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People tend to think of the proof you are thinking of as easier because they are more familiar with it. There are some subtle things that the proof you mention

People tend to think of the proof you are thinking of as easier because they are more familiar with it. There are some subtle things that the proof you mentioned glosses over. Namely, if p|A and p|B then p|(A-B). Now this is not a deep concept but it is something which people who are not accustomed to thinking of proofs tend to gloss over. Namely, they will see the part that if P = p1 p2 p3...pn + 1 and p|P then p|1 and not think that this does itself require justification. Personally, I really like the proof using sines. I won't claim it is easier or simpler but it is very nice. I've never thought to prove the infinitude of primes this way and it uses facts from trigonometry. I will now incorporate this into my trig classes. I think it is quite straightforward and easy for one to understand. Students in trigonometry and calculus generally do not have experience in proving statements about arithmetic. Here is an excellent way to show a connection between concepts that seemingly have nothing to do with each other. It is this that mathematicians really like.

JadeNB 10y ago
> Namely, they will see the part that if P = p1 p2 p3...pn + 1 and p|P then p|1 and not think that this does itself require justification. While your point that every step of a proof requires justification is a good one, this is really, really easy to verify. That doesn't excuse one from actually giving that justification, in the course of a formal proof; but there's at least as much implicit knowledge bound up in the linked proof. (For example, one needs the existence of sine, its 2π-periodicity, the fact that it is positive from 0 to π/2 ….)
- yequalsx 10y ago
  Yes, definitely there is more information implicitly wrapped into the sine proof. That's why it's not simpler. However, it is not a worse proof either. It's just a different one that is clever and can help to reinforce concepts from trigonometry.
  - bonoboTP 10y ago
    There's not much trigonometry in this proof, just the fact that sine has period 2pi is zero at pi. Any function that has these two properties could directly replace sine in this proof. And actually any periodic function that is zero somewhere could be used in a tiny bit modified version of this proof.
    - yequalsx 10y ago
      That is true but in Trigonometry Sine the function one studies and so using this result in a trigonometry course is nice.
    - JadeNB 10y ago
      > And actually any periodic function that is zero somewhere could be used in a tiny bit modified version of this proof. I think that, for the modification to be tiny, you need something like: there is an interval on which the function is positive, and on the boundary of which it vanishes. (Some sign condition is necessary; taking the constant function at 0 doesn't work! Of course, this particular condition is automatically satisfied for any continuous, periodic function that vanishes somewhere, but is not identically 0.)

JadeNB 10y ago

> Namely, they will see the part that if P = p1 p2 p3...pn + 1 and p|P then p|1 and not think that this does itself require justification. While your point that every step of a proof requires justification is a good one, this is really, really easy to verify. That doesn't excuse one from actually giving that justification, in the course of a formal proof; but there's at least as much implicit knowledge bound up in the linked proof. (For example, one needs the existence of sine, its 2π-periodicity, the fact that it is positive from 0 to π/2 ….)

yequalsx 10y ago
Yes, definitely there is more information implicitly wrapped into the sine proof. That's why it's not simpler. However, it is not a worse proof either. It's just a different one that is clever and can help to reinforce concepts from trigonometry.
- bonoboTP 10y ago
  There's not much trigonometry in this proof, just the fact that sine has period 2pi is zero at pi. Any function that has these two properties could directly replace sine in this proof. And actually any periodic function that is zero somewhere could be used in a tiny bit modified version of this proof.
  - yequalsx 10y ago
    That is true but in Trigonometry Sine the function one studies and so using this result in a trigonometry course is nice.
  - JadeNB 10y ago
    > And actually any periodic function that is zero somewhere could be used in a tiny bit modified version of this proof. I think that, for the modification to be tiny, you need something like: there is an interval on which the function is positive, and on the boundary of which it vanishes. (Some sign condition is necessary; taking the constant function at 0 doesn't work! Of course, this particular condition is automatically satisfied for any continuous, periodic function that vanishes somewhere, but is not identically 0.)