Prove that there is an irrational number a such that $2\lt a \lt 3$. Prove also that there is an irrational number $b$ such that $2 \lt b^2 \lt 3$

Question

I tried to prove a is irrational by subtracting two all three sides so I get 0 is less than a-2 less than 1. From there I proved that a-2 is irrational by saying a-2 is equal to rational which contradicts my assumption. Therefore a must be irrational such that 2

Answer 1

$a = 2 + \dfrac{1}{\sqrt{2}}, b = \sqrt{2} + \dfrac{\sqrt{3}+\sqrt{2}}{2}$ will do.