1.2 The Fundamental Theorem of Arithmetic
Theorem 1.1
Fundamental Theorem of Arithmetic
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.The prime factorisation of a natural number is unique, except for the order of its factors.
Prime numbers upto 100 are
$2,\:3,\:5,\:7,\:11,$
$13,\:17,\:19,\:23,\:29,$
$31,\:37,\:41,\:43,\:47,$
$53,\:59,\:61,\:67,\:71,$
$73,\:79,\:83,\:89,\:97$
Tests of divisibility
Divisibility by 2 :unit place ends with 0, 2, 4, 6 or 8.
Divisibility by 3 :
sum of all digits is a multiple of 3.
Divisibility by 4 :
last two digits is divisible by 4.
Divisibility by 5 :
unit place ends with 0 or 5.
Divisibility by 6 :
satisfies condition for 2 and 3.
Divisibility by 7 :
Double the unit digit and subtract from the remaining digits and repeat the same until you get a smaller value. If the final value is divisible by 7, the given number is divisible by 7.
Aliter:
Divide the given number in blocks of 3 from the right. If the difference between sum of odd blocks and sum of even blocks is 0 or a multiple of 7, then the given number is divisible by 7.
Divisibility by 8 :
last three digits is divisible by 8.
Divisibility by 9 :
sum of all digits is a multiple of 9.
Divisibility by 10 :
unit place ends with 0.
Divisibility by 11 :
Difference between sum of digits at odd places and sum of digits at even places is 0 or a multiple of 11.
Divisibility by 12 :
satisfies condition for 3 and 4.
Divisibility by 13 :
Divide the given number in blocks of 3 from the right. If the difference between sum of odd blocks and sum of even blocks is 0 or a multiple of 13, then the given number is divisible by 13.
Divisibility by 14 :
satisfies condition for 2 and 7.
Divisibility by 15 :
satisfies condition for 3 and 5.
$Product \: of \: numbers(a,b)=LCM(a,b) \times HCF(a,b)$
$a \times b=LCM(a,b) \times HCF(a,b)$
$a=\frac{LCM(a,b) \times HCF(a,b)}{b}$
$b=\frac{LCM(a,b) \times HCF(a,b)}{a}$
$LCM(a,b)=\frac{a \times b}{HCF(a,b)}$
$HCF(a,b)=\frac{a \times b}{LCM(a,b)}$
Example 1 :
Consider the numbers $4^n$, where n is a natural number. Check whether there is any value of n for which $4^n$ ends with the digit zero.
Solution:
If a number ends with '0', it will have the prime factors 2 and 5.
$4^n=(2^2)^n=2^{2n}$
$4^n$ does not have the factor 5.
Therfore, $4^n$ does not end with '0' for any value of n.
Example 2 :
Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Solution :
$6=2^1 \times 3^1$
$20=2^2 \times 5^1$
$LCM(6,20)=2^2 \times 3^1 \times 5^1 = 60$
$HCF(6,20)=2^1=2$
Example 3:
Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.
Solution :
$96=2^5 \times 3^1$
$404=2^2 \times 101^1$
$HCF(96,404)=2^2=4$
$LCM(96,404)=\frac{96 \times 404}{HCF(96,404)}=\frac{96 \times 404}{4}=96 \times 101 =9696$
Example 4 :
Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
Solution :
$6=2^1 \times 3^1$
$72=2^3 \times 3^2$
$120=2^3 \times 3^1 \times 5^1$
$HCF(6,72,120)=2^1 \times 3^1=2 \times 3=6$
$LCM(6,72,120)=2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5=360$
EXERCISE 1.1
1. Express each number as a product of its prime factors:(i) 140
Solution :
$140=2^2 \times 5^1 \times 7^1$
(ii) 156
Solution :
$156=2^2 \times 3^1 \times 13^1$
(iii) 3825
Solution :
$3825=3^2 \times 5^2 \times 17^1$
(iv) 5005
Solution :
$5005=5^1 \times 7^1 \times 11^1 \times 13^1$
(v) 7429
Solution :
$7429=17^1 \times 19^1 \times 23^1$
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
Solution :
$26=2^1 \times 13^1$
$91=7^1 \times 13^1$
$HCF(26,91)=13^1=13$
$LCM(26,91)=2^1 \times 7^1 \times 13^1 = 2 \times 7 \times 13=182$
$26 \times 91=2366$
$LCM \times HCF=182 \times 13 = 2366$
Hence verified.
(ii) 510 and 92
Solution :
$510=2^1 \times 3^1 \times 5^1 \times 17^1$
$92=2^2 \times 23^1$
$HCF(510,92)=2^1=2$
$LCM(510,92)=2^2 \times 3^1 \times 5^1 \times 17^1 \times 23^1= 4 \times 3 \times 5 \times 13 \times 23=23460$
$510 \times 92=46920$
$LCM \times HCF=23460 \times 2 = 46920$
Hence verified.
(iii) 336 and 54
Solution :
$336=2^4 \times 3^1 \times 7^1$
$54=2^1 \times 3^3$
$HCF(336,54)=2^1 \times 3^1=2 \times 3=6$
$LCM(336,54)=2^4 \times 3^3 \times 7^1 = 16 \times 27 \times 7=3024$
$336 \times 54=18144$
$LCM \times HCF=336 \times 54 = 18144$
Hence verified.
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
Solution :
$12=2^2 \times 3^1$
$15=3^1 \times 5^1$
$21=3^1 \times 7^1$
$HCF(12,15,21)=3^1=3$
$LCM(12,15,21)=2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7=420$
(ii) 17, 23 and 29
Solution :
$17=17^1$
$23=23^1$
$29=29^1$
$HCF(17,23,29)=1$
$LCM(17,23,29)=17^1 \times 23^1 \times 29^1 = 17 \times 23 \times 29=11339$
(iii) 8, 9 and 25
Solution :
$8=2^3$
$9=3^2$
$25=5^2$
$HCF(8,9,25)=1$
$LCM(8,9,25)=2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25=1800$
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution :
$LCM(306, 657)=\frac{306 \times 657}{HCF(306, 657)}=\frac{306 \times 657}{9}=34 \times 657 =22338$
5. Check whether $6^n$ can end with the digit 0 for any natural number n.
Solution :
If a number ends with '0', it will have the prime factors 2 and 5.
$6^n=(2 \times 3)^n=2^n \times 3^n$
$6^n$ does not have the factor 5.
Therfore, $6^n$ does not end with '0' for any value of n.
6. Explain why $7 × 11 × 13 + 13$ and $7 × 6 × 5 × 4 × 3 × 2 × 1 + 5$ are composite numbers.
Solution :
$7 × 11 × 13 + 13=13(7\times11+1)$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=13 \times (77+1)$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=13 \times 78$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=1 \times 13 \times 78$
Therefore, it is a composite number.
$7 × 6 × 5 × 4 × 3 × 2 × 1 + 5=5(7 × 6 × 4 × 3 × 2 × 1 + 1)$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=5(1008+1)$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=5 \times 1009$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=1 \times 5 \times 1009$
Therefore, it is a composite number.
7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution :
$18=2^1 \times 3^2$
$12=2^2 \times 3^1$
$LCM(18,12)=2^2 \times 3^2= 4 \times 9=36$
Therefore, after 36 minutes, they will meet again at the starting point.
1.3 Revisiting Irrational Numbers
A number ‘s’ is called irrational, if it cannot be written in the form , $\frac{p}{q}$ where p and q are integers and $q \ne 0$.
Theorem 1.2 :
Let p be a prime number. If p divides $a^2$, then p divides a, where a is a positive integer.Theorem 1.3 : $\sqrt{2}$ is irrational.
Proof:Let us assume that $\sqrt{2}\:$ is a rational number of the form $\frac{a}{b}$, $b\ne 0$, where a and b are coprimes.
$\sqrt{2}=\frac{a}{b}$
squaring on both sides, $(\sqrt{2})^2=(\frac{a}{b})^2$
$2=\frac{a^2}{b^2}$
$b^2=\frac{a^2}{2}$ ---(1)
2 divides $a^2$, so, 2 divides a.
$\implies$ a=2c.
(1) $\implies$ $b^2=\frac{(2c)^2}{2}$
$b^2=\frac{4c^2}{2}$
$b^2=2c^2$
$c^2=\frac{b^2}{2}$
2 divides $b^2$, so, 2 divides b.
$\implies$ a and b has a common factor 2.
This contradicts the fact that a and b are coprime.
Our assumption is wrong.
$\implies \sqrt{2}\:$ is irrational.
Example 5 : Prove that $\sqrt{3}$ is irrational.
Solution:Let us assume that $\sqrt{3}\:$ is a rational number of the form $\frac{a}{b}$, $b\ne 0$, where a and b are coprimes.
$\sqrt{3}=\frac{a}{b}$
squaring on both sides, $(\sqrt{3})^2=(\frac{a}{b})^2$
$3=\frac{a^2}{b^2}$
$b^2=\frac{a^2}{3}$ ---(1)
3 divides $a^2$, so, 3 divides a.
$\implies$ a=3c.
(1) $\implies$ $b^2=\frac{(3c)^2}{3}$
$b^2=\frac{9c^2}{3}$
$b^2=3c^2$
$c^2=\frac{b^2}{3}$
3 divides $b^2$, so, 3 divides b.
$\implies$ a and b has a common factor 3.
This contradicts the fact that a and b are coprime.
Our assumption is wrong.
$\implies \sqrt{3}\:$ is irrational.
Example 6 : Show that $5 – \sqrt{3}$ is irrational.
Solution:Let us assume that $5-\sqrt{3}\:$ is a rational number of the form $\frac{a}{b}$, $b\ne 0$, where a and b are coprimes.
$5-\sqrt{3}=\frac{a}{b}$
$\sqrt{3}=5-\frac{a}{b}$
$\sqrt{3}=\frac{5b-a}{b}$
$\frac{5b-a}{b}$ is rational.
$\implies$ $\sqrt{3}$ is also rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
Hence our assumption is wrong.
$\implies 5-\sqrt{3}\:$ is irrational.
Example 7 : Show that $3\sqrt{2}$ is irrational.
Solution:Let us assume that $3\sqrt{2}\:$ is a rational number of the form $\frac{a}{b}$, $b\ne 0$, where a and b are coprimes.
$3\sqrt{2}=\frac{a}{b}$
$\sqrt{2}=\frac{a}{3b}$
$\frac{a}{3b}$ is rational.
$\implies$ $\sqrt{2}$ is also rational.
But this contradicts the fact that $\sqrt{2}$ is irrational.
Hence our assumption is wrong.
$\implies 3\sqrt{2}\:$ is irrational.
EXERCISE 1.2
1. Prove that $\sqrt{5}$ is irrational.
Solution:Let us assume that $\sqrt{5}\:$ is a rational number of the form $\frac{a}{b}$, $b\ne 0$, where a and b are coprimes.
$\sqrt{5}=\frac{a}{b}$
squaring on both sides, $(\sqrt{5})^2=(\frac{a}{b})^2$
$5=\frac{a^2}{b^2}$
$b^2=\frac{a^2}{5}$ ---(1)
5 divides $a^2$, so, 5 divides a.
$\implies$ a=5c.
(1) $\implies$ $b^2=\frac{(5c)^2}{5}$
$b^2=\frac{25c^2}{5}$
$b^2=5c^2$
$c^2=\frac{b^2}{5}$
5 divides $b^2$, so, 5 divides b.
$\implies$ a and b has a common factor 5.
This contradicts the fact that a and b are coprime.
Our assumption is wrong.
$\implies \sqrt{5}\:$ is irrational.
2. Prove that $3+2\sqrt{5}$ is irrational.
Solution:Let us assume that $3+2\sqrt{5}\:$ is a rational number of the form $\frac{a}{b}$, $b\ne 0$, where a and b are coprimes.
$3+2\sqrt{5}=\frac{a}{b}$
$2\sqrt{5}=\frac{a}{b}-3$
$2\sqrt{5}=\frac{a-3b}{b}$
$\sqrt{5}=\frac{a-3b}{2b}$
$\frac{a-3b}{2b}$ is rational.
$\implies$ $\sqrt{5}$ is also rational.
But this contradicts the fact that $\sqrt{5}$ is irrational.
Hence our assumption is wrong.
$\implies 3+2\sqrt{5}\:$ is irrational.
3. Prove that the following are irrationals :
$(i) \: \frac{1}{\sqrt{2}}$Solution:
Let us assume that $\frac{1}{\sqrt{2}}\:$ is a rational number of the form $\frac{a}{b}$, $b\ne 0$, where a and b are coprimes.
$\frac{1}{\sqrt{2}}=\frac{a}{b}$
$\sqrt{2}=\frac{b}{a}$
$\frac{b}{a}$ is rational.
$\implies$ $\sqrt{2}$ is also rational.
But this contradicts the fact that $\sqrt{2}$ is irrational.
Hence our assumption is wrong.
$\implies \frac{1}{\sqrt{2}}\:$ is irrational.
$(ii)\: 7\sqrt{5}$
Solution:
Let us assume that $7\sqrt{5}\:$ is a rational number of the form $\frac{a}{b}$, $b\ne 0$, where a and b are coprimes.
$7\sqrt{5}=\frac{a}{b}$
$\sqrt{5}=\frac{a}{7b}$
$\frac{a}{7b}$ is rational.
$\implies$ $\sqrt{5}$ is also rational.
But this contradicts the fact that $\sqrt{5}$ is irrational.
Hence our assumption is wrong.
$\implies 7\sqrt{5}\:$ is irrational.
$(iii)\: 6+\sqrt{2}$
Solution:
Let us assume that $6+\sqrt{2}\:$ is a rational number of the form $\frac{a}{b}$, $b\ne 0$, where a and b are coprimes.
$6+\sqrt{2}=\frac{a}{b}$
$\sqrt{2}=\frac{a}{b}-6$
$\sqrt{2}=\frac{a-6b}{b}$
$\frac{a-6b}{b}$ is rational.
$\implies$ $\sqrt{2}$ is also rational.
But this contradicts the fact that $\sqrt{2}$ is irrational.
Hence our assumption is wrong.
$\implies 6+\sqrt{2}\:$ is irrational.