Question

prove that root 3 is irrational number​

Answer 1

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{blue}{..!!Hello Dear!!...}}}}$$\huge {\mathcal{\blue{..!!!!!!!A}\red{ns}\purple{wer}\green{!!!!!!!..}}}$

━━━━━━━━━━━━━━━━━━━━━━

$\huge\bold\red{{solution:-}}$

we have to prove $\huge\bold\purple{ \sqrt{3}}$ is a irrational.

let us assume the opposite

i.e,${ \sqrt{3}}$is rational

hence, ${ \sqrt{3}}$ can be written in the form of $\frac{a}{b}$

✰where a and b ( b≠0) are co- prime( no common factor other than 1)

hence,${ \sqrt{3}}=\frac{a}{b}$

${ \sqrt{3}}=a$

squaring both side

${\sqrt{3}}^{2}={a}^{2}$

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

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

hence,3 divide${a}^{2}$

✰By theorem:-if p is a prime number and p divides${a}^{2}$ then p divides a, where "a" is a positive number.

so, 3 shall divide also.............................(1)

hence,

we can say

$\frac{a}{3}=c$ is some integer

so,

${a=3c}$

now we know that

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

putting ${a=3c}$

$3{b}^{2}=(3{c})^{2}$

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

${b}^{2}=\frac{1}{3}×9{c}^{2}$

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

${b}^{2}/3={c}^{2}$

hence, 3 divides${b}^{2}$

✰By theorem :- If p is a prime number and p divides${a}^{2}$ then p divides a, where "a" is a positive number.

so, 3 divides b also...........................(2)

By (1) and (2)

3 divides both a and b

hence 3 is the factor of a and b

so, a and b have a factor 3

Therefore, a and b are not co-prime .

our assumption is wrong

Therefore,

By contradiction,

$\huge\bold\red{ \sqrt{3}}$ is a irrational.

━━━━━━━━━━━━━━━━━━━━━━