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Prove that, if $f$ is a function of one real variable such that for every real number $y$, there is a unique real number $x$ such that $f(x) = y$, then the function $f$ is one-to-one.


Proposition: If $f$ is a function of one real variable such that for every real number $y$, there is a unique real number $x$ such that $f(x) = y$, then the function $f$ is one-to-one.

A (hypothesis): $f$ is a function of one real variable such that for every real number $y$, there is a unique real number $x$ such that $f(x) = y$.

B (conclusion): The function $f$ is one-to-one.


My Work

B1: For all real number $x$ and $y$ with $x \not = y$, $f(x) \not = f(y)$.

A1: Let $x, y \in \mathbb{R}$ and $x \not = y$.

A2: $f(x) = y$

$f(y) = x$

A3: $x \not = y$

$\therefore f(x) \not = f(y)$

Now I prove uniqueness:

A4: Let $x$ = $t$

A5: $f(t) = y$

A6: $f(x) = y = f(t)$

$\implies f(x) = f(t)$

$Q.E.D.$


I would greatly appreciate it if people could please take the time to review my proof for correctness. If there are any errors, please explain why and what the correct procedure is.

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  • $\begingroup$ A2 doesn't follow. It's not true that $f(x)=y$ for arbitrary $x,y$ as defined in A1. $\endgroup$ – dxiv Feb 20 '17 at 4:14
  • $\begingroup$ The statement to be proved seems like a tautology. What is your definition of "one-to-one function"? $\endgroup$ – bof Feb 20 '17 at 4:16
  • $\begingroup$ @bof The definition is B1. $\endgroup$ – The Pointer Feb 20 '17 at 4:17
  • $\begingroup$ @dxiv A says that, for every real number $y$, there is a unique real number $x$ such that $f(x) = y$. Since it's part of the hypothesis, we can work forward using it? $\endgroup$ – The Pointer Feb 20 '17 at 4:18
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    $\begingroup$ @ThePointer The $x,y$ you work with in A2 are presumably those you defined just before that in A1: "let $x,y \in \mathbb{R}$ and $x\ne y$". A1 says nothing about $f(x)=y\,$ (and, even if it did, it couldn't say that $f(y)=x$ at the same time, which you use next). My advice would be that you first clarify for yourself the line of proof, maybe write it down in plain English, first, then only worry about formalizing it afterwards. What you wrote thus far is not just wrong, but doesn't even give a hint of what the logic/flow of the proof was meant to be. $\endgroup$ – dxiv Feb 20 '17 at 4:24
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Your proof is wrong for the reason pointed out earlier.

You are trying to prove $A_1$. You should first have a method of proof in mind. I do: by contradiction.

You should say: Suppose that $x \neq y$, then if , by contradiction, $f(x)=f(y)$, then let $z=f(x)$.

Then, because of the hypothesis, there exists a unique $u$ such that $f(u) = z$. However, since we have already seen that $f(x) = f(y) = z$, it follows that $u=x=y$. This is the contradiction, and hence $f(x) \neq f(y)$. This is your condition $A_1$, your definition of injectivity.

Please practice these kind of proofs often. I am writing complete proofs so that you can take notice of careful constructions and flow of logic.

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The $x$ and $y $ in the hypothesis are not the same $x$ any why in the conclusion or proof. They are just variables. So in the hypothesis you say $f (x)=y $. But when you say in A1) $x\ne y $ these have nothing to do with the $x,y $ in B1. So A2) $f (x)=y $ is idiotic.

Try this:

Hypothesis: for every $y $ there is a unique $x $ so that $f (x)=y $.

A1: let $w\ne u $. Let $f (w)=z $ and $f (u)=t $

Then try to prove $f (w) \ne f (u) $.

Hint: if $f (w)=f (u)=z=t$ then $f (w)=z $ and $w $ is unique. And $f (u)=t $ and $u $ is unique.

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