Let $f(x) = x^{2}$ for all $x \in \Bbb R$. Show that $f[\Bbb Q] \subset \Bbb Q$

Question

We know that $f[\Bbb Q]$ is the set of all values that $f$ takes on given points in $\Bbb Q$, i.e. $f[\Bbb Q] = \{f(x):x\in \Bbb Q\}$.

But how do I show that every $f(x)$ is in $\Bbb Q$?

Thanks!

That's equivalent to asking "How do I know $q^2$ is rational if q is". Can you do that? If so, simply state "Let $x\in \mathbb Q$. Then $f(x) = x^2 \in \mathbb Q$. So $f(\mathbb Q) \subset \mathbb Q$. — fleablood, Apr 12 '16 at 23:08

score 6 · Answer 1 · answered Apr 12 '16 at 22:58

6

Let $x \in \mathbb{Q}$, then $x=\frac{a}{b}$, $a,b \in \mathbb{Z}$, $b \not= 0$. $f(x)=\frac{a^{2}}{b^{2}} \in \mathbb{Q}$, so $f(\mathbb{Q}) \subseteq \mathbb{Q}$

answered Apr 12 '16 at 22:58

mich95

8,713

1

An you can drop the potentially equal, since nothing in $\Bbb Q$ squares to $2$. – snulty Apr 13 '16 at 17:49

score 3 · Answer 2 · answered Apr 12 '16 at 22:58

3

It helps to formalize what $\mathbb{Q}$ is.

$\mathbb{Q}=\{\frac{a}{b}|a,b\in\mathbb{Z}\text{ and }b\not=0\}$.

Let $q\in\mathbb{Q}$. Then, there are integers $a$ and $b$ such that $q=\frac{a}{b}$. Hence, $f(q)=q^{2}=\left(\frac{a}{b}\right)^{2}=\frac{a^{2}}{b^{2}}\in\mathbb{Q}$.

answered Apr 12 '16 at 22:58

ervx

12,208

Ah that does it. I guess its never a bad idea to break everything down as much as possible! – kanker7 Apr 12 '16 at 23:02
Right. Glad to help. – ervx Apr 12 '16 at 23:02

score 1 · Answer 3 · answered Apr 12 '16 at 23:43

Other two answers showed that if $x$ is rational then $x^2$ is rational.

You can use induction and show that if $x$ is rational then $x^n$ is rational for all natural numbers $n$.

Since the finite addition/multiplication of rationals are rationals too, we can conclude that fon any polynomial $P$ over $\mathbb Q$ we have $$P(\mathbb Q) \subset \mathbb Q.$$

Let $f(x) = x^{2}$ for all $x \in \Bbb R$. Show that $f[\Bbb Q] \subset \Bbb Q$

3 Answers3