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I was pondering, could one define a function such that

$$T(a,b)^{T(c,d)}=T(a^c,b^d)$$

$∀a,b,c,d$?

I asked the same question some minutes ago, and someone was able to find an example, that of $T(x,y)=1, ∀x,y$. And as much as this is completely correct, I was wondering if more not-so-trivial examples exist.

I'll add a couple of questions to the post then.

  1. Could one define a function as the latter which maps $T(a,b)$ to a number $w$ of our choosing?

For instance, would it be possible to have $T(8,25) = e$? Notice that by defining $T(8,25)$, one is also in turn defining $T(8^8,25^{25})$, among other infinitely many other pairs of values on $T$. I'm not sure however, if by defining $T(8, 25)$ one is also defining $T(2,2)$, for example, which leads to my next question.

  1. Assuming that the answer to question $1$ is "yes", is it then possible to map, again, another pair of elements in $T$ to a number of our choosing, say...$T(2,2)=π$? Or would this lead to a contradiction?

$T(1,1)$ would definitely have to be equal to $1$, since $$T(a,b)^{T(1,1)}=T(a^1,b^1)$$

I would truly appreciate any help/thoughts!

Sam
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1 Answers1

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If $T(1,b)\ne1$ then $T(c,d)=\log(T(1,b^d))/\log(T(1,b))$ is independent of $c$. That gives $T(c,d)=T(d)$. In the same way, there is $T(c,d)=T(c)$.

Otherwise, assume $T(a,1)=T(1,b)=1$.

Let $T(2,2)=X$. Then $T(4,4)=X^X$, and $2^4=4^2$, so $T(16,16)=X^{(X^X)}=(X^X)^X=X^{(X^2)}$, so either $X=1$ or $X^X=X^2$ so $X=2$.

If there is any value $T(a,b)>0,T(a,b)\neq1$ then $$(T(a,b)^{T(c,d)})^{T(e,f)}=T(a^c,b^d)^{T(e,f)}=T(a^{ce},b^{df})=T(a,b)^{T(ce,df)}$$ so $$T(c,d)T(e,f)=T(ce,df)\\T(c,d)=T(c,1)T(1,d)=1$$

To conclude, either $T(a,b)=T(a)$ or $T(a,b)=T(b)$.

Suppose $T(2)=2$. Then $T(4)=T(2^2)=2^2=4$. If $T(k)=k$ then $$. 2^{2k}=4^k=T(4)^{T(k)}=T(4^k)=T(2^{2k})=2^{T(2k)}$$ so $T(2k)=2k$. By induction, $T(2^n)=2^n$ for all $n\in\mathbb{N}$, so $2^{T(n)}=2^n$ and $T(n)=n$ for all $n\in\mathbb{N}$.

Empy2
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  • Thank you. You showed various properties that I didn't know. I have a couple of questions though. First, what do you mean by "independent"? Secondly, when you refer to "$T(x)$", are you abbreviating "$T(x,x)$"? – Sam Feb 22 '18 at 01:32
  • First, you get the same answer for T(c,d) no matter what c is, because it doesn't appear on the right-hand side. Second, yes, I mean T(x,x). – Empy2 Feb 22 '18 at 01:36
  • How exactly do you get $T(c, d)=T(d)$ by assuming $T(1,b)≠1$. – Sam Feb 22 '18 at 03:18
  • $T(1,b)^{T(c,d)}=T(1,b^d)=T(1,b)^{T(1,d)}$, and the base $T(1,b)$ is not 1, so the exponents are the same. – Empy2 Feb 22 '18 at 03:26