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Given $n\in\mathbb N$, there exists a non-negative integer $a$ and an odd integer $b$ such that $n=2^a\cdot b$

Base Case: $n=1 =2^0\cdot1$

I.H: Assume its true for $n=1,2,\ldots,k$

How would i prove this?

Bill Dubuque
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TheGamer
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3 Answers3

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It's a rather straight forward induction.

Base Case : $n=1$ is odd.

$n = 2 = 2^1 \times 1$, if you want another.

Now, let $1,...,k$ be covered in the hypothesis. Look at $k+1$.

Either $k+1$ is odd or even. If it is odd, then $k+1 = 2^0 (k+1)$, where $k+1$ is odd (of course). Hence, we are done in this case.

If $k+1$ is even, then $\frac{k+1}{2} < k+1$ is an integer which has been covered in the inductive hypothesis, say $\frac{k+1}2 = 2^ab$, where $b$ is odd. Transposing, $k+1 = 2^{a+1}b$, where $b$ remains odd. Hence, $k+1$ can always be written in the desired form.

This completes the induction.

Sarvesh Ravichandran Iyer
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Firstly note that this statament is reasonable: every number, odd or even could be written in this form. When $n$ is odd $a$ must be $0$. When $n$ is even $a \ge1$.

So, more formally: take an arbitrary $n \in \mathbb{N}$.

  • If $n$ is odd we can say that $n=2^0n$
  • If $n$ is even consider the prime factorization of $n$, $n=2^{\alpha_1}*3^{\alpha_2}*...*p_n^{\alpha_n}$ note that $ 2^{p_1}|n \space$ and $\space 2^{p_1+1} \not| n$ $\implies \exists b\in \mathbb{N}: n=2^{p_1}b$.
ictibones
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  • Use \mid ($a \mid b$) and \not\mid ($a \not\mid b$) rather than | ($a|b$) and \not| ($a \not| b$). – Trevor Gunn Jul 17 '17 at 17:46
  • This is not a valid proof by induction. Moreover, you are essentially assuming what needs to be proved. – Bill Dubuque Jul 17 '17 at 17:49
  • Why? I take a as the exponent of 2 in the prime factorization of n – ictibones Jul 17 '17 at 17:50
  • This simple induction does not need strong results like prime factorization (and using them may be circular). In any case, I don't see any (strong) induction above, so it does not answer the question. – Bill Dubuque Jul 17 '17 at 17:52
  • While one may argue that using the prime factorization theorem is circular, I think that it is just far too strong in the case above. First of all, a traditional proof of the prime factorization theorem, involves strong induction itself of a very similar form (if $n$ is prime, done, otherwise $n=ab$, where $1<a,b < n$ etc.) just replace the word prime by odd and $a$ by $2$, and apply the induction hypothesis, and we are done. Instead, going by the prime factor theorem is like proving a large theorem and taking a small corollary out of it. Like an ordinary couple living in a 100 bedroom house. – Sarvesh Ravichandran Iyer Jul 17 '17 at 18:40
  • @астонвіллаолофмэллбэрг See this answer for a proof along those lines, i.e. similar to the inductive proof that naturals $> 1$ can be written as a product of primes (i..e irreducibles). This question is a dupe of that one. – Bill Dubuque Jul 17 '17 at 19:00
  • @BillDubuque Thanks, Honestly I write this because it was the first think that "translates" my intuition (i.e you can divide a number by 2 finitely many times). But surely is not an economic way to solve this problem – ictibones Jul 17 '17 at 19:05
  • @Leonardo The point is that results like this are usually used to motivate more general results such as the (strong) inductive proof of existence of prime factorizations. By deriving it as a corollary of the more general result the proof is no longer a (strong) inductive proof, so it no longer serves as a good example for learning (strong) induction - which is usually the reason for posing such exercises. – Bill Dubuque Jul 17 '17 at 19:15
  • @BillDubuque It's good we got it cleared up. Thank you for the link. – Sarvesh Ravichandran Iyer Jul 17 '17 at 19:16
  • @астонвіллаолофмэллбэрг Indeed, thanks for contributing. – Bill Dubuque Jul 17 '17 at 19:17
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In your comments you inquired how we could discover the proof that I hinted. The idea is simply that every natural $>0$ can be generated by repeatedly multiplying its odd part $\,\color{#c00}b\,$ by a power of $\,\color{#0a0}2.\,$ Said inductively: the set $\,S\,$ of naturals $> 0\,$ writable in form $\,\color{#0a0}2^a \color{#c00}b\,$ satisfies the following.

Lemma $ $ If $\,S\subseteq \Bbb N_{>0}$ contains all $\color{#c00}{\rm odds}$ and is closed under $\rm\color{#0a0}{multiplication\ by\ 2}\,$ then $\,S = \Bbb N_{>0}$

Proof $ $ Suppose for strong induction that $S$ contains every natural $< n.\,$ If $\,n\,$ is odd then $\,n\in S\,$ by $\rm\color{#c00}{hypothesis}$. Else $n = 2k\,$ for $k< n$ so $\,k\in S\,$ by induction, so $n = 2k\in S\,$ by $\rm\color{#0a0}{hypothesis}$.

Remark $ $ More generally, replacing $\,2\,$ by any natural $\,c > 1,\,$ and "odd" by nonmultiple of $\,c,\,$ the same proof shows that any natural $>0\,$ can be written in form $\,c^k b,\ c\nmid b\,$ (the above induction amounts to pulling out as many factors of $\,c\,$ as possible, i.e. until $\,c\nmid b).\,$ This representation is unique: $ $ if $\,c^kb = n = c^K d,\,$ wlog $\,K > k,\,$ then $\,b = c^{K-k}d\Rightarrow c\mid b \,\Rightarrow\!\Leftarrow$

More generally see here, which shows how this is a special case of inductive generation of multiplicatively closed sets (here uniqueness may fail).

Beware $ $ If $\,c\,$ is not prime then the cofactor $b$ need not be coprime to $\,c\,$ so powers of $\,c\,$ need not combine additively under multiplication, e.g. for $\,c = 6\,$ we have $\,(6^2\cdot 2)(6^3\cdot 3) = 6^6\,$ so we get an extra power of $\,6\,$ from $\,bb' = 2\cdot 3,\,$ which can't occur for prime $\,c=p\,$ by $\,p\nmid b,b'\Rightarrow p\nmid bb'$ by Euclid's Lemma.

Bill Dubuque
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