Question

Prove that a nonzero integer b has only finitely many divisors.

Answer 1

For a number to be a divisor of b, it has to be smaller than b.

If b is some huge number, it will have a lot of divisors. Let's assume every number smaller than this huge number is a divisor (this isn't true, but it helps illustrate the point). If that's the case, then sure we have a lot of divisors, but it's not an infinite number of them. It's still finite.

If there were an infinite number of divisors, then you'd have to throw in decimal values too. But divisors can only be whole numbers. So this is where the contradiction occurs.

Answer 2

I got the idea but do you know if we can solve it mathematically?

Answer 3

Let D be the set of divisors of b

So D is set up like so

D = {d | b/d is an integer}

D is the set of numbers d such that b/d is a whole number (ie integer)

------------------------------------------------

Claim: D is an infinite set

If the claim were true, then D would be filled with numbers that go on forever. But they can't go on forever because d has to be smaller than b. Also, we cannot use nonwhole numbers (eg: 3.67) because d has to be an integer

So D is actually a finite set. Even if b is a very very large number, D would still be finite (even if D is a very large set)

I'm not sure how else to write this out but someone may have a different approach

Answer 4

Thanks