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In this link Prove that every positive integer $n$ has a unique expression of the form: $2^{r}m$ where $r\ge 0$ and $m$ is an odd positive integer we have a proof without induction (and I understand it). But my teacher wants us to use induction to solve this problem.

How do I go from this: $n = 2^a.b$, to this: $n + 1 = 2^a.b + 1 = 2^c.d$ for some $c$ and $d$?

  • Have you learned about strong induction? – Another User Mar 23 '23 at 13:11
  • no, I haven't. Do you have a pdf? a link? –  Mar 23 '23 at 13:14
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    Yeah, you'll need strong induction, not regular induction. In strong induction, you still prove for $1$ then assume the propositions is true for all $n\leq m,$ and prove it is true for $m+1.$ Strong induction really is just regular induction. Starting with your proposition, $P(n),$ we construct a new proposition $Q(n)$ which means "For all $m\leq n,$ $P(n).$ Then strong induction is just proving $Q(n)$ by regular induction. But proving $Q(n)$ for all $n$ all proves $P(n)$ for all $n.$ – Thomas Andrews Mar 23 '23 at 14:26

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It is better if the induction is a different one, I mean, suppose that for any number less than $n$, the property holds, then show that the property also holds for $n$. In this case, you can divide it into two cases. First, if $n$ is odd, you are done. Secondly, if $n$ is even. So $n=2k$ for some $k<n$, now apply the property to $k$, which says that $k=2^am$, where $m$ is an odd number, and then $n=2^{a+1}m$ and you get what you want. I hope this proof provides a tool for settling this question using induction.

Thomas Andrews
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sti9111
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You can prove by induction. When $n$ is odd, you are done. When $n$ is even, you factor a 2 out, and represent $n/2$ by induction.

Connor
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