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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.

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    $\begingroup$ For b. consider $y=2\,$ for example. $\endgroup$ – dxiv Feb 2 '17 at 0:16
  • $\begingroup$ MathJax note: for sets in math mode use \{ \} not { }. $\endgroup$ – David Feb 2 '17 at 0:19
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    $\begingroup$ @dxiv Oh yeah I forgot about that. Thanks! $\endgroup$ – readytorockandorroll Feb 2 '17 at 0:22
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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?

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  • $\begingroup$ It is not true for part b bcause if we take $y=2$ there is no rational expression expression for $x$. $\endgroup$ – readytorockandorroll Feb 2 '17 at 0:41
  • $\begingroup$ That's right... though my main point is that for a proper proof of (a) you really should include the red bit. $\endgroup$ – David Feb 2 '17 at 0:53

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