-2

I recently posted a question (Solving $f(x+f(y)) = f(x) + y$). However, I can't wrap my head around how to prove this function is injective. I tried using the answers in the responses but my I don't understand how "Let's start by plugging in x=0. We have f(f(y))=f(0)+y. This tells us that f is injective, since if $f(a)=f(b)$, then $f(0)+a=f(f(a))=f(f(b))=f(0)+b$ and hence $a=b$." works. If anyone can give a basic explanation that would be great.

Thanks in advance.

Somos
  • 35,251
  • 3
  • 30
  • 76
user687894
  • 89
  • 5
  • 2
    Why not ask the users who posted answers to the original post? The logic is spelled out quite clearly in those responses....it's easy save for the technical fact that you first must argue that $f(x)$ is rational for all $x$. Personally, I think that should just have been stated as an assumption....the "proof" that it is so comes down to "well, if it is not so, then the question does not make any sense". – lulu Aug 17 '21 at 13:28
  • Exactly where did you get stuck? If $f(a) =f(b) $, then $f(f(a))=f(f(b))$. Does that help? – Naj Kamp Aug 17 '21 at 13:30
  • 1
    This should be a comment on the answer that you don't understand, not a separate question. – Rob Arthan Aug 17 '21 at 13:32
  • The definition of "injective" is: for all $a, b$, if $f(a) = f(b)$ then $a = b$. I proved that. So I'm not sure why you're confused. – Mark Saving Aug 17 '21 at 16:22

1 Answers1

0

Recall that all that is needed to show injectivity is to show that the implication "if the function values are equal, then the function arguments must be equal" holds.

So, to check injectivity, the simplest test is: take equal function values $f(a)=f(b)$, and show that this already implies $a=b$.

This is exactly what this answerer did. Fix $f(a)=f(b)$, then (a bit longer) $$f(f(a))=f(f(b))$$ so you can write $$f(f(a))=f(0+f(a))=f(0)+a=f(0)+b=f(0+f(b))=f(f(b)).$$ Looking at the middle, you can see that $a=b$ holds, which is all we needed to test for.

Ruben Kruepper
  • 2,256
  • 5
  • 14