0
$\begingroup$

Building upon this question, I would like to extend the unanswered question in the comments.

That is, do there exist $g,h: \mathbb{N} \rightarrow \mathbb{N}, y \in \mathbb{N}$, such that the following system holds? \begin{align*} (g \circ h \circ h )(y)= 3y \\ (h \circ g \circ h )(y)= 5y \\ (h \circ h \circ g )(y)= 7y \end{align*} Clearly using the provided method fails as these numbers no longer follow the same coincidental pattern ($4 = 2 \times 2$). However, we still have the equations $h(3y)=5h(y)$ and $h(5y)=7h(y)$ after plugging in $h$ to the second and third equations. As stated in the previous question the equations of the form $f(ax)=bf(x)$ only have one solution over the reals ($f(x)=cx^{\log_{a}b}$), but "many over the naturals". Is it still correct to say that these equations can not hold simultaneously? if so, why?

$\endgroup$
4
  • $\begingroup$ Does your definition of $\mathbb{N}$ include $0$? $\endgroup$ Commented Oct 24 at 13:05
  • $\begingroup$ no, it is just the positive integers. i didn't know $\mathbb{N}$ could also include $0$ $\endgroup$ Commented Oct 24 at 13:06
  • $\begingroup$ That's the normal definition, but you never know... I've seen different. $\endgroup$ Commented Oct 24 at 13:07
  • $\begingroup$ USAMTS Year 37 Round 2 $\endgroup$ Commented Nov 3 at 15:24

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.