# Prove that $f(x) = x^a$ is an injective function when a is a real number except 0.

The teacher said there's a mistake in my proof:

Let me start by the definition : $$f$$ would be an injective function if for any $$x,y$$ with $$x\neq y$$, $$f(x) \neq f(y)$$. $$x$$ and $$y$$ are in the domain of the function.

Let $$x$$ and $$y$$ be two real numbers where $$x \neq y$$ and let's show that $$x^a \neq y^a$$ .

Suppose that $$x^a=y^a$$.

$$\ln(x^a)=\ln(y^a)$$

$$a\ln(x)=a\ln(y)$$.

This means that $$x=y$$.

End of proof

If that's the end of the proof, then the proof would be correct for $$a = 0$$. But is the assertion true for $$a = 0$$? Why or why not? What did you forget?

Post-comment remarks If a proof of a statement doesn't use the fact that $$a \ne 0$$, then the proof also proves the statement in the case where $$a = 0$$. (The proof leads from a set of assumptions to a conclusion; the only thing needed to justify the conclusion is the set of assumptions.)

But your proof goes from $$a \ln x = a \ln y$$ (presumably, although it's omitted) to $$\ln x = \ln y$$ presumably by "dividing through by $$a$$"; that operation is really "multiplying through by $$a^{-1}$$, which is valid only if $$a^{-1}$$ exists, which is true only for $$a \ne 0$$.

Anyhow, you've now added that assumption, so you can legitimately get to $$\ln x = \ln y.$$ Now the question becomes "How, from knowing this, do you know that $$x = y$$?" You didn't provide any justification for that step, so your proof is incomplete.

There's an even earlier problem, which is that $$x^a$$ may not be well-defined when $$x < 0$$ and $$a < 1$$. What, for instance, is $$(-1)^\frac12$$?

And then (assuming that your teacher really meant to ask you this only for $$a \ge 1$$), there's the further problem that $$\ln x$$ isn't defined for $$x \le 0$$, so the very first step, where you take log of both sides...that's not valid either.

Writing complete proofs can be hard work. I strongly recommend two-column proof format until you get really good; it helps to keep you honest.

I suggest that you consider the following claim:

For $$a \ne 0$$, let $$f_a : \Bbb R^{+} \to \Bbb R^{+} : x \mapsto x^a.$$ Then $$f_a$$ is injective as a function on the positive real numbers.

That claim is actually true, and a proof very similar to what you wrote will actually prove it.

• Sorry my fault, I changed the title because I forgot to mention that the question said a ≠ 0. I also am not sure about what you mean by "the proof would be correct for a=0. But is the assertion true for a=0? " Nov 19, 2020 at 22:14
• See post-comment additions. Nov 19, 2020 at 23:06