# Mathematical induction

So this formula right over here, this expression it worked for 1, so we have proved our base case. Math induction is just a shortcut that collapses an infinite number of such steps into the two above. So this 2 is this 2 right over there and this k is this k right over there. Let me colour code those. Plus k plus 1.

## Mathematical induction problem solver

That proves to us that it works for all positive integers. This variant goes by the name of Complete Induction or Strong Induction. So it definitely will work for 2. But since we can assume it works for 2 we can now assume it works for 3. This is what I'm assuming. So we are going to have k times k plus 1 plus 2 times k plus 1. We've just added all of them, it is just 1. Times k plus 1 plus 1. Or everything above some threshold. So if you assume it worked for 1 then it can work for 2. And the reason why this is all you have to do to prove this for all positive integers it's just imagine: Let's think about all of the positive integers right over here.

Therefore, P n is true for all n starting with 1. This expression worked for the sum for all of positive integers up to and including 1.

Why is that?

### Principle of mathematical induction proof

The induction hypothesis was also employed by the Swiss Jakob Bernoulli , and from then on it became more or less well known. Another Frenchman, Fermat , made ample use of a related principle, indirect proof by infinite descent. So let's take the sum of, let's do this function on 1. We are assuming that we know that. Therefore P 2 is true. The result is then proved for all n from n0 on. Well we are assuming that we know what this already is. Math induction is just a shortcut that collapses an infinite number of such steps into the two above. History[ edit ] In BC, Plato 's Parmenides may have contained an early example of an implicit inductive proof. Well we have just proven it. Proof by induction math Video transcript I'm going to define a function S of n and I'm going to define it as the sum of all positive integers including N. So I will assume true for some number k. So if you assume it worked for 1 then it can work for 2.

So the base case we're going to prove it for 1. Now spoken in generalaties let's actually prove this by induction.

### Mathematical induction tool

There are other examples proved by MI:. Therefore P 3 is true which implies P 4 and so on. This expression worked for the sum for all of positive integers up to and including 1. Well we have already proven that it works for 1 so we can assume it works for 1. And now we can prove that this is the same thing as 1 times 1 plus 1 all of that over 2. In problem solving, mathematical induction is not only a means of proving an existing formula, but also a powerful methodology for finding such formulas in the first place. We are assuming that we already have a formula for this. So we are going to have k times k plus 1 plus 2 times k plus 1. Needless to say nothing can be proved this way. Intuitively, the inductive second step allows one to say, look P 1 is true and implies P 2. So it definitely will work for 2. This is equal to. Why is this interesting to us? You see how this induction step is kinda like a domino, it cascades and we can go on and on forever so it works for all positive integers.

The induction hypothesis was also employed by the Swiss Jakob Bernoulliand from then on it became more or less well known. Or everything above some threshold.

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