If $\alpha$ is an algebraic number, then prove that $\frac{2}{3}\alpha$ is also algebraic. 
If $\alpha$ is an algebraic number, then prove that
  $\frac{2}{3}\alpha$ is also algebraic. Note: $\alpha \in \mathbb{R}$

I know that if $\alpha$ is algebraic, then there exists some $f(x) \in \mathbb{Q}[x]$ such that $f(\alpha)=0$. But I don't know how to conclude that there exists some $g(x)\in \mathbb{Q}[x]$ such that $g(\frac{2}{3}\alpha)=0$. I thought I could use some kind of property of functions, but I can't find one.
Is there a theorem or something that I'm just completely missing?! THANKS!
 A: First:
$$
\mathbb{Q}[X] = \{ \sum_{i=0}^{n} a_iX^i: a_i\in \mathbb{Q}, n\in \mathbb{N}_0\}.
$$
If $\alpha$ is algebraic, then there is a $f \in \mathbb{Q}[x]$ such that $f(\alpha) = 0$. If you want to show that $\frac{2}{3}\alpha$ is algebraic you just have to find a $g\in \mathbb{Q}[x]$ with $g(\frac{2}{3}\alpha) = 0$. As pointed out by @Med above you can simply take $g(X) = f(\frac{3}{2}X)$. You should, of course, still convince yourself that $g\in \mathbb{Q}[x]$, but this is not hard because if:
$$
f(X)  = \sum_{i=0}^{n} a_iX^i,
$$
then
$$
g(X) = f(\frac{3}{2}X) = \sum_{i=0}^{n} a_i(\frac{3}{2}X)^i = \sum_{i=0}^{n} a_i\frac{3^i}{2^i}X^i = \sum_{i=0}^{n} b_iX^i
$$
where
$$
b_i = a_i\frac{3^i}{2^i} \in \mathbb{Q}.
$$
A: If you want a theorem -
If $\eta$ is an algebraic number over $F$ then $F(\eta)$ has no numbers transcendental over $F.$ A corollary is that the set of all algebraic numbers over $F,$ called $\mathbb{A},$ forms a field.
Let $F(\alpha, \beta)$ be an extension of a number field $F$ and $\alpha$ and $\beta$ be algebraic over $F.$ That is, $\alpha,\beta\in\mathbb{A}.$ Then $\alpha, \beta\in F(\alpha, \beta)$ and since $F(\alpha, \beta)$ is a field it follows that $\alpha\pm\beta, \alpha\beta, \alpha/\beta\in F(\alpha,\beta),$ and, there is also some $\gamma\in F(\alpha,\beta)$ such that you could set $F(\alpha, \beta)=F(\gamma)$ where $\gamma$ is also algebraic over $F.$ There are no numbers in $F(\gamma)$ transcendental over $F,$ so $\alpha\pm\beta, \alpha\beta, \alpha/\beta\in\mathbb{A}.$
Next, let $m_{\alpha, F}(x)$ be the minimal polynomial for $\alpha$ and let its degree be $n.$ Then $m_{\alpha, F}(-x)$ has $-\alpha$ as a zero and $x^nm_{\alpha, F}(1/x)$ has $1/\alpha$ as a zero. $\mathbb{A}$ thus has inverses. Commutativity and associativity come from $\mathbb{C},$ which completes the proof.
Of course, if you just wanted to finish your homework you wouldn't bother to kill a fly with a hammer, so there are simpler ways as noted above. But this is more general, if you are interested.
