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Can someone guide me in the right direction on this question?

Prove that every $n$ in $\mathbb{N}$ can be written as a product of an odd integer and a non-negative integer power of $2$.

For instance: $36 = 2^2(9)$ , $80 = 2^4(5)$ , $17 = 2^0(17)$ , $64 = 2^6(1)$ , etc...

Any hints in the right direction are appreciated (please explain for a beginner). Thanks.

Bill Dubuque
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1337holiday
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3 Answers3

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To prove something by strong induction, you have to prove that

If all natural numbers strictly less than $N$ have the property, then $N$ has the property.

is true for all $N$.

So: our induction hypothesis is going to be:

Every natural number $k$ that is strictly less than $n$ can be written as a product of a power of $2$ and an odd number.

And we want to prove that from this hypothesis, we can conclude that $n$ itself can be written as the product of a power of $2$ and an odd number.

Well, we have two cases: either $n$ is odd, or $n$ is even. If we can prove the result holds in both cases, we'll be done.

Case 1: $n$ is odd. Then we can write $n=2^0\times n$, and we are done. So in Case 1, the result holds for $n$.

Case 2: If $n$ is even, then we can write $n=2k$ for some natural number $k$. But then $k\lt n$, so we can apply the induction hypothesis to $k$. We conclude that ...and you should finish this part...

So we conclude that the result holds for all natural numbers by strong induction.

Arturo Magidin
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  • So if k < n then by induction hypothesis k can be written as a product of a power of 2 and an odd number? Then that would imply that n itself follows from the hypothesis? – 1337holiday Mar 09 '11 at 05:52
  • @1337holiday: I assume you are talking about case 2. Not "if", but since $k\lt n$, then $k$ can be written as a product of a power of $2$ and an odd number (by the induction hypothesis). So $k=2^r\times s$, where $s$ is odd. What does that tell you about $n$? – Arturo Magidin Mar 09 '11 at 05:53
  • So since n = 2k and k = (2^r(s)). It means that n = 2(2^r(s)) or n = (2^(r+1))(s) and therefore it is true by the hypothesis? – 1337holiday Mar 09 '11 at 06:00
  • @1337holiday: Essentially yes: though I would phrase it as "and therefore, $n$ can be written as the product of a power of $2$ and an odd number." The induction hypothesis has already been invoked, no need to remember her yet again in the coda. (-: – Arturo Magidin Mar 09 '11 at 06:02
  • THIS IS AMAZING! I didnt understand this at all but now im starting to get it. Thanks so much! – 1337holiday Mar 09 '11 at 06:06
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This follows from a more general result that's both simpler to prove and more insightful, viz. the result follows immediately by this frequently applicable multiplicative form of induction, which shows that for multiplicative sets we need only check the generators (here $1$ and primes).

Lemma $\rm\, \mathbb N$ is the only set of naturals containing $\color{#c00}1$ and $\rm\color{#c00}{all\ primes}$ and $\rm\color{#0a0}{\text {closed under product}}$.

Proof $\ $ Suppose $\rm\!\ S\subset \mathbb N\,$ has said properties. $ $ We prove by strong induction every natural $\rm\!\ n\in S.\, $ If $\rm\,n\,$ is $\,\color{#c00}1\,$ or $\color{#c00}{\rm prime}$ then by hypothesis $\rm\,n\in S.\,$ Else $\rm\,n\,$ is composite, $ $ hence $\rm\ n = j k\ $ for $\rm\, j,k < n.\,$ By strong induction the smaller $\rm\ j,k\in S,\,$ $\rm\color{#0a0}{thus}$ $\rm\, n = jk\in S.\ $ $\small\bf QED$

This yields the sought result. Let $\rm\!\ S\!\ $ be the set of naturals that have the form $\rm\,2^{\,j}\!\ n\,$ for odd $\rm\,n\in \mathbb N.\ $ $\, 1\!\ $ and all primes $\rm\,p\,$ are in $\rm\!\ N,\, $ by $\, 2 = 2^{\!\ 1}\!\cdot\! 1\,$ and $\rm\, p = 2^{\!\ 0}\!\cdot\! p\,$ for odd $\rm\,p\,$ (and $1).$ $\rm\,S\!\ $ is closed under multiplication by $\rm\, (2^{\,j}\!\ m) (2^{\,k}\!\ n) = 2^{\,j+k}\!\ m\!\ n,\, $ with $\rm\!\ m\!\ n\!\ $ odd by $\rm\!\ m,n\!\ $ odd. $\!\ $ So $\rm\ S = \mathbb N\ $ by Lemma.

Corollary $\ $ Every natural $> 0\,$ is a product of primes (i.e. irreducibles).

Proof $\, $ It is clear for $1$ and primes, and products of primes are closed under multiplication.

Remark $ $ We could also deduce the sought result from the prior Corollary. Then it reduces to (inductively) proving that a product of $n$ odds remains odd. But this way is not an instructive example of strong induction, since the strong induction is hidden in the proof of the corollary.

Bill Dubuque
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Proof:

By the fundamental theorem of algebra, every integer $N$ can be uniquely factored as $\prod^{n}_{i=1}p_{i}^{a_{i}}$. Now, mark $2=p_{1}$, note $a_{i}$ can take value of $0$. You got the theorem.

For the "inductive" proof, suppose for $n<k$ this is true. For $n+1$ its factors must be in previous $n$ numbers. Hence $n+1=\prod n_{i}$. Decompose $n_{i}$ by induction hypothesis you get the statement.

Kerry
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    @user7887: The Fundamental Theorem of Algebra says that every nonconstant polynomial with complex coefficients has at least one complex root; you mean the Fundamental Theorem of Arithmetic. – Arturo Magidin Mar 09 '11 at 05:37
  • now I understand. – Kerry Mar 09 '11 at 05:49
  • @user7887: Your inductive proof has two problems: first, the OP asked for a proof by strong induction, not regular induction. Second, your argument is incorrect as given (it's possible for $n+1$ to be prime, after all). – Arturo Magidin Mar 09 '11 at 05:50
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    I don't think $n+1=\prod n_i$ is true (should be $p_i$), and if $n+1$ is prime its factor(s) are not in the previous $n$ numbers. You are close-if $n+1$ is prime, it is odd. – Ross Millikan Mar 09 '11 at 05:52
  • I guess answering more such questions can help me realize my own weakness. Thanks for the comments. – Kerry Mar 09 '11 at 05:54
  • @Ross: Unless $n=1$... there's a reason why this proof is so much easier with "true for all $k\lt n\Rightarrow$ true for $n$"-induction than with "true for $n\Rightarrow$ true for $n+1$"-induction. – Arturo Magidin Mar 09 '11 at 05:55
  • @Arturo: You are right about the advantage of strong induction over (weak, simple-which is your favorite?) induction. I was focused on the case where the number in question was prime, where you can't factor it into smaller numbers. Your comment on this subject came in while I was writing mine. – Ross Millikan Mar 09 '11 at 06:05
  • @Ross: I call it "regular induction", which is probably just as big a misnomer as calling the other one "strong". (-: – Arturo Magidin Mar 09 '11 at 06:09
  • @Arturo: works for me. – Ross Millikan Mar 09 '11 at 06:12