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My understanding of the reason for using proof by induction is to see if the expression used to calculate what the nth term in a sequence is, always holds or not.

A proof by induction requires a basis step. It's not explicitly stated why the basis step is important when learning this. I hear analogies that proof by induction is like a ladder, like dominoes, like stairs, so I think to myself what is similar about those objects. The segments of a ladder, or stairs all look identical to each other going all the way up the ladder or all the way up the stairs.

This leads me to believe there is an assumption that an equation performs an identical action on each of the numbers inputted into it. Which seems reasonable to me. An equation performs the same action on the number 2, whether it be scaling it, adding to it, etc, that it would perform on the next number, say 3.

Some expressions are hard to see exactly what the pattern would be, but by looking at the a few terms in the pattern we notice certain pattern, sometimes that pattern does break and we discover the actually equation that would hold that pattern forever is different from what we thought it was originally.

So this is where the distinction that we assume the expression we are given originally is correct, in the induction hypothesis we use the logical expression know as implication, "If p then q" if you recall the truth table for that expression it can only be proven to be false when p is true and q is false. So the truth of p is actually irrelevant, we are checking to see that if p were true than would q hold.

We test the induction hypothesis by setting the original equation on one side of an equals symbol, adding the k+1 last term to it, then we put the expression with the k+1 replacing every instance of k. We massage the equations to see if they look identical, and if they do we can see our equality holds.

I'm not really sure why we bother doing all of this in the first place, If we are assuming our propositional statement is true to begin with, and if we know from the onset that our equation behaves like a ladder or stairs, can't we just infer from the very beginning that k+1 holds . .

I'm not too certain what the point of the proof really is. It still seems circular to me. I must be missing some really important insight. I don't want to just route memorize this. I get some of the basic ideas of the proof and I think I understand what it's trying to accomplish, it just doesn't seem rigorous like proof by contradiction or proof by contra positive.

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    $\begingroup$ Does this answer your question? Dominoes and induction, or how does induction work? $\endgroup$ – Brian Jan 27 at 17:26
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    $\begingroup$ You might be over thinking this. Induction is just logic. You want to prove that a set $S$ is actually $\{1,2,3, \ldots\}$, the natural numbers. So you prove $1 \in S$. Then you show that if $1 \in S$, then $2 \in S$. So $2 \in S$. Then you show that if $2 \in S$, then $3 \in S$, so $3 \in S$. And so on. Eventually you get tired of this, so you show that in general, if $n \in S$ then $n+1 \in S$. In this way you show every natural number is in $S$. $\endgroup$ – Jair Taylor Jan 27 at 17:35
  • $\begingroup$ Suppose you want to prove a collection of propositions $p_0,p_1,p_2,...$. Suppose you know that $p_0$ is true and also that if $p_n$ is true then $p_{n+1}$ is also true. Then you know all of them are true. Since $p_0$ is true then the rule tells us that $p_1$ is true, and so $p_2$ is true and so on. $\endgroup$ – copper.hat Jan 27 at 17:41
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    $\begingroup$ Induction is not just about equations! $\endgroup$ – Arturo Magidin Jan 27 at 17:46
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    $\begingroup$ @AndrewFerro Induction is of course rigorous. Every positive integer is finite and can be reached by counting , beginning with $0$. So, that there is no largest number is no problem. Induction also is obvious intuitively, and in this case, the intuition is right. Your question mentions some very important conditions. Another issue appears in the fake proof that in a collection of numbers, all the numbers must be equal. In this case the step from $n=1$ to $n=2$ is the error. $\endgroup$ – Peter Jan 27 at 17:57
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You have implied in your post that you are happy with proof by contradiction.

You can think of any proof by induction as a form of proof by contradiction.

For example, suppose you are trying to prove that $\sum_{i=1}^n i=\frac{1}{2}n(n+1)$ for all positive integers $n$.

A proof by contradiction might go like this:-

Suppose the result is false.

Then there is some positive integer $n$ for which $\sum_{i=1}^n i\ne\frac{1}{2}n(n+1)$. Since $1=\frac{1}{2}\times 1\times2$, the result is true for $n=1$ so suppose the smallest $n$ for which the result is false is $n=k+1$.

The result is true for $n=k$ and therefore $\sum_{i=1}^k i=\frac{1}{2}k(k+1)$. Then $$\sum_{i=1}^{k+1} i=\frac{1}{2}k(k+1)+(k+1)=\frac{1}{2}(k+1)(k+2).$$

The result is true for $n=k+1$ after all, a contradiction. We conclude there are no counterexamples.

In the above you should be able to spot the analogue of the base case and the inductive step. So, if you're happy with proof by contradiction, you can also be happy with induction.

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induction follows:

Base case

This is used to establish that there is a case where it's true.

Induction step

This is used to show if there is a general case that's true ( see for example the base case), it leads to another case that's true (the next case we are considering hopefully).

The reason we need both parts, is because either the induction step can fail, or if it suceeds, there need not be a base case where it's true.

for example, lets hypothesize that if $2^n-1$ is prime $2^{n+2}-1$ is prime, well $2^n-1$ is prime for $n=2$ (it's equal to 3, base case established) but $2^{n+2}-1=4(2^{n}-1)+3$ which when the parenthesized expression is 3 or a multiple of it, makes the whole expression have a factor of 3 ( clearly composite unless 3 is of form $4k+1$), this means our induction step would fail as it's not generally true.

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Try thinking about the domino metaphor. We have infinitely many dominoes and we want to prove that they all fall down. When we prove the base-step we have proven that the first domino falls. When we prove the induction step we prove that when the nth domino falls, the (n+1)th domino falls as well. As the first one falls, the second must also fall down and as the second falls down, the third falls down and so on and therefore all of the dominoes fall.

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