Prove that√8 is a irrational?

Let us assume that <m:omath><m:r>√8</m:r></m:omath> is rational.

That is, we can find integers a and b (<m:omath><m:r>≠0)</m:r></m:omath> such that a and b are co-prime

{tex}\style{font-family:Arial}{\begin{array}{l}\sqrt8=\frac ab\\b\sqrt8=a\\on\;squaring\;both\;sides\;we\;get\\8b^2=a^2\end{array}}{/tex}

Therefore, a² is divisible by 8,

 it follows that a is also divisible by 8.

So, we can write a = 8c for some integer c.

Substituting for a, we get 8b² = 64c², that is, b^{​​​​​​2} = 8c²

This means that b² is divisible by 8, and so b is also divisible by 8

 Therefore, a and b have at least 8 as a common factor.

But this contradicts the fact that a and b are co-prime.

This contradiction has arisen because of our incorrect assumption that <m:omath><m:r>√8</m:r></m:omath> is rational.

So, we conclude that <m:omath><m:r>√8</m:r></m:omath> is irrational.

Firsty let root 8 is rational Then root8/1=p/q(p and q r coprime anf q is not equal to 0) Then cross multiplication then we obtain P=root8q Square on both side P2=(root8q)2 Then we obtain P2=8q2 because root se square cancel ho jayga Then we can say that p2 divide 8exactly and p will also divide 8 exaclty Then by euclid division lemma P=8q+0 Then square on both side P2=(8q)2 Then put the value of p2 8q2=64q2 8q2=8(8q2) Then 8 se 8 cancel ho jayga Then q2=8q2 Then we can say that q2 will divide 8 exactly and q will divide exactly Then the common factor of p and p is 8 then it is contraduction to our supposition. So, our supposition is wrong . Hence root 8 is irrational no.