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PROOF 77

Prove.

A fraction terminates exactly when its denominator's prime factors are 2 and 5

On the rational numbers ℚ.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

Definition D1

x is a terminating decimal if x = m / 10ⁿ for some integer m and some integer n ≥ 0.

Proof.

A notation ending at the n-th place after the point means the same as a fraction with denominator 10ⁿ.convention

∎

Definition D2

p/q is in lowest terms if p and q share no prime factor. Below, p and q are natural numbers.

Proof.

That is, no prime divides both p and q.convention, A3

∎

Lemma L1

If N = d·e, then the prime factors of N are exactly the prime factors of d together with those of e.

Proof.

(1)If d > 1, then d is a product of primes. If d = 1, regard it as the empty product.A1, convention
(2)The same holds for e.A1, convention
(3)Concatenating the two products writes N = d·e as a product of primes.(1), (2), A6, A7, E
(4)If N > 1, that decomposition is unique, so (3) is the prime factorization of N.(3), A1
(5)Therefore the prime factors of N are exactly the prime factors of d and of e.(4)

∎

Lemma L2

For every integer n ≥ 0, the only prime factors of 10ⁿ are 2 and 5.

Proof.

(1)If n = 0, then 10⁰ = 1 and there are no prime factors.convention
(2)10 = 2 · 5convention
(3)10ⁿ = (2·5)ⁿ = 2ⁿ5ⁿ(2), A5, E
(4)2ⁿ5ⁿ is a product of primes composed of 2 and 5 only.(3)
(5)By uniqueness this is the prime factorization of 10ⁿ.(4), A1

∎

Theorem T1

If q = 2ᵃ5ᵇ, then p/q is a terminating decimal.

Proof.

(1)Set n = max(a, b); then n − a ≥ 0, n − b ≥ 0convention
(2)k = 2ⁿ⁻ᵃ · 5ⁿ⁻ᵇ is a natural number.(1)
(3)q·k = 2ᵃ5ᵇ · 2ⁿ⁻ᵃ5ⁿ⁻ᵇ = 2ⁿ5ⁿ(2), A4, A6, A7
(4)2ⁿ5ⁿ = (2·5)ⁿ = 10ⁿA5
(5)q·k = 10ⁿ(3), (4)
(6)p · 10ⁿ = p(qk) = q(pk)(5), A6, A7, E
(7)p/q = (pk) / 10ⁿ(6), A2
(8)Therefore p/q is a terminating decimal.(7), D1

∎

Theorem T2

If a fraction p/q in lowest terms is a terminating decimal, then the only prime factors of q are 2 and 5.

Proof.

(1)There are a natural number m and an integer n ≥ 0 with p/q = m / 10ⁿ.assumption, D1
(2)p · 10ⁿ = m · q(1), A2
(3)q | p · 10ⁿ(2), A3, A7
(4)Let r be a prime factor of q.assumption
(5)There is a natural number e with p · 10ⁿ = q · e.(3), A3
(6)r is a prime factor of q, hence a prime factor of p · 10ⁿ.(4), (5), L1
(7)A prime factor of p · 10ⁿ is a prime factor of p or of 10ⁿ.L1
(8)The only prime factors of 10ⁿ are 2 and 5.L2
(9)If r were a prime factor of p, then r would be a common prime factor of p and q, contradicting lowest terms.(4), D2
(10)Therefore r = 2 or r = 5.(6), (7), (8), (9)
(11)The only prime factors of q are 2 and 5, so the prime factorization of q has the form q = 2ᵃ5ᵇ.(10), A1

∎

Corollary C

For a fraction p/q in lowest terms, being a terminating decimal and having q = 2ᵃ5ᵇ are equivalent.

Proof.

(1)T1 gives one direction.T1
(2)T2 gives the other direction.T2
(3)Both directions hold, so the two conditions are equivalent.(1), (2)

∎

Remark

Can A1 be removed from this proof?

Check which of the two directions uses the uniqueness of prime factorization.