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Prove the function $f:\mathbb R \setminus\{0\} \to\mathbb R \setminus \{0\} : x \mapsto \frac1 x$ is a bijection.

Surjective?

Let y ∈ ℝ ∖ {0} such that y = 1 / x.

Notice $f(1/y) = 1 / x = y$.

∴ Surjective.

Injective?

Test if $f(x_1) = (fx_2)$, then $x_1 = x_2$.

$1 / x_1 = 1 / x_2$

$x_2 = x_1$

∴ Injective.

∴ Bijective.

I wanted to prove this by parts. Can anyone spot anything wrong with this logic? It seems sufficient to me. No, I don't want to use the inverse.

Jennifer
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    It's helpful when asking questions like this to give some indication of what you've tried to do to solve the problem. That way we can tailor our help to exactly where you are stuck. Please could you edit the question to tell us what you've tried? – Mathmo123 Aug 28 '14 at 12:31
  • There is my attempt. I'd like verification on this proof. Surely, this question is fine now? – Jennifer Aug 28 '14 at 14:32
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    The "Surjective" part does not seem to verify anything at all... What are you trying to say in this part exactly? – Did Aug 28 '14 at 15:06
  • Sounds like you've answered your question. :) Here's a small linguistic note for future reference: When you write "such that...", you need to give a condition to be imposed on the subject of the sentence. What you have, "$y$ such that $y = 1/x$", isn't completely suitable because you haven't said what $x$ is. What you meant, presumably, is "Let $y$ be a non-zero real number, and let $x = 1/y$. Notice $f(x) = 1/x = y$...." – Andrew D. Hwang Aug 28 '14 at 16:36

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Hint: A function is a bijection if and only if it has an inverse. Can you show that $f^2$ is the identity (i.e. $f^2(x) = f(f(x)) = x$ for all $x$)?

Mathmo123
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