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In practice, do you do some work on the inductive step and then reverse your steps?

For example. Say you have this recurrence:

$f(n+1) = 2f(n) + 1$ with $f(0) = 0$

This creates the sequence $0, 1, 3, 7, 15, 31, \cdots$

"Huh this looks like it could be $f(n) = 2^n-1$, so I'll try to prove it with induction."

Is this the right way to do it:

Known as true: $f(n+1) = 2f(n) + 1$ with $f(0) = 0$.

Assumption: $f(n) = 2^n-1$

Base case: $f(0)= 2^0-1 = 0$ which matches the known formula's result, so the base case is true for our assumption.

Induction procedure:

If $f(n) = 2^n-1$ for some $n$, we need to show that $f(n+1) = 2^{n+1}-1$. Since $f(n+1) = 2f(n) + 1$ is true, it suffices to show that $f(n+1) = 2f(n) + 1 = 2^{n+1}-1$.

$2f(n) + 1 = 2^{n+1}-1$

$2f(n) = 2^{n+1} - 2$

$f(n) = 2^n - 1$, our assumption, so the induction step is true.

My question:

Did I do this correctly? Is an inductive proof true if you can make some assumption, state what you want to prove, and then prove it in terms of your assumption? Or do you have to somehow prove it "in terms of" what is known to be true (the original recurrence)? Did I do it backwards (I had this problem in this question here and still don't fully understand why that one was backwards)?

I'm not really sure where I'm supposed to start and how exactly I know when I've proven what I want to prove.

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AJJ
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  • With $P$ as a statement and $T$ as a tautology (always true), the following are not equivalent. $P\Rightarrow T~$ and $~T\Rightarrow P$. Start with what is known to be true and show that it implies what you hope is true. Do not start with what you hope to be true and reach a tautology because you might have made a step in there that doesn't work backwards. E.g. $x=y\Rightarrow \frac{x}{x}=\frac{y}{y}\Rightarrow 1=1$. However, $1=1\not\Rightarrow x=y$. – JMoravitz Jan 05 '16 at 18:00
  • @JMoravitz Are you suggesting that my inductive step is backwards (as it was in the other link)? – AJJ Jan 05 '16 at 18:00
  • This looks perfectly to me. The only minor change I would suggest is in the third last equation. When you try to prove an identity, do not start with the identity and derive backwards. You should proceed like this: $2f(n)+1=2(2^n-1)+1=2^{n+1}-1$. – KittyL Jan 05 '16 at 18:02
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    Yes. should be first$\rightarrow f(n+1)=2f(n)+1=2(2^n-1)+1=2^{n+1}-1\leftarrow$ should be last. – JMoravitz Jan 05 '16 at 18:02
  • @JMoravitz I want to show that $T \implies P$, right? – AJJ Jan 05 '16 at 18:02
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    In practice would you "solve it backwards" as I have done (which really feels like solving it forwards to me) and then reverse the steps? – AJJ Jan 05 '16 at 18:04
  • If it is a difficult to spot step, absolutely. Just be very wary of whether certain steps cannot be reversed. For example, division by an unknown. $a=b\Rightarrow ax=bx$ however $ax=bx\not\Rightarrow a=b$. – JMoravitz Jan 05 '16 at 18:06
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    I recommend reading Proof writing: how to write a clear induction proof and am labeling this as a duplicate of that. – JMoravitz Jan 05 '16 at 18:07

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