********************
log x - log y = log x
y
**********************
=> log 1125
32
Exercise -1.5
1) Write the following in logarithmic form.
1) 3^5 =243
Sol) Apply "log" on both side ,We get
log 3^5 = log 243
-------------------------------------------------------------------------------------------------------------------
2). Write the following in exponential form.
3) Determine the value of the following :-
1) and 2)
For "5" we can write 5^2 *1/2 = 25 ^1/2
similarly for "3" => 3^4 * 1/4 => 81^1/4
------------------------------------------------------------------------------------------------
3) and 4)
5) and 6)
7) and 8)
----------------------------------------------------------------------------------------------------
Formulas :-
1) log a + log b = log ( a * b )
2) log a - log b = log (a/b)
3) a log b = log b^a
4) log (a *b* c) = log a +log b +log c
4) Write each of the following expressions as log N. determine the value of N (You can assume the base is 10, but the results are identical which ever base used).
(1) log 2 + log 5
Sol) log a +log b = log (a*b)
=> log ( 2 * 5)
=> log (10) = log N
(2) log 16 - log 2
Sol) log a - log b = log (a/b)
=> log (16/2)
=> log 8 = log N
(3) 3log4
sol) => log 4^3
=> log 64 = logN
(4) 2log3 - 3log2
sol) log3^2 - log2^3
=> log9 - log8
=> log (9/8) = log N
(5) log 243 + log 1
sol) => log ( 243 * 1)
=> log 243 = log N
(6) log10 + 2 log 3 - log 2
Sol) => log 10 + log 3^2 - log 2
=> log 10 + log 9 - log 2
=> log (10 * 9) - log2
=> log (90) - log2
=> log 90/2
=> log 45 = log N
-------------------------------------------------------------------------------------------
5. Expand the following.
1) log 1000
Sol) = log 10^3
= 3log10
= 3log( 2*5)
= 3 {log2 + log5}
2) log 128/625
sol) => log 128 - log 625
=> log 2^7 - log 5^4
=> 7log2 - 4log5
3) log x^2 . y^3 .z^4
sol) log ( a *b* c) = log a +log b + log c
=> log x ^2 + log y^3 +log z^4
= 2logx +3logy +4logz
4) log p^2 q^3
r
Sol) log (a/b) = log a - log b
=> log (p^2 . q^3) - logr
= log p^2 + log q^3 - log r
= 2logp + 3logq -logr
5) log _/x^3
_/y^3
sol)
(*) Optional Exercise
1) Can the number 6^n, n being a natural number, end with digit 5? Give reason.
sol) Given: n is a natural number.
Natural Number :- it includes 1,2,3,4.....(also called the positive numbers) or 0,1,2,3.....(also called the non-negative integers or whole numbers)
The number 6^n, never end with "5"
Reason:- lets take "n= 1,2,3,4......"
6^1 = 6
6^2= 6 * 6 =36
6^3= 6 * 6 * 6 = 216 etc...
all ending with "6"
6 is an even number
6^n will also be even, it cannot end with "5"
or
if a number is ending with "5" it should contain "5" in its factors
example :- for 25 = 5*5, 35 = 7*5
but for 6^n = (2*3)^n
it does not contain "5" as factor so for any natural number "n" it cannot end with "5"
*****************************************
2) Is 7*5*3*2+3 a composite number? Justify your answer.
sol) Composite number : A whole number that can be formed by multiplying other whole numbers.
example : 6 = 2*3, 8 =2*4 etc...
If we can express 7*5*3*2+3 as a product of two numbers then we can call it as composite numbers.
we can write the above numbers as
taking "3" as common
3 *(7*5*2+1)
as "7*5*3*2+1" is expressed as product of 2 numbers we can call it as composite numbers.
********************************************************
3) Check whether 12^n can end with digit "0" for any natural number "n" ?
sol) we know that
2 12
2 6
3 3
1
12^n = (2*2*3)^n
=> 2^n * 2^n * 3^n
If the number 12^n ends with the digit "0", then it is divisible by "5"
example:- 10 = 2*5, 100= 2*2*5*5
.^. , the prime factorization of 12^n should contain "5".
This is not possible because the prime factorization of "12" are "2" and "3".
So, the uniqueness of Fundamental Theory of Arithmetic guarantees there are no other primes in the factorization of "12"
Hence, there is no natural number "n" for which "12^n" ends with the digit zero.
********************************************************
4) Show that one and only one out of n, n+2 or n+4 is divisible by 3, where n is any positive integer.
sol) We apply Euclid Division algorithm i.e
a = bq + r where , 0<r<3
put a= n and b = 3 we get
1) n = 3q + 0 = 3q
2) n = 3q+1
3) n = 3q + 2
Now,
condition (1) when n = 3q
n= 3q is clearly divisible by "3"
n+2 = 3q + 2 is not divisible by "3" as remainder is "2"
n+4 = 3q + 4 = 3q + 3 + 1
is not divisible by "3" as remainder is "1"
.^. , in this case one out of n, n+2, n+4 is divisible by "3"
similarly we can prove
condition (2) : when n = 3q +1
(*) n+2 = 3q+1+2 = 3q + 3 is divisible by "3"
condition(3) : when n = 3q+ 2
(*) n+4 = 3q+2+4 = 3q + 6 is divisible by "3"
********************************************************
5) prove that ( 2 √3 + √5 ) is an irrational number. Also check whether ( 2 √3 + √5) ( 2 √3 - √5 ) is rational or irrational..
sol) R.T.P:- ( 2 √3 + √5 ) is an irrational number
By using proof of contradiction,let us assume the contrary.i.e,
( 2 √3 + √5 ) is rational
If it is rational then, there must exists two integers "p" and "q"(q=/=0) such that
p = ( 2 √3 + √5 )
q
squaring on both sides,
p^2 = ( 2 √3 + √5 )^2
q^2
(a)^2 + (b)^2 + 2 (a * b )
p^2 = (2√3)^2 + (√5)^2 + 2 (2√3 *√5)√√√√√
q^2
p^2 = 12 + 5 + 4√15
q^2
p^2 = 17 + 4√15
q^2
p^2 -17q^2 = √15
4q^2
p^2 -17q^2 =Rational
4q^2
√15 = Irrational
Since , rational =/= Irrational
This is a contradiction
.^. Our assumption is incorrect
Hence, ( 2 √3 + √5 ) is an irrational number
Hence proved.
**************************************************
!!) ( 2 √3 + √5) ( 2 √3 - √5 )
sol) Its in (a+b) (a-b) = a^2 - b^2 form
=> (2 √3)^2 - (√5)^2
=> 4 * 3 - 5
=> 12 - 5 = 7
we can write "7" as
7 = p
1 q
both 'P' and 'q' are co-prime numbers.
Hence, ( 2 √3 + √5) ( 2 √3 - √5 ) is rational Number
*********************************************************
6) Without actual division, find after how many places of decimals in the decimal expansion of the following rational numbers terminates. verify by actual division. What do you infer?
Solution :- We know that if the denominator of a rational number had no prime factors other than 2 or 5, then it is expressible as a terminating,
otherwise it has non-terminating repeating decimal representation. Thus, we will have to check the prime factors of the denominators of each of the given rational numbers.
or
p is terminating if
q
(*) p & q are co-prime
and
(*) q or denominator is of the form 2^n * 5^m
Where "n" and "m" are non-negative integers
(!) 5
16
sol) Denominator : 16 = 2*2*2*2 = 2^4
So we can write denominator in the form = 2^4 * 5^0(5^0 =1)
where n= 4 and m = 0
Thus, it is a terminating decimal
Division:-
5 = 0.3125 terminating
16
*****************************************************
(!!) 13
2^2
sol) Denominator : 2^2
we can write it in the form : 2^2 * 5^0
where n= 3, m = 0
Thus its a terminating decimal.
(*) Division :
13 = 3.25
2^2
*******************************************************
(!!!) 17
125
sol) Denominator : 125 = 5*5*5 = 5^3
Its in the form : 2^0 * 5^3
n = 0 m = 3
Thus, its terminates
(*) Division :-
17 = 0.136
125
(!v) 13
80
sol) Denominator : 80 = 2*2*2*2*5 = 2^4 * 5^1
Here n = 4 , m = 1
Thus, it terminates
(*) Division :-
13 = 0.1625
80
******************************************************
(v) 15
32
sol) denominator : 32 = 2*2*2*2*2 = 2^5 * 5^0
Here, n = 5, m = 0
Thus, its terminates
(*) Division :-
15 = 0.46875
32
******************************************************
(V!) 33
2^2 * 5
sol) Denominator : 2^2 * 5 = 2^n * 5^m
Here, n = 2 m = 1
Thus, its terminates
(*) Division :-
33 = 1.65
2^2 * 5
*******************************************************
7) If x^2 +y^2 = 6xy, prove that 2 log(x+y) = logx + logy + 3log2
sol) Given :- x^2 + y^2 = 6xy -----(1)
we know (a+b)^2 = a^2 + b ^2 + 2ab
**********************
(x+y)^2 = x^2 + y ^2 + 2xy
(x+y)^2 - 2xy = x^2 + y^2 ------(2)
***********************
substitute (2) in (1) we get,
(x+y)^2 - 2xy = 6xy
(x+y)^2 = 8xy
(x+y)^2 = 2^3 * x * y
Applying log on both sides
log (x+y)^2 = log (2^3 * x *y)
*******************************
Log (x+y)^2 : Log (a)^b = b log a
Log (2^3 * x * y) :Log (a*b*c) = Log a + Log b + Log c
*****************************
2log(x+y) = log 2^3 + log x + log y
2log(x+y) = 3log2 + log x + log y
Hence proved.
****************************************************
8) Find the number of digits in 4^2013, if log10 2 = 0.3010
sol) Let x = 4^2013
=> x = (2^2)^ 2013
=> x = 2^ 2*2013
=> x = 2^ 4026
Apply log on both sides we get
log x = log 2^4026
log x = 4026 log 2
log x = 4026 * 0.3010
log x = 1211.826
*****************************
1211 : characteristic
826 : mantisa
***********************
No. of digits in 4^2013 : characteristic +1
1211 + 1 = 1212
******************************************************
Application of Logarithm
Example-15. The magnitude of an earthquake was defined in 1935 by Charles richer with the expression M = log I
S
where I is the intensity of the earthquake tremor and S is the intensity of a "threshold earthquake".
(a) if the intensity of an earthquake is 10 times the intensity of a threshold earthquake, then what is its magnitude?
sol) Let the intensity of the earthquake be I, then we are given
I = 10 S
The magnitude of an earthquake is given by-
M = log 10S
S
.^. The magnitude of the earthquake will be
M = log I
S
= log 10
= 1
**************************************************
(b) Let x be the number of times the intensity of the earthquake to that of a threshold earthquake. So the intensity of earthquake is-
I = xS
We know that
M = log I
S
So, the magnitude of the earthquake is
M = log xs
s
M = log x
We know that M = 10
So log x = 10 and
.^. x = 10^10
Thanks for providing useful information about Form 2290 Schedule 1 proof
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