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How can I prove by induction that for any natural number $n$ there exists integers $a,b$ so that $2a+3b=n$

I can prove the base case, and I can imagine why it works but how can I prove it mathematically via induction?

mathse
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Ogen
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3 Answers3

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Because $gcd(2,3)=1$ there are integers $r,s$ with $2r+3s=1$ by the Euclidean algorithm. Now multiply by $n$ to obtain $$ n=2(rn)+3(sn)=2a+3b, $$ with integers $a=rn$ and $b=sn$. Of course, here we may take $r=-1$, $s=1$.

Dietrich Burde
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    I dont think this is proof by induction, where is the base case and the inductive step? – Ogen May 11 '14 at 13:10
  • @DietrichBurde But he is a beginner, and he wants a proof by induction :) – mathse May 11 '14 at 13:14
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    Yes, he has a proof by induction (see below). Now its time to learn something more ! – Dietrich Burde May 11 '14 at 13:15
  • @mathse: I just love seeing different solutions to a same problem. Though OP is asking for a proof by induction, he has been given that already, so why not answer a generalization of the problem? It's a knowledge-sharing community after all. – Deathkamp Drone May 11 '14 at 13:28
  • @DeathkampDrone Alright, my comment was half-earnest, half-joking. (But I think the induction proof was chronologically only second). – mathse May 11 '14 at 13:30
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The easiest answer is that $a=-n,b=n$ yields $2a+3b = n$.

The induction proof. For $n=1$, $a=-1,b=1$.

Assuming $2a+3b=n$, we need to find $a',b'$ so that $2a'+3b'=n+1$. You can choose $a'=a-1$ and $b'=b+1$.

Thomas Andrews
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  • That makes sense but how did you get to it. I can't just say make a = a -1 and b = b + 1. I'd like to understand how you got to that – Ogen May 11 '14 at 13:13
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    Oh wait, minusing a 2 and adding a 3 will add an extra 1. I understand. Thanks a lot. – Ogen May 11 '14 at 13:14
  • @Clay See my answer where I elaborate on intuition behind the proof . – Bill Dubuque May 11 '14 at 17:21
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In a comment you asked for some intuition on the source of the proof. Below I show that it is a special case of the fact that a set of integers containing $1$ and closed under addition must also contain $\,n = 1+\cdots+1,\,$ by an obvious inductive proof (see below). Let's consider your case.

Note that the set $S$ of integers of form $\,2a+3b,\,\ a,b\in\Bbb Z,\,$ is $\rm\color{#0a0}{closed\ under\ addition}$

$$\begin{eqnarray} n = 2a+3b \in S\\ \hat n = 2\hat a+3\hat b\in S\end{eqnarray}\ \ \Rightarrow\ \ \color{#0a0}{n+\hat n} = 2(a\!+\!\hat a) + 3(b\!+\!\hat b)\color{#0a0}{\in S}$$

Thus, since $\ \color{#c00}{1 = 2(-1)+3(1)\in S}\ $ we infer by the inductive proof below that $\, \Bbb N\subseteq S$

Lemma $\ $ If $\,S\subset \Bbb Z\,$ is $\,\rm\color{#0a0}{closed\ under\ addition}$ and $\,\color{#c00}{1\in S}\,$ then $\, \Bbb N\subseteq S$

Proof $\ $ We prove by induction on $\,n\,$ that $\,n\in S.\,$ Base case $\,n=1\in S\,$ is true by hypothesis. Suppose for induction that $\,n\in S.\,$ Then since, by hypothesis, $\,S\,$ is closed under addition, $\,n,1\in S\,\Rightarrow\, n+1\in S,\,$ yielding the inductive step, so completing the induction. $\ \ $ QED

Remark $\ $ More generally we can prove by a similar induction that if $\,d\,$ is the least positive element of $\,S\,$ then $\,d\,\Bbb N\subseteq S.\,$ Further, if $\,S\,$ is closed under negation (so subtraction) then $\, S = d\,\Bbb Z,\,$ which yields a conceptual inductive proof of Bezout's identity for the gcd, which immediately yields Euclid's Lemma

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Bill Dubuque
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