Prove that .root 3 is irrational …

Let us assume that <m:omath><m:r>√</m:r></m:omath>3 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}\sqrt3=\frac ab\\b\sqrt3=a\\on\;squaring\;both\;sides\;we\;get\\3b^2=a^2\end{array}}{/tex}
Therefore, a² is divisible by 3,
 it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b² = 9c², that is, b^{​​​​​​2} = 3c²
This means that b² is 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 co-prime.
This contradiction has arisen because of our incorrect assumption that <m:omath><m:r>√</m:r></m:omath>3 is rational.
So, we conclude that <m:omath><m:r>√</m:r></m:omath>3 is irrational.