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Is it correct to define the positive real numbers as $\{f(x) = x^2\mid x \in \mathbb R\}$?

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    $\begingroup$ You probably mean $\{x^2\mid x\in\mathbb{R}\}$. That is a description of the set of non-negative reals. $\endgroup$ – André Nicolas Oct 21 '14 at 3:12
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    $\begingroup$ The image of $f(x)=\dfrac{1}{|x|}$ will give you all positive reals. $\endgroup$ – David Peterson Oct 21 '14 at 3:16
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    $\begingroup$ @David: Of course this wasn't mentioned in the question, but if you're going to start nitpicking and remove points from the domain of the function, why not the identity function on the positive reals and that's it? $\endgroup$ – Asaf Karagila Oct 21 '14 at 3:20
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Why don't use such that $\{x\in \Bbb R: x>0\}$?

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$$ \{y^2:y \in \mathbb{R} \backslash \{x^2:x \in \mathbb{R}\}\} $$

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  • $\begingroup$ Maybe try to explain why this is an answer to the question? $\endgroup$ – Eric Stucky Oct 21 '14 at 4:31
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    $\begingroup$ @David-Holden This is clever. $\endgroup$ – J. David Taylor Oct 21 '14 at 4:49
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Try the exponential function. Paul, he wanted to do it using a function.

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  • $\begingroup$ f(x)=x is also a function ;) $\endgroup$ – Paul Oct 21 '14 at 3:24
  • $\begingroup$ Why is this half a comment? Comments go to the comments, answers go to the answers. $\endgroup$ – Asaf Karagila Oct 21 '14 at 3:24
  • $\begingroup$ Also, it's not clear that E wanted to do it using a function. Here is a reason why E might have not: $x^2$ has the advantage over $e^x$ that it is defined completely algebraically, showing that we can build the topology from the algebraic structure. $\endgroup$ – Eric Stucky Oct 21 '14 at 4:32

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