Question

examine whether the following numbers are rational or irrational
6\2√3​

Answer 1

I assume what you meant when you wrote the question is $\frac{6}{2\sqrt{3}}$

If we simplify the equation, we get:

$\frac{3}{\sqrt{3}}$

$\sqrt{3}$

It is proven that for all n is an integer, but not a perfect square, the square root of n is irrational; therefore, the number is irrational.

The proof is by contradiction.