2

I have the following exercise for discrete mathematics:

Show that $f(x)=x^3$ (real-valued) is a bijection. So I have to show that the function is both surjective and injective. So, I know how to do this but I was thinking about an alternative way to show these properties. Are they alright?

Injective

$f'(x)=3x^2$ whenever $x > 0$, $ \text{ } f'(x) > 0$ so it is increasing or decreasing. Whenever $x < 0, \text{ }$ $f'(x) > 0$ so its either increasing or decreasing. Only when $x = 0, f(x) = 0$. So it is impossible that a value in the image is mapped to more than once.

Surjective

$\lim_{x \rightarrow \infty} f(x) = \infty$ and $\lim_{x \rightarrow -\infty} f(x) = -\infty$ so it must be surjective, since it will reach all values in the codomain.

K Abdulahi
  • 23
  • You're making this harder than it needs to be. What's the definition of injectivity? Of surjectivity? – Ben W Dec 28 '18 at 17:01
  • I'm just curious. For me this is high school material that has been drilled in. The way our professor taught us in the introductory discrete math class in university is the way presented in the book 'discrete mathematics and its applications', but I find these methods rather vague. I will of course study them and use those on the exam. – K Abdulahi Dec 28 '18 at 17:05
  • $f'(x)=3x^2$, not $2x^3$. – 2'5 9'2 Dec 28 '18 at 17:10

3 Answers3

0

Taking @Yanko 's comments into consideration, I am editing.

Edit: (for surjectivity)

WTS $\forall a \in \mathbb{R}$ (range), $a$ is in the image of $f$.

Because $f$ is continuous and necessarily non-decreasing (as discussed in a possible proof for injective) then there must exist some $k_1, k_2 \in \mathbb{R}$ s.t. $k_1 < a$ and $a < k_2$. It follows from the intermediate value theorem that $f$ takes the value $a$ for some $x_0$ in the domain.

Previously:

Hint (for surjectivity):

Let $a \in \mathbb{R}$. Can we solve $a = x^3$ for $x$? What does this tell you about the mapping $f$?

T. Fo
  • 652
  • this is not that trivial. I mean, how can you solve $a=x^3$? You probably mean that $x=\sqrt[3]{a}$ but the fact that $\sqrt[3]{}$ exists requires that $f(x)=x^3$ is surjective (and injective). – Yanko Dec 28 '18 at 17:18
  • Isn't it okay to assume that $\sqrt[3]{a}$ exists for all $a$? – T. Fo Dec 28 '18 at 17:22
  • 1
    a function $f$ has an inverse if and only if it is injective and surjective. So if you assume that, you also assume the answer to the question. – Yanko Dec 28 '18 at 17:24
0

Your injectivity proof looks good (except for the typo on the derivative), but you need to say something about $f$ being continuous on $\mathbb{R}$ if you want to use the fact that the limits go to $\pm\infty$ in order to conclude that $f$ is surjective.

DudeMan
  • 111
  • Isn't that trivial for $f$? – K Abdulahi Dec 28 '18 at 17:07
  • Well yes, but so is bijectivity. The point of the exercise is to gain understanding of the definitions. – Ben W Dec 28 '18 at 17:08
  • @KAbdulahi Depends on the level of the course I guess. The point is that having limits that go to $\pm\infty$ do not promise that you hit every intermediate value; there are hyperbolas with slant asymptotes that miss certain values. He clearly wants to invoke the IVT, but he didn't explicitly verify the hypotheses. – DudeMan Dec 28 '18 at 17:09
0

Your answers are good but here are some things I would add/correct:

Injective:

The derivative of $f$ is not $2x^3$, it is $3x^2$. So $f'(x)>0$ for all $x\not = 0$ and so it is always increasing. Note that a function can increase then decrease while not being injective. (for example $f(x)=x^2$ decrease then increase but is not injective because $f(-1)=f(1)$).

Surjective:

You need to mention that you're using the Intermediate Theorem.

Yanko
  • 13,758
  • Yes! Of course that's the derivative. How silly of me. I have never heard of the Theorem you mention, but I will look in to it. Thank you. – K Abdulahi Dec 28 '18 at 17:10
  • @KAbdulahi Ok. The Theorems says that a continuous function $f$ which takes the values $a,b$ takes also all the values in between. – Yanko Dec 28 '18 at 17:11