Determining function is onto a. Consider the  set $ℝ^+ = \{x∈ℝ|x>0\}$ together. Let $f:ℝ^+→ℝ^+$ be the function  given by $f(x) = x^2.$ Is $f$ onto?
b. Consider the  set $ℚ^+ = \{x∈ℚ|x>0\}$ together. Let $f:ℚ^+→ℚ^+$ be the function  given by $f(x) = x^2.$ Is $f$ onto?
Workings:
a. Let $y \in \mathbb{R}^+$. Let $x = \sqrt{y}$.
Then we have
$f(x)=x^2 = (\sqrt y)^2 = y$
Therefore $f$ is onto.
What I am wondering is. If the same would follow b. Any help will be appreciated.

Your complete proof for (a) should be as follows (the red is the bit you left out).

Let $y\in{\Bbb R}^+$. Let $x=\sqrt y$.  Then $\color{red}{x\in{\Bbb R}^+}$ and we have
  $$f(x)=x^2=(\sqrt y)^2=y\ .$$
  Therefore f is onto.

The corresponding proof for (b) would be:

Let $y\in{\Bbb Q}^+$. Let $x=\sqrt y$.  Then $\color{red}{x\in{\Bbb Q}^+}$ and we have
  $$f(x)=x^2=(\sqrt y)^2=y\ .$$
  Therefore f is onto.

Can you decide whether or not this is correct?
