proof that τ⁢(n) is the number of positive divisors of n


The following is a proof that τ counts the positive divisorsMathworldPlanetmathPlanetmathPlanetmath of its input (which must be a positive integer).

Proof.

Recall that τ behaves according to the following two rules:

  1. 1.

    If p is a prime and k is a nonnegative integer, then τ⁢(pk)=k+1.

  2. 2.

    If gcd⁡(a,b)=1, then τ⁢(a⁢b)=τ⁢(a)⁢τ⁢(b).

Let p be a prime. Then p0=1. Note that 1 is the only positive divisor of 1 and τ⁢(1)=τ⁢(p0)=0+1=1.

Suppose that, for all positive integers m smaller than z∈ℤ with z>1, the number of positive divisors of m is τ⁢(m). Since z>1, there exists a prime divisorPlanetmathPlanetmathPlanetmath p of z. Let k be a positive integer such that pk exactly divides z. Let a be a positive integer such that z=pk⁢a. Then gcd⁡(a,p)=1. Thus, gcd⁡(a,pk)=1. Since 1≤a<z, by the induction hypothesis, there are τ⁢(a) positive divisors of a.

Let d be a positive divisor of z. Let y be a nonnegative integer such that py exactly divides d. Thus, 0≤y≤x, and there are k+1 choices for y. Let c be a positive integer such that d=py⁢c. Then gcd⁡(c,p)=1. Since c divides d and d divides z, we conclude that c divides z. Since c divides pk⁢a and gcd⁡(c,p)=1, it must be the case that c divides a. Thus, there are τ⁢(a) choices for c. Since there are k+1 choices for y and there are τ⁢(a) choices for c, there are (k+1)⁢τ⁢(a) choices for d. Hence, there are (k+1)⁢τ⁢(a) positive divisors of z. Since τ⁢(z)=τ⁢(pk⁢a)=τ⁢(pk)⁢τ⁢(a)=(k+1)⁢τ⁢(a), it follows that, for every positive integer n, the number of positive divisors of n is τ⁢(n). ∎

Title proof that τ⁢(n) is the number of positive divisors of n
Canonical name ProofThattaunIsTheNumberOfPositiveDivisorsOfN
Date of creation 2013-03-22 13:30:24
Last modified on 2013-03-22 13:30:24
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 11
Author Wkbj79 (1863)
Entry type Proof
Classification msc 11A25