Question

Proof root 3 is irrational

Answer 1

hey mate..

Sol: Let us assume that √3 is a rational number.That is, we can find integers a and b (≠ 0) such that √3 = (a/b) Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime. √3b = a ⇒ 3b2=a2 (Squaring on both sides) → (1) Therefore, a2 is divisible by 3 Hence ‘a’ is also divisible by 3. So, we can write a = 3c for some integer c. Equation (1) becomes, 3b2 =(3c)2 ⇒ 3b2 = 9c2 ∴ b2 = 3c2 This means that b2is divisible by 3, and so b is also divisible by 3.Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that √3 is rational. So, we conclude that √3 is irrational.

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Answer 2

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PROOF:

. let us assume that$\sqrt{3}$ is not an irrational.

$\sqrt{3}$ is rational.

$\sqrt{3}$ = $\frac{p}{q}$ where p,q= 0 q is $\cancel{=}$ 0. p,q are co - primes.

squaring on both sides.

$(\sqrt{3})^2$ = ($\frac{p}{q})$${^2}$

3= $\frac{p{^2}}{q{^2}}$

${p}^2$= ${3×{q}^2}$

${p}^2$ is divisible by 3.

${p}$ is also divisible by 3

$\boxed{let p= 3k.....1}$

from equation 1

$({3k}^2)$ = ${3×{q}^2}$

$\cancel{9}$${k}^2$= $\cancel{3}$${q}^2$

${3{k}^2}$ = ${q}^2}$

= ${q}^2}$ =${3{k}^2}$

${q}^2$ is divisible by 3 .

${q}$ is also divisible by 3.

3 is the common factor for both p,q but p,q are co - primes

It is contradiction to our assumption

oure assumption is wrong.

$\sqrt{3}$ is not a rational

$\sqrt{3}$ is an irrational.

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