0

A textbook claims that, if $f$ is invertible, if $h \circ f = f \circ k$, then $h = k$ is false. I've been trying to disprove this by counterexample. But, confusingly, I've been unable to find anything – every example I come up with seems to show that this is true! I then ask GPT4 to show that it is false, but then it tells me that there does not exist any maps $f$, $h$, and $k$ where this is false; GPT4 claims that the invertibility condition on $f$ means/forces that $h \circ f = f \circ k$ means that $h = k$ must be true. Are both GPT4 and I getting this wrong, or is this an error in the textbook?

The Pointer
  • 4,182
  • 2
    Suppose these are bijections. Suppose $h$ and $f$ do not commute, then let $k = f^{-1}\circ h \circ f$. Then we have $h \ne k$. So: find two bijections that do not commute. – GEdgar Sep 20 '23 at 21:42
  • 1
    @MW I do it as a last resort before bothering the good people on this site ;) – The Pointer Sep 20 '23 at 21:45
  • 1
    GPT does not know anything about mathematics. All it can do is spew random text that sounds similar to text it has seen before. This is good enough to sound persuasive, but not good enough to produce correct answers, except by accident. – MJD Sep 20 '23 at 21:52
  • 1
    For example it once said to me that “Any number with four or more digits can be less than 1000, depending on the specific digits that are used. For example, the number 9991 is a four-digit number that is less than 1000.” https://blog.plover.com/tech/gpt/four-digit-numbers.html – MJD Sep 20 '23 at 21:55
  • For a broader, related discussion see also https://math.stackexchange.com/q/4132091/169085 – Alp Uzman Sep 20 '23 at 21:58
  • 1
    don't mention what gpt claims because it usually isn't relevant. If it cannot back up its claim with a proof, when it comes to pure math, you should not assume what it said was true or use that to signal mine. That is how as a society we overextend our trust – Snared Sep 20 '23 at 22:02
  • The statement would become true if either $h \circ f$ is replaced with $f \circ h$ or $f \circ k$ is replaced with $k \circ f$. – Geoffrey Trang Sep 20 '23 at 22:07
  • I think we'd rather be bothered with your questions than have to read sketchy GPT "analyses" lol – H. sapiens rex Sep 20 '23 at 22:57

1 Answers1

3

Here's a counterexample: Let $f,h,k:\mathbb R\to\mathbb R$ be defined by $$f(x)=x^3,\ h(x)=x+1,\ k(x)=\left(x^3+1\right)^{1/3}.$$ Then $$h(f(x))=h(x^3)=x^3+1=k(x)^3=f(k(x)),$$ so $h\circ f=f\circ k$. But $h$ and $k$ are not the same function.

Carl Schildkraut
  • 32,931
  • GPT4: "Yes, I apologize for the oversight in my earlier response. You're right. Given the functions $f$, $h$, and $k$ you provided, they indeed satisfy $h \circ f = f \circ k$ with $f$ being invertible, and $h$ and $k$ being distinct.

    Thank you for pointing it out and providing a concrete counterexample."

     – The Pointer Sep 20 '23 at 21:40
  • @ThePointer :) I guess GPT4 can't do the abstract stuff just yet. Probably best not to rely on it as an arbiter of mathematical truth. – Carl Schildkraut Sep 21 '23 at 03:16