# Fixed: Is this set empty? $S = \{ x \in \mathbb{Z} \mid \sqrt{x} \in \mathbb{Q}, \sqrt{x} \notin \mathbb{Z}, x \notin \mathbb{P}$ }

This question has been "fixed" to reflect the question that I intended to ask

Is this set empty? $S = \{ x \in \mathbb{Z} \mid \sqrt{x} \in \mathbb{Q}, \sqrt{x} \notin \mathbb{Z}, x \notin \mathbb{P}$ }

Is there a integer, $x$, that is not prime and has a root that is not irrational?

• Just take $x=\frac14$, for instance. – Brian M. Scott Jan 8 '13 at 19:03
• Over what domain are you quantifying $x$? If $x$ is allowed to be rational, then trivially $x=\frac14$ works; if $x$ is supposed to be integral, then its square root is either integral or irrational (this is a nice and standard exercise in a beginning number-theory class). – Steven Stadnicki Jan 8 '13 at 19:03
• How do I rewrite the question to say X is integer and not prime? Is there a notation for the prime numbers? – Leonardo Jan 8 '13 at 19:04
• Just say that $x\in\Bbb Z$; the fact that it’s not prime is irrelevant. – Brian M. Scott Jan 8 '13 at 19:07
• I want to know if any numbers greater than 1 that are not prime have a root that is not irrational. I will try and make a new question that is better. – Leonardo Jan 8 '13 at 19:08

Edited in response to the latest change to the question

The answer is ‘no’. As you are considering $\sqrt{x}$, we must look at $x \geq 0$.

Suppose that $\sqrt{x} \in \mathbb{Q} \setminus \mathbb{Z}$. Let $\sqrt{x} = \dfrac{p}{q}$, where $p \in \mathbb{N}_{0}$, $q \in \mathbb{N}$ and $\gcd(p,q) = 1$. This yields $$q^{2} x = p^{2}.$$ By way of contradiction, assume that $x$ is an integer. Then by the identity above, $q$ must divide $p^{2}$. However, $\gcd(p,q) = 1$, so this means that $q = 1$. Hence, $\sqrt{x} = p$, which is a contradiction because we started our argument with $\sqrt{x} \in \mathbb{Q} \setminus \mathbb{Z}$.

Conclusion: $S = \varnothing$.

If you see it this way: $S=\{x^2|x\in Q,x\not\in Z\}$, then every a²/b² is valid where b doesn't divide a.

To answer according to the last edit: Yes.

Let $a\in\mathbb{Z}$ and consider the polynomial $x^{2}-a$.

By the rational root theorm if there is a rational root $\frac{r}{s}$ then $s|1$ hence $s=\pm1$ and the root is an integer. So $\sqrt{a}\in\mathbb{Q}\iff\sqrt{a}\in\mathbb{Z}$ .

Since you assumed that the root is not an integer but is a rational number this can not be.