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I've been playing with Peano Arithmetic and I've got multiplication, division, exponentiation, and logarithms. I can't figure out root extraction but I have a stab at it.

Exponentiation: $a^0 = 1, a^{S(b)} = a\cdot a^b$ Logarithm: $\log_a(1) = 0, \log_a(b) = 1 + \log_a(b/a)$

I'm finding root extraction much more complicated to work out a clean recursive definition for.

This is what I have so far, and I know it's wrong:

$\sqrt[b]{a} = f(1)$, $f(n) = n$ if $n^b = a$, otherwise $f(n+1)$

David Bandel
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  • Could you exmplify what you get wehen trying to compute $\sqrt[2]5$? (Or $\sqrt[2]9$ if you require the root to exist in $\mathbb N$) Your formulas don't make full sense for me ... – Hagen von Eitzen Aug 08 '13 at 14:04
  • Well for this, I'm only concerned with powers of natural numbers. I think I could put in some truncated stuff to deal with things like $\sqrt[2]{5}$ etc. but I'm only concerned with natural roots of natural powers of natural numbers. – David Bandel Aug 08 '13 at 14:13
  • The answer is yes, we can define "$b$-th root extraction" of numbers of the form $n^b$, and we can define a function that returns $0$ or $1$, returning $1$ iff the number has such a form. – Andrés E. Caicedo Aug 08 '13 at 14:17
  • I'm curious about what you mean by "using Peano arithmetic". By the way, your recursive description of exponentiation has a typo, and the one for logarithm is incorrect. But more seriously, what does this have to do with Peano arithmetic, how are you actually defining these functions once you have the recursive descriptions? I assume by "defining using Peano arithmetic" you mean in particular that definitions are first order formulas. (And this can be done, it is only that there is no hint in what you wrote that you see how.) – Andrés E. Caicedo Aug 08 '13 at 14:21
  • To illustrate the difficulties, the link discusses how to define exponentiation in Peano arithmetic. – Andrés E. Caicedo Aug 08 '13 at 14:23
  • Yep I've been all over that page in the past couple days. I fixed the typo in exponentiation's definition.

    By using Peano arithmetic, I'm referring (probably incorrectly) to defining addition as $a + 0 = a$ and $a + S(b) = S(a + b)$ and similarly defining subtraction, and then building it up into multiplication, division, exponentiation, root extraction, and logarithms.

    I'm not sure I see how the one for logarithms is incorrect. Assuming that there is some natural number $k$, such that $a^k = b$, then $log_a(b)$ should equal $1 + log_a(b/a)$

     – David Bandel Aug 08 '13 at 14:27

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