Rather than trying to guess and check, we can actually construct a counterexample to the statement.
So, what is an irrational number? The prefix "ir" means not, so we can say that an irrational number is something that's not a rational number, right? Since we know a rational number is a ratio between two integers, we can conclude an irrational number is a number that's not a ratio of two integers. So, an easy way to show that not all square roots are irrational would be to square a rational number then take the square root of it. Let's use three halves for our example:
[tex] \sqrt{(\frac{3}{2})^2}=\\\sqrt{\frac{9}{4}}=\\\frac{3}{2} [/tex]
So clearly 9/4 is a counterexample to the statement. We can also say something stronger: All squared rational numbers are not irrational number when rooted. How would we prove this? Well, let [tex] \frac{a}{b} [/tex] be a rational number. That would mean, [tex] \frac{a^2}{b^2} [/tex], would be a/b squared. Taking the square root of it yields:
[tex] \sqrt{\frac{a^2}{b^2}}}=\\ \frac{\sqrt{a^2}}{\sqrt{b^2}}=\\ \frac{a}{b} [/tex]
So our stronger statement is proven, and we know that the original claim is decisively false.
