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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 is both simpler to prove and more insightful, viz. it follows immediately by the below $\color{#0a0}{\style{font-family:inherit;}{\text{multiplicative}}}\:\!$ form of induction, which shows that for $\color{#0a0}{\style{font-family:inherit;}{\text{multiplicative}}}$ sets we need only check the (base-case) $\color{#c00}{\style{font-family:inherit;}{\text{generators}}}$ (here $\color{#c00}1$ and $\color{#c00}{\style{font-family:inherit;}{\text{primes}}}$).

Lemma $\, \mathbb N$ is the only set of naturals containing $\color{#c00}1$ and $\color{#c00}{\style{font-family:inherit;}{\text{all primes}}}$ and $\color{#0a0}{\style{font-family:inherit;}{\text {closed under product}}}$.

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

This yields the sought result. Let $\!\ S\!\ $ be the set of naturals that have the form $\,2^{\,j}\!\ n\,$ for odd $\,n\in \mathbb N.\, $ $\, \color{#c00}1\!\ $ and $\color{#c00}{\style{font-family:inherit;}{\text{all primes}}}$ $\,p\,$ are in $\!\ S\, $ by $\, p = 2^{\!\ 0}\!\cdot p\,$ for odd $\,p\,$ (and $1)\,$ and $\, 2 = 2^{\!\ 1}\!\cdot 1.\,$ $S\!\ $ is closed under $\color{#0a0}{\style{font-family:inherit;}{\text{multiplication}}}$ by $\, (2^{\,j}\!\ m) (2^{\,k}\!\ n) = 2^{\,j+k}\!\ m\!\ n,\, $ with $\!\ m\!\ n\!\ $ odd by $\!\ m,n\!\ $ odd. $\!\ $ So $\ S = \mathbb N\ $ by Lemma.

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

Proof $\ $ It is true for the base cases $\color{#c00}1$ and $\color{#c00}{\style{font-family:inherit;}{\text{primes}}}\,$ (since they are empty and singleton products), and prime products are closed under $\color{#0a0}{\style{font-family:inherit;}{\text{multiplication}}},\,$ so applying the Lemma concludes the proof.

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