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I know it is a supposedly trivial proof and that of course there are a lot of similar topics, but I'm unclear about whether I could logically exhausts the demonstration (I'm also not very clear about the process of proving by induction, say, if I took the steps properly). So I'm showing my demonstration and asking for critiques on it, hoping you won't find it annoying to have someone asking this again here. It goes like that:

Proposition: $1 + 2 + 3 + \dotsb+ n = \frac{n(n + 1)}{2}$

Proof:

Verifying the expression for the base case $n = 1$: $$\frac{1(1 + 1)}{2} = 1$$ which is correct. So we can assume the proposition is true for any $k \leq n$. Now let's verify if it holds for $n + 1$: $$\sum_{i=1}^{n+1} i = \frac{(n + 1)[(n + 1) + 1]}{2} = \frac{(n + 1)(n + 2)}{2} = \frac{n^2 + 3n + 2}{2}.$$

Now, subtracting $(n + 1)$ from both sides of the equation yields $$\sum_{i=1}^{n} i = \frac{n^2 + 3n + 2 - 2n - 2}{2} = \frac{n^2 - n}{2} = \frac{n(n + 1)}{2}$$ which is true and confirms the hypothesis for $n+1$ also.

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    $\begingroup$ Eh, it's backwards. You should go from $n(n+1)/2$, add $(n+1)$ and show that it is $(n+1)(n+2)/2$. When you said "$\sum_{i=1}^{n+1}i=(n+1)[(n+1)+1]/2$", it's not necessarily true. In fact, it's precisely what you're trying to prove. In other words, you started at your conclusion and ended at your assumption. $\endgroup$ – JasonM Aug 5 '16 at 6:31
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    $\begingroup$ You seem to have shown that if the proposition holds at a particular number $n+1$, then it holds at $n$. The induction step should run in precisely the other direction. $\endgroup$ – André Nicolas Aug 5 '16 at 6:32
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    $\begingroup$ First of all you supposed the hypothesis for strong induction $k \leq n$ but in fact you used the simple induction $k=n$. Moreover you made a disaster during computation, just write $n(n+1)/2 + (n+1)$ and regroup $(n+1)$. $\endgroup$ – Maffred Aug 5 '16 at 6:32
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    $\begingroup$ The key is to write $\sum_{i=1}^{n+1} i = (n+1) + \sum_{i=1}^n i$ and then use the induction hypothesis :) $\endgroup$ – Zubzub Aug 5 '16 at 6:35
  • $\begingroup$ Many thanks for the answers :) $\endgroup$ – izzorts Aug 5 '16 at 6:44
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Your induction step consists only of a remark that subtracting $\,n\!+\!1\,$ for both sides of $\,S(n\!+\!1) = T(n\!+\!1)\,$ yields $\,S(n) = T(n).\,$ However, for the induction step we need $\,S(n) = T(n)\,\Rightarrow\, S(n\!+\!1) = T(n\!+\!1).\,$ We can deduce that by reversing your your observation, i.e. it follows by adding $\,n+1\,$ to both sides. But it is essential to explicitly state this derivation of the inductive step $\,P(n)\,\Rightarrow\,P(n\!+\!1)\,$ in order to convince the reader that you understand how to correctly present a proof by induction.

Remark $\ $ Your observation is that $\,S(n\!+\!1)-S(n) = n\!+\!1 = T(n\!+\!1)-T(n),\,$ therefore $\,S\,$ and $\,T\,$ both satisfy the recurrence $\,F(n\!+\!1)-F(n) = n+1,\ F(1) = 1.\,$ The inductive proof is essentially proving the uniqueness theorem for such recurrences, i.e that any two solutions $\,S,T\,$ that start off equal at the initial point $\,n=1\,$ must remain equal for all $\,n\ge 1.\,$ With no extra effort you can prove this for any similar recurrence, which yields a more conceptual view of the matter.

Alternatively you can view this type of induction as a special case of a telescopic sum. See my posts on telescopy for further discussion of this viewpoint.

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  • $\begingroup$ See here for an example of such an inductive uniqueness proof for recurrence solutions. $\endgroup$ – Bill Dubuque Jul 11 '17 at 14:42
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First of all, you're using strong induction unnecessarily. Here's the basics of how weak induction works:

Say there's a statement $S(n)$ that you want to prove for all whole numbers $n\geq 1$. Then you first need to prove it for $n=1$ (which you did), and then you assume $S(n)$ is true for some value $n$ (because you just proved it was). The "inductive step" tries to prove $S(n+1)$ is true. If it is true, then it will be true for the next value, and the next value, and the next value, ... etc (dominoes are the common analogy), so we will be done.

The problem with your proof is that it assumed $S(n+1)$ was true in the first place, and then reasoned that $S(n)$ was true. It should be the other way around. Just because you come to a true statement at the end of an argument doesn't mean you came to the required statement.

Now it is often helpful to start with the solution and work backwards, but keep in mind it must be in the right order when you write down a proof.

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  • $\begingroup$ Thanks for the answers, it really clarified most of my doubts. One still bothers me, nonetheless. By going backwards as I've done, I commit the logic fallacy of assuming that if A implies B, B also necessarily implies A? I mean, I've proven that $P(n)$ is true, and was expected to prove that this implies $P(n+1)$ is also true. Instead I've assumed $P(n+1)$ is true and shown that this implies a known truth i.e. $P(n)$. But were $P(n +1)$ false, it could also imply a true proposition, therefore this deduction is logically flawed. Am I correct? $\endgroup$ – izzorts Aug 6 '16 at 0:30
  • $\begingroup$ @izzorts If you had instead remarked that the implication reverses by adding $\,n+1\,$ then the proof would be fine. The problem is that generally such implications need not reverse, so you need to explicitly say why it does in order to obtain the sought inductive step. $\endgroup$ – Bill Dubuque Aug 6 '16 at 0:44
  • $\begingroup$ @izzorts For example, see this answer where I do a similar "backward analysis" to see what is needed to make the induction step work, but I do it bidirectionally (using $\!\iff\!)$ so that I get the inference in both directions. That answer also explains how telescopy applies to your case. $\endgroup$ – Bill Dubuque Aug 6 '16 at 0:48
  • $\begingroup$ @izzorts Well we know $P(n+1)$ is true, because there is a correct way of proving it for this particular problem. But make no mistake, showing $P(n+1) \implies P(n)$ feels like a legitimate way to prove it because each step could easily be reversed. However, it's not always the case that such steps can be reversed, so that was a BIG assumption you were making. $\endgroup$ – JasonM Aug 6 '16 at 0:49

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