I think that , in general, one can find the inverse relation of a function $y = f(x)$ by exchanging $x$'s and $y$'s places, which gives : $x = f(y)$.
Note : I'm not talking of " inverse function" but simply of " inverse relation". I do not ask for a relation that would necessarily be also a function.
But how to prove formally that this method is correct?
In particular, how to prove formally that the inverse relation of
$\{ (x,y) \mid y = x² \} $
is
$\{ (x,y) \mid x = y² \} $?
How to prove that the ordered pairs belonging to the second relation are the inversed couples of the first relation?
A desired proof would look like this . It would begin with
Inverse of R = { (y,x) | y= f(x) }
and end with
Inverse of R = { (x,y) | x = f(y)} .
My question is : what syntactical "manoeuvres" allow to exchange x's and y's place on the right of the vertical bar?
Best Answer
First you need a definition of what an inverse function is. Then it's almost certain $y=x^2$ fits the definition.
Actually I suppose we also have to know what you mean when you say $x= y^2$ or $y=x^2$ is a relation.
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When you say $x=y^2$ is a relation you are referring to a set $\{(x,y)| x= y^2\} \subset \mathbb R\times \mathbb R$. We can call that set/relation $R$. And when we refer to $y=x^2$ as a relation we are referring to the set, $\{(x,y)| y = x^2\}$. We can call that set $S$.
Now we need a definition of "inverse relation". I'm not sure what your text uses but I think a definition would be
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Now the proof should be straight forward:
If $R = \{(x,y)\in \mathbb R\times \mathbb R|x=y^2\}$ and $S=\{(x,y)\in \mathbb R\times \mathbb R|y=x^2\}$ then
$x=y^2 \iff (x,y) \in \{(x,y) \in \mathbb R\times \mathbb R|x = y^2\} = R$.
$x=y^2 \iff (y,x) \in \{(y,x) \in \mathbb R\times \mathbb R|x = y^2\} =\{(w,z)\in \mathbb R\times \mathbb R|z=w^2\} =\{(x,y)\in \mathbb R\times \mathbb R|y=x^2\}=S$ (variables are nothing but place holders)
So $R^{-1} = S$.