proof that τ(n) is the number of positive divisors of n
The following is a proof that τ counts the positive divisors of its input (which must be a positive integer).
Proof.
Recall that τ behaves according to the following two rules:
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1.
If p is a prime and k is a nonnegative integer, then τ(pk)=k+1.
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2.
If gcd(a,b)=1, then τ(ab)=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 divisor p of z. Let k be a positive integer such that pk exactly divides z. Let a be a positive integer such that z=pka. 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=pyc. Then gcd(c,p)=1. Since c divides d and d divides z, we conclude that c divides z. Since c divides pka 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(pka)=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 |
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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 |