Question

please answer my question : prove that √2 is irrational​

Answer 1

Step-by-step explanation:

$\bold\red{Proof:}$

This type of proofs are done from Contradiction.

So,

Let us assume that √2 is a rational number.

then,

as we know that,

A rational number should be in the form of $\bold{\frac{p}{q}}$

where p and q are co- prime number.

So,

=> √2 = p/q { where p and q are co- prime}

=> √2q = p

Now,

by squaring both the side

we get,

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

$2{q}^{2} = {p}^{2}$ ....... ( i )

So,

if 2 is the factor of ${p}^{2}$

then, 2 is also a factor of p .... ( ii )

=> Let p = 2m { where m is any integer }

Again,

squaring both the sides,

ws get,

${p}^{2} = {(2m)}^{2}$

${p}^{2} = 4{m}^{2}$

Now,

putting the value of ${p}^{2}$ in equation ( i ),

we get,

$=>{2q}^{2} = {p}^{2}\\\\=>2{q}^{2} = 4{m}^{2}\\\\=>{q}^{2} = 2{m}^{2}$

So,

if 2 is factor of $\bold{{q}^{2}}$

then, 2 is also factor of q

Since

2 is factor of p & q both

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

Hence,

√2 is an irrational number

Thus,

Proved

Answer 2

Hey your answer is in the given attachment

Hope it helps.

❤#thank you#❤