# Prove that given primes in ascending order where $p_1<p_2<,\ldots$, the following is true: $p_{n+1} \leq p_1p_2…p_n + 1 \quad \forall n \in \Bbb N$.

I have absolutely no idea where to start, since I hardly ever have to use inequalities when it comes to working with primes. If anyone could help, I would really appreciate it. An example that shows what the inequality is trying to prove is: $$p_1=2, \, p_2=3, \, p_3=5 \implies p_3 = 5 \leq p_1p_2+1 = 2.3+1$$

Edit: It was very confusing and badly worded so I fixed it.

• Are you familiar with Euclid’s proof of infinitely many primes? – J. W. Tanner Jul 2 '20 at 23:59
• Hi, to me it does not seem clear what the actual inequality you want to show is. – Watercrystal Jul 3 '20 at 0:00
• @Watercrystal: the OP is trying to show $$p_{n+1}\leq p_1\cdots p_n+1$$ where $p_n$ represents the $n^{th}$ prime. – Clayton Jul 3 '20 at 0:09
• Ah, I read $p_n$ rather than $p_{n+1}$, my bad. – Watercrystal Jul 3 '20 at 0:11
• @Watercrystal Yes, that's it – Kevin Catino Jul 3 '20 at 0:31

$$p_1p_2\dots p_n+1$$ cannot be divisible by $$p_1, p_2, ...,$$ or $$p_n$$,

because it leaves remainder $$1$$ when divided by any of those primes.

Therefore, $$p_1p_2\dots p_n+1$$ is prime or divisible by a prime greater than $$p_n$$.

In any case, there is another prime besides $$p_1,p_2,...,p_n$$ that is at most $$p_1p_2\dots p_n+1$$.

Therefore, $$p_{n+1}\le p_1p_2\dots p_n+1$$.

This is how Euclid proved there are infinitely many primes.

$$p_1p_2\dots p_n$$ is called the primorial.

• Very clear explanation. Thank you so much! – Kevin Catino Jul 3 '20 at 0:52

This result also follows from Bertrand's postulate (now proved) which states that in the interval between any integer $$k$$ and $$2k$$ there is at least one prime. This also holds when $$k$$ is a prime, for example $$p_n.

So if $$p_1p_2\dots p_{n-1}\ge 2$$ then $$p_n.

• Interesting. Did not know about that. Thank you! – Kevin Catino Jul 3 '20 at 11:40