**Introduction**

**Prime numbers** are numbers with only two factors, $latex 1$ and itself. Eleven is prime since its factors are only $latex 1$ and $latex 11$. Numbers that are not prime are called **composite**. Fifteen is composite because it has more than $latex 2$ factors. The factors of $latex 15$ are $latex 1$, $latex 3$, $latex 5$, and $latex 15$.

As you count farther, you will observe that the number of primes are getting fewer and fewer. There are $latex 168$ primes between $latex 1$ and $latex 1000$, $latex 135$ primes between $latex 1000$ and $latex 2000$, $latex 127$ primes between $latex 2000$ and $latex 3000$, and $latex 120$ primes between $latex 3000$ and $latex 4000$. Now, the question is, “Is there a largest prime?” The equivalent to this question is, “Are the number of primes finite?” In this post, we are going to prove intuitively that the number of primes is infinitely many.

**The Proof**

Theorem: There are infinitely many primes.

Proof:

We can prove that there are infinitely prime numbers if we can show that given any finite list of primes, we can always find a new prime not on that list. We will only start with cases, but this is enough for the proof. We will discuss the generalization of the proof with algebraic symbols in a different post.

- Suppose that there are only a finite number of primes and it is contain in the set $latex S_1 = \{2, 3, 5 \}$. If we multiply all the numbers in the set and add 1, we have $latex (2 \times 3 \times 5 + 1 = 31)$, the product can be prime or composite.
- Verifying 31, we see that its is a prime. So, we found a prime not in $latex S_1$.
- Well, maybe $latex S_2 =\{ 2, 3, 5, 31 \}$ are the only primes. Repeating the process, we have $latex 2 \times 3 \times 5 \times 31 + 1 = 931$.
- Now, $latex 937$ is composite since $latex 931 = 19(7)(7)$. Now, we can see that $latex 19$ and $latex 7$ are primes, but they are not on the list.
- Note that the process above can be repeated infinitely.

**Why the Proof Above Works**

In the case above, we have shown that given a finite list of primes, we can always find a prime not on that list. By multiplying all the numbers on the list, you can have two cases: the product can be prime or composite.

If the product of all the numbers added to 1 is prime, then the product is a prime number not on the list. Therefore, we found another prime not on the list.

If the product of all the numbers added to 1 is composite, then it must be a product of primes. But the product gives a remainder of 1 if you divide it with any of the factors (Why?). Therefore, one of its factors is a prime not on the list. Again, we found another prime not on the list.

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In steps 3 and 4 of the proof, “937” should be “931”. Also, in step 3, a “+1” is missing in the last equation.

Thanks. I changed it already. 🙂

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By multiplying all the numbers on the list, you can have two cases: the product can be prime or composite.

“Case 1: The product is prime.

If the product of all the numbers added to 1 is prime, then the product is a prime number not on the list. Therefore, we found another prime not on the list.”

Are you sure? I think your word order might be out.

If the product of all the numbers plus 1 is prime, then the product is prime itself not on the list?

2*3*5 + 1 = 31 {correct prime but}

2*3*5 = 30 {Not prime}

my mistake.

Typo, step 4 reads 937 instead of 31. best regards

Typo, step 4 reads 937 instead of 931. best regards (someone already commented on this)