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Let $f:\mathbb{R}\to\mathbb{R}$ be defined as $f(x)=|x|$. How to prove that $f$ is not surjective?

I need some help with this exercise.

Suppose $f$ is surjective. Be $a\in\mathbb{R}$ , $\exists\;b\in\mathbb{R} $ such that $f(b)=a$, well. i consider two cases.

Case 1: $b>0$, then $f(b)=b=a$

Case 2: $b<0$, then $f(b)=-b=a\Rightarrow b=-a$ Okay, in this case i have a problem because i didnt see the contradiction, can someone help me?

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    $|x|\geq 0$ for all $x\in\mathbb{R}$, right? So no negative number is in the range of $f$. – 2017-01-16
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    okay @carmichael561 then the contradiction is in $b=-a$? – 2017-01-16
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    Yep. @FlybyNight sorry – 2017-01-16
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    Don't be afraid to give a *specific* counterexample, as in Jack's answer. To show that a function is not surjective, it is enough to show a *single* element of the range that the function cannot reach. – 2017-01-16
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    The key issue here is that you've focused on the input, $b$, rather than the output, $a$ of the function. You can't arrive at a contradiction because there *are* some $a \in \Bbb R$ that are the output of $f(x)$. – 2017-01-16

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Your cases should depend on the sign of $a$, since for a given $a$ you're trying to find $b$ such that $f(b)=a$.

If $a\geq 0$ then taking $b=a$ gives $f(b)=a$.

But if $a<0$ then there is no $b\in\mathbb{R}$ such that $|b|=a<0$. Indeed, if $b\geq 0$ then $|b|=b\geq 0$, and if $b<0$ then $|b|=-b>0$. In both cases $|b|$ is non-negative, so cannot be equal to $a$.

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You are making things complicated. Look at the definition of "surjective" again:

A function $f:\mathbb{R}\to\mathbb{R}$ is surjective if (and only if) for every $y\in\mathbb{R}$, there exists $x\in\mathbb{R}$ such that $f(x)=y$.

There is no $x\in\mathbb{R}$ such that $f(x)=-1$.