PROOF 194

Prove.

0.999… = 1 is a question of limits

On the real numbers ℝ.

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

0.999… denotes the limit of the sequence of partial sums (sₙ).

Proof.

(1)Let sₙ = 9·10⁻¹ + 9·10⁻² + ⋯ + 9·10⁻ⁿ. It is a finite decimal with n nines.convention

(2)0.999… is a symbol whose digits continue without end, and its value is defined as lim sₙ. Which number this symbol denotes is settled only once the limit is settled.convention

∎

Definition D2

lim sₙ = L means: for every ε > 0 there is an N such that |sₙ − L| < ε for all n ≥ N.

Proof.

The word 'approaches' is replaced by this single inequality. Every claim about limits is verified only through this definition.convention

∎

Lemma L1

For every n, sₙ = 1 − 10⁻ⁿ

Proof.

(1)n = 1: s₁ = 9·10⁻¹ = 0.9 = 1 − 10⁻¹D1, N

(2)Assuming sₙ = 1 − 10⁻ⁿ, sₙ₊₁ = sₙ + 9·10⁻⁽ⁿ⁺¹⁾assumption, D1

(3)　　　 = 1 − 10·10⁻⁽ⁿ⁺¹⁾ + 9·10⁻⁽ⁿ⁺¹⁾ = 1 − 10⁻⁽ⁿ⁺¹⁾(2), A4, N, E

(4)By induction it holds for every n.(1), (3), A3

∎

Lemma L2

For every n, 10⁻ⁿ ≤ 1/n

Proof.

(1)10ⁿ = (1 + 9)ⁿ ≥ 1 + 9n > nA2, N

(2)0 < n < 10ⁿ, so 10⁻ⁿ ≤ 1/n(1), A5

∎

Theorem T1

lim sₙ = 1

Proof.

(1)Let ε > 0 be arbitrary.assumption

(2)By the Archimedean property there is a natural number N with 1/N < ε.(1), A1

(3)|sₙ − 1| = |−10⁻ⁿ| = 10⁻ⁿL1, E, N

(4)If n ≥ N, then 10⁻ⁿ ≤ 1/n ≤ 1/N < εL2, A5, (2), A6

(5)Therefore |sₙ − 1| < ε for all n ≥ N. The condition of D2 is met.(3), (4), D2

(6)lim sₙ = 1(5), D2

∎

Corollary C

0.999… = 1. This equality is a consequence of the definition of the notation and of the limit.

Proof.

(1)0.999… = lim sₙD1

(2)lim sₙ = 1T1

(3)0.999… = 1(1), (2), E

(4)The manipulation of setting x = 0.999… and computing 10x − x (Proof 06) presupposes the existence of the number x denotes. That existence and that value have been settled here.(3)

∎

Remark

Can A1 be removed from this proof?

That 10⁻ⁿ decreases and that it falls below every positive number are different statements.