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$s : \mathbb{R} \to \mathbb{R}$ given by $s(x)$ = $x^2$.

In this instance why is the function $s$ not onto?

ASB
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user123
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  • because s(1)=s(-1)=1 - two answers to 'if s(x)=1, what is x?' – JMP Mar 10 '15 at 12:15
  • @JonMarkPerry You just explained why the function is not injective (one-to-one). The question is asking why the function is not surjective (onto), meaning that there is a real number that is not in the range. – N. F. Taussig Mar 10 '15 at 12:17
  • i know, just testing - whats bijective - onto-to-onto? – JMP Mar 10 '15 at 12:19
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    @JonMarkPerry They usually write it out as one-one and onto. I have no idea why one wouldn't use the much clearer terms injective, surjective and bijective though. – AlexR Mar 10 '15 at 12:25

3 Answers3

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Hint: What real number has square $-1$?

N. F. Taussig
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$s$ being "onto" means for all $x\in\mathbb R$ there is a $y$ s.t. $s(y)=x$. If $x<0$ then no such $y$ exists. Therefore $s$ cannot be onto.

Gregory Grant
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I'm agree with your answer. If you select a negative number then there is no valu in the domain so that $s(x)=y$

user222469
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