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I can write $x$ is a power of $2$ as: $$(\forall p)(1 < p \wedge (\exists z) x =p\times z \implies (\exists k) k+k = p)$$ but if I wanted to write $x = 2^y$ for some positive integer $y$ would this be the same as the above? There is no mention of $y$ in the above but it seems like $x$ is a power of $2$ and $x = 2^y$ are equivalent statements so there should be no need to change the above.

$1)$ I think it is necessary to change the above in the second case but why?

$2)$ What should it be changed to in the $x= 2^y$ case?

mtheorylord
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  • The statement you wrote is, indeed, equivalent to the statement

    $$(\exists y)(x=2^y)$$

    but the equivalence is non-trivial. You can't just replace one part of your statement to change into the one I wrote, but it is possible to prove the equivalence of the two statements.

     – 5xum Oct 05 '17 at 07:50
  • I am interested in the proof of this. – mtheorylord Oct 05 '17 at 07:50
  • How formal do you want the proof to be? Does it have to go way down to axioms? – 5xum Oct 05 '17 at 07:51
  • Preferably, yes. Or just give me the outline and I can fill in the details – mtheorylord Oct 05 '17 at 07:53
  • What is the statement $x =2^y$ in first order logic? – mtheorylord Oct 05 '17 at 07:53
  • as I wrote, $\exists y: x=2^y$. – 5xum Oct 05 '17 at 07:56
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    https://math.stackexchange.com/questions/312891/how-is-exponentiation-defined-in-peano-arithmetic – mercio Oct 05 '17 at 07:56
  • But as far as I know you cant use $2^y$ in first order number theory until you define $2^y$ which is non-trivial. – mtheorylord Oct 05 '17 at 07:56
  • Your statement is basically, in English: "The only (positive) odd divisor of $x$ is $1$". Going from there to "$x$ is a power of $2$" requires the fundamental theorem of arithmetic, which means that it's basic number theory, but perhaps not so simple to state the proof using pure logical notation. – Arthur Oct 05 '17 at 08:33
  • Easier to say "every nontrivial ($\ne 1$) factor of $x$ is even". A proof would probably use strong induction and the transitive nature of divisibility, that is $a|b \land b|c \implies a|c$ – DanielV Oct 05 '17 at 08:39
  • I would suggest establishing equivalency by showing both statements are equivalent to "the only prime factor of $x$ is 2". – DanielV Oct 05 '17 at 08:54
  • To prove equivalence, you need existence and uniqueness of prime factorization, or at least of a unique factorization as $2^ku$, where $u$ is odd. –  Oct 05 '17 at 09:54

1 Answers1

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Here's one way to approach proving the equivalence (and contrary to claims in the comments, we don't have to use the fundamental theorem of arithmetic anywhere):

Suppose $x$ satisfies your statement; we want to show that $x$ is a power of $2$.

Let $y$ be the least natural number satisfying $2^{y+1}\not\vert x$; proving that such a $y$ exists is easy (clearly $2^{x+1}\not\vert x$ - now apply infinite descent). I claim $2^y=x$.

For suppose $2^y\not=x$. Then we have $x=2^y\cdot a$ for some $a>1$. This $a$ can't be even, since otherwise $2^{y+1}\vert x$, and we can't have $a=1$ since $2^y\not=x$. But this contradicts the assumption that every divisor of $x$ greater than $1$ is even.

Noah Schweber
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