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I try to prove that:

Statement (S): $\frac{1}{n-1}(1-\frac{1}{n}) = \frac{1}{n}$

by induction for all $\forall n \in \mathbb{N} \setminus$ {$ 1 $}.

Here is my solution:

Base case: S(2) $ = \frac{1}{2-1} (1 - \frac{1}{2}) = \frac{1}{1} - \frac{1}{2} = \frac{1}{2}$, which is true.

Assuming S(n) is true, then: S(n+1) $ = \frac{1}{n} (1 - \frac{1}{n+1}) = \frac{1}{n}(\frac{n+1}{n+1} - \frac{1}{n+1}) = \frac{1}{n}(\frac{n}{n+1}) = \frac{1}{n+1}$,

which concludes the proof. Is the solution correct, or is there an error in my argument.

user2550228
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    Why do you want to do this with induction when it is a direct and simple consequence of the laws of algebra? – Lee Mosher Mar 14 '20 at 13:52
  • Training exercise basically. – user2550228 Mar 14 '20 at 13:55
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    Well, someone is trying to train you to use a screwdriver to put in a nail. I suggest you hire another trainer. – Lee Mosher Mar 14 '20 at 13:57
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    As far as I can tell you haven't used the assumption. It seems like you just manipulated the algebra which you could have done originally. – Karl Mar 14 '20 at 13:58
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    The algebraic manipulation an inductive solution would require is basically the same manipulation you'd require to straightaway solve this. So I don't see a reason why this would even make sense. And anyway, your solution doesn't use the hypothesis (the statement $S(n)$ when you try to show that $S(n+1)$ is true). So no, your solution isn't correct either. – oldsailorpopoye Mar 14 '20 at 14:02
  • @LeeMosher I see nothing bad to try the scheme of an induction proof even if there is a much easier solution. It helps to get used to induction. To clarify : I am not the trainer :) – Peter Mar 14 '20 at 14:29
  • @charlesh This is true. If we actually want to use induction, then of course we must use the hypothesis instead of switching to a method that we could have applied from begin. – Peter Mar 14 '20 at 14:35
  • Thank you all for your comments. You have pointed out where mistake was and I really appreciate it. This was my first attempt to prove something, and I made a mistake. So if there was an answer I would upvote charlesh, but karl as well. – user2550228 Mar 14 '20 at 19:09
  • A nice starter proof would be the sum of the 1st $n$ square numbers. This can be derived a few ways but it's a straight forward induction proof if that's the way you went. – Karl Mar 14 '20 at 21:02
  • I would like to clarify something. There is no error in your proof. The error is in your process. Proof by induction requires showing $S(n) \implies S(n+1)$. It does not put requirements on how this is proved. If you show $S(n+1)$ directly without assuming $S(n)$, that still shows $S(n) \implies S(n+1)$. So your proof is technically correct. It is confusing and involves pointless elements, but never-the-less, it qualifies as a proof. Your flaw was only in failing to recognize that since you didn't use $S(n)$ to prove $S(n+1)$, induction wasn't needed and it should be written directly instead – Paul Sinclair Mar 15 '20 at 02:29

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