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I have the following question $h(x)=4-x^2, x\in \Bbb{R}$

From my understanding it is only surjective when for all values in the Range that is in the Co domain there exist a $x$ value in the domain for it, so the co-domain and range must be the same, not surjective I assume that a value exist in the co domain that doesn't exist in the range.

This is what I tried:

$-(\sqrt{y-4})=x$

when y = 3:

$-(\sqrt{1})=x$

Since the co-domain is restricted to Real numbers and that is an imaginary number can I safely say that it is not surjective ?

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  • $\begingroup$ h(1) = 3. But, try to find something that maps to 2. $\endgroup$
    – Doug M
    Nov 9, 2016 at 2:56
  • $\begingroup$ @DougM but is that the mathematical way to solve it , trying values ? i dont think i am doing it properly. $\endgroup$
    – user2861799
    Nov 9, 2016 at 3:05
  • $\begingroup$ You only need one example to show that it is not a subjection. If x is in the integers, there is no x that maps to 2, and you are done. That is the "mathematical" way to solve it. $\endgroup$
    – Doug M
    Nov 9, 2016 at 3:07
  • $\begingroup$ @user2861799 May I know the co-domain is $\Bbb{R}$ or $\Bbb{Z}$? $\endgroup$
    – Alan Wang
    Nov 9, 2016 at 3:16

1 Answer 1

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$h(x) \le 4$ for all $x \in \mathbb R$.

Therefore, there is no $x$ such that $h(x)=5$ for instance. Thus, $h$ is not surjective.

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  • $\begingroup$ just saw an error , its Real numbers, sorry i had Integers $\endgroup$
    – user2861799
    Nov 9, 2016 at 3:25
  • $\begingroup$ Just for clarity , if we had placed a boundary of some sort on the domain , we could make it surjective correct ? @lhf $\endgroup$
    – user2861799
    Nov 9, 2016 at 3:34
  • $\begingroup$ @user2861799, on the domain, no; but on the codomain, yes. $\endgroup$
    – lhf
    Nov 9, 2016 at 3:36
  • $\begingroup$ yea sorry meant codomain thanks $\endgroup$
    – user2861799
    Nov 9, 2016 at 3:41

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