# How do I use the induction method

The required proof is often given by contradiction. I want to do that too first. As a second proof, I will give that by complete induction. It will be seen that the contradiction proof is more laborious. Namely, the contradiction is generated exactly with the constructive idea for the complete induction.

**If there really are an infinite number of prime numbers**, one certainly cannot write down all prime numbers. But you can check the possibility that there are only a finite number of prime numbers and consistently think further about this possibility. At the end of this reflection you will find that something is wrong. And if an end result based on the laws of logic obviously cannot be true, it has been proven that the assumption made at the beginning cannot be true either. According to mathematical logic, something wrong can never follow from something right. This proof technique is called a **Proof of contradiction**.

**Accepted** there would only be a finite number of prime numbers p_{1}, ...., p_{n}.

Then consider the number p = p_{1}* ... * p_{n}+1, which is obviously not due to any of the p_{i}, i = 1, ..., n is divisible. Then p, which of all p_{i} is different, obviously be a prime number.

That is a contradiction to assumption.

So the assumption was wrong that there must be an infinite number of prime numbers.

Instead of proving by contradiction, one would have that too

**direct evidence**being able to lead. It goes like this:

**Let the first n prime numbers be known.** Then consider number q = p_{1}* ... * p_{n}+1, which is obviously not due to any of the p_{i}, i = 1, ..., n is divisible.

We don't know whether q is a prime number, so let's look at both possibilities now. **Case 1: q is a prime number.** Then we found another prime number. **Case 2: q is not a prime number.** Then there is a proper divisor of q. (A real divisor is neither 1 nor q itself).

According to the construction of q, this divisor is not one of the prime numbers p_{1}, ..., p_{n}. So there must be another prime that divides q. This "other" prime number is greater than p_{n}. I will call this new prime p^{*}. p^{*} is not necessarily the n + 1 th prime (there may be other prime numbers between the largest prime number among the first n prime numbers and the new prime number), but the existence of n prime numbers implies the existence of **at least** n + 1 prime numbers. That way of inferring is the **complete induction**. The existence of a prime number suffices as an induction start. Starting from p_{1}= 2 one proves the existence of another prime number.

If you are now wondering whether q is not always a prime number, I will give you a counterexample:

2 * 3 * 5 * 7 * 11 * 13 + 1 = 30031 is not a prime number, because 30031 = 59 * 509.

One must therefore be careful in the induction step.

From the first n prime numbers p_{1}, ...., p_{n} results in the existence of another. The proof does not say what this new prime number is.

And the prime number p^{*} is not necessary the (n + 1) -th prime number. But if it is up to p^{*} there are more than n + 1 prime numbers, then that's enough. One then looks for the first n + 1 from the more than n + 1 prime numbers and can thus carry out the induction step from n + 1 to n + 2.

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