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I am confident that both functions f and g have inverse functions, because they are one-to-one and onto. But I am having trouble proving it.

When I draw the image for f((3,4)), on the left hand side (3,4) points to (4,3) (on RHS) and on the left hand side 4 points to 3 (on RHS). When I draw the inverse function, (4,3) points to (3,4). Each point on the image on the right hand side only has one arrow pointing into it from the left hand side, which is why I believe that it is one-to-one.

The image of f is equal to its domain (3,4) = (3,4) hence I believe that it is onto.

Does my logic and proof sound correct? I have the feeling that it could be explained in a more concrete manner.

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  • $\begingroup$ Hint: $g(2,0)=(1,0)=g(-2,0)$. $\endgroup$ – vadim123 Apr 24 '18 at 3:05
  • $\begingroup$ The proof that the first function is invertible (and hence bijective) should be done much more generally without reference to any specific values of $x$ or $y$. For all pairs of pairs in the domain, if you have $f((x_1,y_1))=f((x_2,y_2))$ does that imply that $(x_1,y_1)=(x_2,y_2)$? If you have a desired pair in the codomain, $(a,b)$ must there exist some pair $(x,y)$ such that $f((x,y))=(a,b)$? $\endgroup$ – JMoravitz Apr 24 '18 at 3:11
  • $\begingroup$ ahh i understand the hint now. since (2,0) and (-2,0) achieve the same result, it is not one to one, therefore the inverse of g cannot exist. thank you for your help! $\endgroup$ – JohnLovesMaths Apr 24 '18 at 4:34
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Your idea on the first one is correct, for the second one it's not so easy. Note that if $g(x,y) = (3,4)$ we clearly have $y=4$ but $x$ could be both $6$ and $-6$. If the domain of $g$ was $\mathbb{Z}^+\times \mathbb{Z}$ it would be a different story, but as stated, is $g^{-1}$ a function?

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