Prove under the root 7 irrational

Let us assume that <m:omath><m:r>√7</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}\sqrt7=\frac ab\\b\sqrt7=a\\on\;squaring\;both\;sides\;we\;get\\7b^2=a^2\end{array}}{/tex}

Therefore, a² is divisible by 7,

 it follows that a is also divisible by 7.

So, we can write a = 7c for some integer c.

Substituting for a, we get 7b² = 49c², that is, b^{​​​​​​2} = 7c²

This means that b² is divisible by 7, and so b is also divisible by 7

 Therefore, a and b have at least 7 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>√7</m:r></m:omath> is rational.

So, we conclude that <m:omath><m:r>√7</m:r></m:omath> is irrational.