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?

  • 3
    $\begingroup$ Just take $x=\frac14$, for instance. $\endgroup$ – Brian M. Scott Jan 8 '13 at 19:03
  • 1
    $\begingroup$ 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). $\endgroup$ – Steven Stadnicki Jan 8 '13 at 19:03
  • $\begingroup$ How do I rewrite the question to say X is integer and not prime? Is there a notation for the prime numbers? $\endgroup$ – Leonardo Jan 8 '13 at 19:04
  • $\begingroup$ Just say that $x\in\Bbb Z$; the fact that it’s not prime is irrelevant. $\endgroup$ – Brian M. Scott Jan 8 '13 at 19:07
  • $\begingroup$ 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. $\endgroup$ – 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.


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