PROOF 06
Prove.
0.999… = 1
On the reals 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
The value of 0.999… is the limit of the sequence of partial sums (Sₙ).
Proof.
(1)Set Sₙ = 9·10⁻¹ + 9·10⁻² + … + 9·10⁻ⁿ, the finite decimal with n nines.convention
(2)0.999… is a symbol with digits written on without end. Its value is fixed only by this convention.convention
∎
Lemma L1
For all n, Sₙ = 1 − 10⁻ⁿ
Proof.
(1)Sₙ = 9(10⁻¹ + 10⁻² + … + 10⁻ⁿ)D1, A7
(2)10Sₙ = 9(1 + 10⁻¹ + … + 10⁻⁽ⁿ⁻¹⁾)(1), A6, A7, E
(3)10Sₙ − Sₙ = 9(1 − 10⁻ⁿ)(1), (2)
(4)10Sₙ − Sₙ = 9SₙA7
(5)9Sₙ = 9(1 − 10⁻ⁿ)(3), (4)
(6)Sₙ = 1 − 10⁻ⁿ(5), E
∎
Lemma L2
(Sₙ) is increasing and bounded above by 1, hence converges.
Proof.
(1)Sₙ = 1 − 10⁻ⁿL1
(2)10⁻⁽ⁿ⁺¹⁾ < 10⁻ⁿA4
(3)1 − 10⁻ⁿ < 1 − 10⁻⁽ⁿ⁺¹⁾, that is, Sₙ < Sₙ₊₁(1), (2), A5
(4)0 < 10⁻ⁿA4
(5)1 − 10⁻ⁿ < 1 − 0 = 1, that is, Sₙ < 1(1), (4), A5, A8
(6)Being a bounded increasing sequence, (Sₙ) has a limit.(3), (5), A1
∎
Theorem T
0.999… = 1
Proof.
(1)(Sₙ) converges.L2
(2)0.999… = lim SₙD1, (1)
(3)Sₙ = 1 − 10⁻ⁿL1
(4)lim Sₙ = lim(1 − 10⁻ⁿ)(3), E
(5) = 1 − lim 10⁻ⁿA2
(6) = 1 − 0A3
(7) = 1A8
(8)0.999… = 1(2), (7)
∎
Remark
Which steps of this chosen proof route need new justification if A1 is removed?
Distinguish this chosen route from other proofs. A3 and L1 remain available.