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In actual induction, I know that we are supposed to start with $n+1$ and find $n$ within it, assuming $n$ is true. But I was hoping if anyone could give an explanation or analogy on why we aren't able to start with $n$ and go from $n$ to $n+1$ in the inductive step. Thank you!

Samah Samiee
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    The point is you "assume for $n$ and show for $n+1$" doesn't matter where you start as long as you end up using your assumption on $n$. – user8469759 Jan 09 '20 at 00:02
  • And show your statement is true for some integer. – Chris Leary Jan 09 '20 at 00:05
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    "I know that we are supposed to start with n+1 and find n within it" Who told you that? The induction step requires you assume something is true for $n$ and prove that therefore it is true of $n+1$. However you want to do it (so far as it is legitimate) you can. I have never heard anyone say start with $n+1$ and find $n$ within it. And I think 90% of the time was start with $n$ get to $n+1$. But so long as you show: true for $n \implies$ true for $n+1$ you are fine. – fleablood Jan 09 '20 at 00:21
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    I recommend this answer to anybody asking about induction – Ross Millikan Jan 09 '20 at 00:40
  • @fleablood That was my mistake. I left out the induction hypothesis. What I should have said was that we start out with our induction hypothesis (our assumption for $n$ ), say $P(n)$ is true. Then for the inductive step, we start with $P(n+1)$ and try to use our induction hypothesis in the inductive step. Which is what I meant by "I know that we are supposed to start with n+1 and find n within it." What I meant to ask was say we have our induction hypothesis is true for $n$. In the inductive step, why aren't we able to start with $P(n)$ and go from $P(n)$ to $P(n+1)$. I hope I was clear. – Samah Samiee Jan 09 '20 at 00:41
  • @RossMillikan Thanks for the link, it's a great post! – Samah Samiee Jan 09 '20 at 00:50
  • "Then for the inductive step, we start with P(n+1) and try to use our induction hypothesis in the inductive step" Again I say, where on earth were you told we couldn't? I don't know anyone who starts with $P(n+1)$ and 90 percent of all proof start from $P(n)$ and got to $P(n+1)$. I can't imagine anyone saying we can't. – fleablood Jan 09 '20 at 06:23

3 Answers3

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We can start with $n$ and prove it for $n+1$. It turns out that, in order to do that, it is often useful to look what you have proved for $n$ with the statement concerning $n+1$. For instance, if you want to prove by induction that$$(\forall n\in\mathbb N):1+2+3+\cdots+n=\frac{n(n+1)}2,$$in order to prove that, if it holds for $n$, then it holds for $n+1$, you notice that\begin{align}\overbrace{1+2+3+\cdots+n}^{\phantom{n(n+1)/2}=n(n+1)/2}+(n+1)&=\frac{n(n+1)}2+n+1\\&=(n+1)\left(\frac n2+1\right)\\&=(n+1)\times\frac{n+2}2\\&=\frac{(n+1)(n+2)}2.\end{align}So, as you see, I have isolated the $1+2+3+\cdots+n$ from $1+2+3+\cdots+n+(n+1)$ in order to apply the induction hypothesis.

José Carlos Santos
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  • Thanks for taking the time out of your day to answer my question. But as a followup question, I was wondering if for certain proofs wouldn't this approach cause build-up error? If so, how do we realize that when doing proofs? – Samah Samiee Jan 09 '20 at 01:14
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'starting with $P(n+1)$ and then finding $P(n)$ wihtin it' is exactly the same as assuming $P(n)$ and deriving $P(n+1)$: it may feel different, but in both cases you show $P(n+1)$ on the basis of $P(n)$

Bram28
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  • Thank you for the response. I commented this under @José Carlos Santos post and I'll do it here as well. A follow-up question is that I was wondering if for certain proofs wouldn't this approach cause build-up error? If so, how do we realize that when doing proofs? – Samah Samiee Jan 09 '20 at 01:14
  • @SamahSamiee I am not sure what ypu mena by 'build-up error' ... could you please elaborate a bit? – Bram28 Jan 09 '20 at 01:27
  • The first problem in this link:

    https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/recitations/MIT6_042JF10_rec08_sol.pdf

    explains it really well. But I'm not sure about how we would realize this when doing proofs.

     – Samah Samiee Jan 09 '20 at 01:30
  • @SamahSamiee Thanks for the link! I'd never heard of build-up error before: it's tricky indeed! ... But, I don't see how it is any more or less tricky whether you go from $P(n)$ to $P(n+1)$ or whether you 'try to find $P(n)$ within $P(n+1)$. I mean, if I could have immediately started with $n+1$ nodes, point out that a single node must be connected to one of a subgraph of $n$ nodes, and since by inductive hypthesis that subgraph is fully connected, the whole graph is fully connected as well ... and I would have made the exact same mistake!! – Bram28 Jan 09 '20 at 01:50
  • Hmm yeah, that's true. I think that makes sense I'll let you know if I have any other questions. Thanks again for the help, much appreciated! – Samah Samiee Jan 09 '20 at 02:05
  • That's the old "all horses are the same color error". When you go from $P(n)$ to $P(n+1)$ you can't assume anything about $n$ than what you know to be true for you base case. In this case $P(1)$ is vacuously true. So when you assume $P(n)$ you can assume it is true but can't assume it is true in any non-vacuous sense. You can't assume there are any vertices with positive degree because there weren't any in your base case. – fleablood Jan 09 '20 at 06:34
  • @fleablood Ahh yeah, that's true. Thanks! – Samah Samiee Jan 09 '20 at 08:24
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As fleablood comments, you should not be starting with $P(n+1)$ and working your way towards $P(n)$.

The major pitfall here is that you might end up showing that if $P(n+1)$ is true, then $P(n)$ is true, not the other way around. Unless you also show that the steps you took to get to $P(n)$ are reversible, you cannot conclude that $P(n)$ leads to $P(n+1)$.

For this reason, it's much safer to start with $P(n)$ and then manipulate it until you get $P(n)$ instead.

Example:

Prove that $n^2\le n$ for all $n\in\mathbb N$.

Clearly this is false, but we will prove it by induction your way.

Trivially the base case holds.

Suppose $n^2\le n$ for some $n=k$. Then we have

\begin{align}&(n+1)^2\le n+1\\\iff{}&n^2+2n+1\le n+1\tag{Expand $(n+1)^2$}\\\iff{}&n^2\le-n\tag{subtract $2n+1$}\\\implies{}&n^2\le n\tag{$-n\le n$}\end{align}

Notice that the last step is not reversible, which is why this proof fails.

Simply Beautiful Art
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