Question

25.
Prove that root3 is an irrational number.​

Answer 1

Sorry for the untidy answer

Step-by-step explanation:

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Answer 2

Assume that

$\sqrt{3} \: is \: irrationl$

$\sqrt{3 = a} \div b$

a, b are cooprimes and b is not equal to 0

$a = b \sqrt{3}$

squaring on both sides , we have

$a {2} = 3b {2}$

Thus 3 is a factor of a square and in result 3 is also a factor of a

Let a =3c where c is some integer

$a 2\ = 9c {2}$

substitute

$a {2} = 9b {2} \: we \: have$

$3b {2} = 9c {2}$

$b {2} \: = \: 3c {2}$

Thus 3 is factor of b2 and in result 3 is also factor of b

thus 3 is a common factor of a and b.But this contraticts thr fact that a and b ate co primes .Thus our assumption is wrong

$hence \: \sqrt{3 \: \: is \: \: irrational}$

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