The domain of $\sqrt x $ is $0\le x<\infty$

And its co-domain is $0\le y<\infty$

We know that the square root function is one-to-one since there exist only a unique $y$ such that $y^2=x$ since $y^2$ is a strictly increasing function. That is, $ z>y \implies z^2>y^2$.

But my question is whether the function $y=\sqrt x$ is surjective (onto) or not. Please explain it. Thanks.

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    $\begingroup$ I do not understand what is being asked. Could you elaborate on what you mean by "whether the function is onto its domain or not (cis)?" $\endgroup$ – Simply Beautiful Art Aug 28 '16 at 12:27
  • $\begingroup$ Well, the domain is $[0, \infty)$. Can you show that for all $y \in [0, \infty)$ you can find some $x$ such that $y= \sqrt{x}$? $\endgroup$ – Crostul Aug 28 '16 at 12:28
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    $\begingroup$ The notion of ‘onto’ involves the codomain, not the domain. Simply, it happens, in the present case, that the range, not the codomain, is also the domain. $\endgroup$ – Bernard Aug 28 '16 at 12:30
  • $\begingroup$ Is sqrt function is surjective? $\endgroup$ – Sathasivam K Aug 28 '16 at 12:30
  • $\begingroup$ This depends what you consider the codomain: if it's $\mathbf R$, it's not onto. $\endgroup$ – Bernard Aug 28 '16 at 12:32

For $f:[0,+\infty)\to [0,+\infty), \;\;f(x) = \sqrt x$ to be onto its domain (or "surjective"), you must prove that for every $y \in [0, +\infty)$ (the co-domain of $f$) there is at least one $x \in [0,+\infty)$ (the domain of $f$) so that $f(x) = y$. And indeed, such an $x$ exists, namely $x = y^2$.

  • $\begingroup$ But in the question, codomain is not mentioned. $\endgroup$ – Rajat Aug 28 '16 at 12:34
  • $\begingroup$ The codomain (or target) of $f$ is not the range in general. The domain is the source of $f$. $\endgroup$ – Bernard Aug 28 '16 at 12:34
  • $\begingroup$ In general, codomain and range are different. For onto map, you have to prove indeed both are same. $\endgroup$ – Rajat Aug 28 '16 at 12:35
  • $\begingroup$ OK, I guess I forgot the English terminology. I will correct my answer. $\endgroup$ – mathguy Aug 28 '16 at 12:36

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