Show that the following theorem is true or false: If $x\notin\Bbb Q$ then $x^2\notin\Bbb Q$.

Question

In my approach, I attempted a proof by contrapositive, where it is assumed that $X^2$ is rational and $X$ is also rational.

Therefore, we can represent $X^2$ with a ratio of two integers, $a$ and $b$, where $b$ is not equal to zero: $$X^2 = a/b.$$

Then, by applying the square root to both sides, we can see that $X$ is equal to $\sqrt{a/b}.$

Is this a contradiction to the fact that $X$ is also rational? Is the square root of $a/b$ sufficient to deduce whether or not a number is rational?

Answer 1

Consider $2\in\Bbb Q$. We have $\sqrt{2}\notin \Bbb Q$. Hence the result is false.

Answer 2

Let $x=\sqrt{2} \not \in \mathbb{Q}$. Clearly, $x^2 = 2 \in \mathbb{Q}$. This is a counterexample to the statement.

Answer 3

The fact that there is a simple proof of the obvious counterexample that e.g. $\sqrt 2$ is not rational but $2$ is rational demonstrates that your supposed proof of an assumption that $x^2$ is rational therefore $x$ is rational must be false.

Answer 4

Your theorem has the form

$$(\forall x\in \Bbb R) \;\; (x\notin \Bbb Q\;\; \implies \;\; x^2\in \Bbb Q)$$

its negation

$$(\exists x\in \Bbb R) \; : \; x\notin \Bbb Q \; \text{ and }\; x^2\in \Bbb Q$$

is true if we take $x=\sqrt{3}$.

The theorem is an invalid argument.