Okay, so I know there are plenty of injective/surjective (and thus, bijective) questions out there but I'm still not happy with the rigor of what I have done. So I'd really appreciate some help!

So the question actually asks me to do two things:

(a) give an example of a cubic function that is bijective. Explain why it is bijective.

(b) give an example of a cubic function that is not bijective. Explain why it is not bijective.

So for (a) I'm fairly happy with what I've done (I think):

$$ f: \mathbb R \rightarrow \mathbb R , f(x) = x^3$$

So we know that to prove if a function is bijective, we must prove it is both injective and surjective.

Proof: $f$ is injective

Let: $$x,y \in \mathbb R : f(x) = f(y)$$ $$x^3 = y^3$$ (take cube root of both sides) $$x=y$$

Proof: $f$ is surjective

Let: $$y \in \mathbb R$$

$$x = \sqrt[3]{y}$$

$$f(x) = (\sqrt[3]{y})^3 = y$$

So I believe that is enough to prove bijectivity for $f(x) = x^3$. Keep in mind I have cut out some of the formalities i.e. invoking definitions and sentences explaining steps to save readers time. This is just 'bare essentials'.

So for (b)

$$f: \mathbb R \rightarrow \mathbb R , f(x) = x^3 - x$$

Now I'm just going to try and prove it is NOT injective, as that should be sufficient to prove it is NOT bijective.


$$x,y \in \mathbb R : f(x) = f(y)$$ $$x^3 - x = y^3 - y$$

This is about as far as I get. Send help. Thanks everyone.

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    $\begingroup$ What happens if $x=0$ or $x=1$? $\endgroup$ – Cameron Williams Mar 28 '16 at 2:59
  • 1
    $\begingroup$ To prove that a function $f$ is not injective, you just need to find two different numbers $x$ and $y$ such that $f(x)\not=f(y)$. $\endgroup$ – Christopher Carl Heckman Mar 28 '16 at 3:02
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    $\begingroup$ @Carl you mean two numbers $x,y$ such that $x\neq y$ but $f(x)=f(y)$ $\endgroup$ – JMoravitz Mar 28 '16 at 3:03
  • $\begingroup$ @CarlHeckman as JMoravitz said, you may confuse op... $\endgroup$ – YoTengoUnLCD Mar 28 '16 at 3:07
  • $\begingroup$ Thanks for the help guys, +1's given. $\endgroup$ – Rubicon Mar 28 '16 at 3:53


A function $f$ from $X\to Y$ is said to be injective iff the following statement holds true:

for every $x_1,x_2\in X$ if $x_1\neq x_2$ then $f(x_1)\neq f(x_2)$

The negation of this then yields:

A function $f$ from $X\to Y$ is not injective iff there exists $x_1,x_2\in X$ such that $x_1\neq x_2$ but $f(x_1)=f(x_2)$

In the case of the cubic in question, it is an easily factorable polynomial and we can find multiple distinct roots. $f(x)=x^3-x=x(x^2-1)=x(x+1)(x-1)$

We know that a root of a polynomial is a number $\alpha$ such that $f(\alpha)=0$

So, $f(1)=f(0)=f(-1)=0$ despite $1,0,-1$ all being distinct unequal numbers in the domain. This shows that it is not injective, and thus not bijective.

As an aside, one can prove that any odd degree polynomial from $\Bbb R\to \Bbb R$ must be surjective by the fact that polynomials are continuous and the intermediate value theorem.

  • $\begingroup$ This is great. I think it has helped my understanding significantly. So what you're sort of saying is because f(1) and f(-1) equal 0, there is more than one-to-one mapping to the elements in the co-domain (hence not injective)? So in this case the set might even look like this: {(1,0) , (-1,0)} (just as a way to visualise it)? $\endgroup$ – Rubicon Mar 28 '16 at 3:53

Regarding (a), when you say "take cube root of both sides" you are (at least implicitly) assuming that the function is injective -- if it were not, the cube root would not be a well-defined operation. Depending on what you are / are not allowed to assume, that could be regarded as a circular argument.

However, there is another approach. Beginning with the assumption that $x^3=y^3$, you can manipulate the expression as follows: $$x^3-y^3=0$$ $$(x-y)(x^2+xy+y^2)=0$$ Now if you can find a convincing argument that the second factor, $x^2+xy+y^2$, is always positive for any choice of $x$ and $y$, then it follows that the only way $x^3=y^3$ can be true is if $x=y$.

The factorization $x^3-y^3=(x-y)(x^2+xy+y^2)$, by the way, could also give you some insight into how to handle (b).

  • $\begingroup$ $$x^2+xy+y^2 = {x^2+y^2\over 2} + {1\over 2}(x^2 + 2xy+y^2) = {x^2+y^2\over 2} + {1\over2}(x+y)^2 \ge 0.$$ The only way to get equality is to have $x=0$ and $y=0$, but then $x=y$ in that case as well. $\endgroup$ – Christopher Carl Heckman Mar 29 '16 at 6:30
  • $\begingroup$ @CarlHeckman Yes, I know. Also $x^2+xy+y^2 = (x + \frac{1}{2}y)^2 + \frac{3}{4}y^2$. I was intentionally leaving that out to leave the OP something to finish. $\endgroup$ – mweiss Mar 29 '16 at 14:16

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