# 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 $2 \lt b^2 \lt 3$

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

• You don't know anything about $a$ until you define it. What have you tried? What irrational number might be between 2 and 3? Do you know any irrational numbers? Is the sum of a rational number and an irrational number rational or irrational? Can the square root of an irrational number be rational? – marty cohen Mar 23 '15 at 5:59
• I first solved that sum of a rational number and an irrational number is always irrational by using contradiction but that still doesn't prove the fact that $a$ is an irrational such that 2<a<3. Can I set $a = \sqrt(5)$ to prove this problem? – strugglinanalysis Mar 23 '15 at 6:09
• If you can prove that $2 < \sqrt{5} < 3$, yes. – marty cohen Mar 23 '15 at 6:14

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