Hello please could you tell me if my proof that the sqrt of 4 is irrational is correct.

$\displaystyle \sqrt{4}=\frac{a}{b}$ therefore $\displaystyle 4=(\frac{a}{b})^{2}$ so $\displaystyle 4b^2=a^2$....1.1

This implies that a is greater than b and that a is even.

a is even so a=2k

put into 1.1 $\displaystyle 4b^2=4k^2$

cancel by 4 $\displaystyle b^2=k^2$

this can be written as $\displaystyle (\frac{2k}{2})^{2}=b^2$ this implies that $\displaystyle (\frac{a}{2})^{2}=b^2$ which implies that $\displaystyle a^2=2b^2$....1.2

Equation 1.2 contradicts equation 1.1. This shows that a and b cannot be whole positive integers.