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Dr. Pinter's "A Book of Abstract Algebra" presents this exercise:

Prove/disprove whether the following function $f$ (with inputs/outputs of real numbers) is injective and/or surjective: $$f(x)=x^{3}+1.$$

Using this helpful answer, I solved for $$f(a)=f(b)\colon$$

$$f(a)=a^{3}+1$$ $$f(b)=b^{3}+1$$ $$a^{3}+1=b^{3}+1$$ $$a^{3}=b^{3} \text{ subtract 1 from each side}$$ $$(a^{3})^{1/3}=(b^{3})^{1/3} \text{ take the cubic root of each side}$$ $$a=b$$

Therefore, $f$ is injective.

Again, following that helpful answer, solve for $x$, and then plug into $f(x)$:

$$y=x^{3}+1$$ $$y-1=x^{3}$$ $$x=(y-1)^{\frac{1}{3}}$$

Now, plug $x$ into $f$, and check if result equals $y$, which would prove the surjective property.

$$f(x)=x^{3}+1$$ $$y=((y-1)^{1/3})^{3}+1 \text{ use $y$ rather than $f(x)$}$$ $$y=y-1+1$$ $$y=y$$

Therefore, it's also surjective.

Is this right?

Community
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Kevin Meredith
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    Yes, it is correct! – wythagoras May 15 '15 at 18:44
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    you need to state what the domain and target are (all real numbers, presumably). also, you write $x^{-3}$ when you mean $x^{1/3}$. – andrewBee May 15 '15 at 18:44
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    Just to reiterate what @andrewBee comment addresses: you must specify the domain and codomain of your mapping. After all, you are trying to prove properties about the domain and codomain. I would not simply say "with inputs/outputs real numbers." I would say something along the lines of "domain and codomain $\mathbb{R}$" or you can do as I have done in my answer(s) and say how the mapping is defined, where the domain and codomain are given: $$f\colon\underbrace{\mathbb{R}}{ \text{domain}}\to\underbrace{\mathbb{R}}{\text{codomain}}.$$ – Daniel W. Farlow May 15 '15 at 19:18

2 Answers2

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As in the answer you linked to (my answer), I would try to structure your proofs more clearly:

Claim: The mapping $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3+1$ is injective.

Proof. Let $x_1,x_2\in\mathbb{R}$ and suppose $f(x_1)=f(x_2)$. Then $$ f(x_1)=f(x_2)\\[0.5em] x_1^3+1=x_2^3+1\\[0.5em] x_1^3=x_2^3\\[0.5em] x_1=x_2. $$ Thus, the mapping is injective. $\blacksquare$

Claim: The mapping $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3+1$ is surjective.

Proof. Suppose $y\in\mathbb{R}$. Then let $x=\sqrt[3]{y-1}$. We have the following: \begin{align} f(x) &= x^3+1\\[0.5em] &= (\sqrt[3]{y-1})^3+1\\[0.5em] &= (y-1)+1\\[0.5em] &= y. \end{align} Thus, the mapping is surjective. $\blacksquare$

Note: As you have seen, writing up proofs in math is not only about squeezing out an accurate proof. You want to write it clearly, concisely, etc. The goal is to be concise but not at the cost of clarity. Oftentimes this is a fine line to walk.

Daniel W. Farlow
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    I think this proof lacks a little bit in precision at the part where $x_{1}^{3}=x_{2}^{3} \Rightarrow x_{1}=x_{2}$ is concluded. It might be the most interesting point of the whole proof. – Celsius May 15 '15 at 19:37
  • @Celsius How else would you go about showing the implication? Since both numbers are being cubed, there is no need to worry about a multi-value problem. – Daniel W. Farlow May 15 '15 at 19:39
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    Yeah, there is still no need to worry about multiple cubic roots. But how would you argue rigorously? Ad hoc I would use the fact that $x^3-p$ splits in $\mathbb{C}$ for all $p \in \mathbb{R}$ because of the algebraic closure of $\mathbb{C}$ and notice that only one of the three roots lies actually in $\mathbb{R}$. But this requires some knowledge in algebra. – Celsius May 15 '15 at 19:55
  • @Celsius This could lead to an endless chain of "why? Why? Why? ..." – wlad May 15 '15 at 21:49
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    It should be quite a natural behaviour of a mathematician to ask qestions why something works fine everywhere you go even if you are confronted with an axiom. At this point you might ask yourself if there would not be a simpler system of axioms you could use for your work or similar things. But normally in a proof, you can stop your chain of conclusions if you reached axioms. In almost all cases this process will terminate - or one gives up before. ;-) – Celsius May 16 '15 at 11:44
  • @Celsius Thanks for asking "why", Celsius:) – Sha Vuklia Sep 07 '19 at 15:12
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Your proof of of $f$ being injective is correct (although when you are at the point $a^3 = b^3$ you want to raise each side to the $\frac{1}{3}$ power, not the $-3$rd power.) As for the surjective part, I would also personally mention that you know $x = \sqrt[3]{y-1}$ is in the domain of $f$ because $\sqrt[3]{y-1}$ exists for every $y \in \mathbb{R}$. After all, this is the key point of a function being surjective. Not just the fact that you can plug in $x$ and get $y=y$. It may be clear to you or others already that $\sqrt[3]{y-1}$ exists for every $y \in \mathbb{R}$, but if you are just starting to work with into/onto properties of functions, I think it is something that should be explicitly stated. All in all, nice job though!

graydad
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