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Prove every nonzero integer has only finitely many divisors

axiom: For $d,n \in \mathbb{Z}$, if $d \mid n$, there $\exists k \in \mathbb{Z}$ such that $d \cdot k = n$

$\forall d > n, \,\, k = \frac nd = \notin \mathbb{Z}$
$\forall d < -n, \,\, k = \frac n{-d} = \notin \mathbb{Z}$
$\longrightarrow d$ is bounded $ \longrightarrow \mid d \mid < \infty$

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Bucephalus
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Let $p_n$ be the prime factors and $q_n$ be the powers of those prime numbers.

We can break a number $r$, into the factored form $p_1^{q_1}p_2^{q^2}p_3^{q_3}$...

And yes, this series has infinite factors.

For example, we have $36=(2^2)(3^2)(5^0)(7^0)(11^0)....$

If we want to construct a divisor of $r$, we pick anywhere from $0$ to $q_1$ factors of $p_1$, we pick $0$ to $q_2$ factors of $p_2$, etc.

For $36$, we have the number of divisors is $(2+1)(2+1)(0+1)(0+1)(0+1)...=9$

Notice that if a number does not contain a particular prime factor, then the power of that prime factor is $0$ in the factorization, and only contributes a factor of $1$ to the product that represents the number of divisors.

Therefore, the number of divisors $r$ has is $\displaystyle (q_1+1)(q_2+1)(q_3+1)...=\prod_{k=1}^\infty(q_n+1)$.

We know that any finite number has a finite number of prime factors, each of which are raised to a finite power.

Therefore, we have that the product of a finite number of finite numbers multiplied by infinite $1s$ is also finite. √

Saketh Malyala
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    ok i did not mean for that last sentence to be a tongue twister – Saketh Malyala Jun 18 '17 at 03:40
  • i don't really understand this. Above my level of comprehension. Particularly, where did you get > for 36 we have the number of divisors is (2+1)(2+1)(0+1).....=9? – Bucephalus Jun 18 '17 at 03:51
  • so let's say we're building a factor of 36. we have two 2s, and two 3s. that means we can pick zero 2s, one 2, or two 2s, so we have three choices for 2s. we can also pick zero 3s, one 3, or two 3s, so we have three choices for 3s. we can also pick zero 5s (since we have none) , so we have one choice for 5s. we can pick zero sevens (since we have none), so we have one choice for 7s. multiplying all the choices together gives 3 x 3 x 1 x 1 x 1 x 1... – Saketh Malyala Jun 18 '17 at 03:55
  • Oh yeah so there are only 9 divisors. So what do I google to find more about these prime numbers and divisors. Is there a certain name for this construct/rule or associated axioms? – Bucephalus Jun 18 '17 at 04:01
  • i have no clue but heres a link i googled http://www.cut-the-knot.org/blue/NumberOfFactors.shtml – Saketh Malyala Jun 18 '17 at 04:05
  • great work, thanks @Saketh – Bucephalus Jun 18 '17 at 04:05