PROOF 114

Prove.

The general term of an arithmetic sequence is linear

On real sequences.

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

A sequence (aₙ) is arithmetic when there is a number d, not depending on n, such that for every natural number n, aₙ₊₁ = aₙ + d.

Proof.

This condition speaks only of two neighboring terms. It says nothing about two distant terms.convention

∎

Definition D2

b − a means b + (−a).

Proof.

Subtraction is not a new operation but the addition of an inverse.convention

∎

Lemma L1

Let P(n) be the statement 'aₙ = a₁ + (n−1)d'. Then P(1) holds.

Proof.

(1)1 − 1 = 1 + (−1) = 0D2, A5

(2)(1−1)d = 0 × d = d × 0 = 0(1), E, A8, P1

(3)a₁ + (1−1)d = a₁ + 0 = a₁(2), E, A4

(4)Hence P(1) holds.(3)

∎

Lemma L2

For every natural number k, if P(k), then P(k+1).

Proof.

(1)Suppose aₖ = a₁ + (k−1)d.assumption

(2)aₖ₊₁ = aₖ + dD1

(3)aₖ₊₁ = (a₁ + (k−1)d) + d(1), (2), E

(4)　　　 = a₁ + ((k−1)d + d)(3), A2

(5)(k−1)d + d = (k−1)d + 1×d = ((k−1)+1)dA7, A8, A6

(6)(k−1) + 1 = (k + (−1)) + 1 = k + ((−1)+1) = k + 0 = kD2, A2, A3, A5, A4

(7)(k+1) − 1 = (k+1) + (−1) = k + (1 + (−1)) = k + 0 = kD2, A2, A5, A4

(8)aₖ₊₁ = a₁ + kd = a₁ + ((k+1)−1)d, so P(k+1) holds.(4), (5), (6), (7), E

∎

Theorem T1

For every natural number n, aₙ = a₁ + (n−1)d

Proof.

(1)P(1) holds.L1

(2)For every k, P(k) ⟹ P(k+1).L2

(3)Hence P(n) holds for every n.(1), (2), A1

∎

Corollary C

aₙ = dn + (a₁ − d). If d ≠ 0, this is a linear expression in n, all points (n, aₙ) lie on the line y = dx + (a₁ − d), and the gap between neighboring points is always (1, d).

Proof.

(1)aₙ = a₁ + (n−1)dT1

(2)(n−1)d = (n + (−1))d = nd + (−1)dD2, A6

(3)(−1)d = −(1×d) = −dA9, A8, A7

(4)aₙ = a₁ + (nd + (−d)) = nd + (a₁ + (−d)) = dn + (a₁ − d)(1), (2), (3), A2, A3, A8, D2

(5)d and a₁ − d do not depend on n, so if d ≠ 0 this is a linear expression in n.(4), D1

(6)The graph of y = dx + (a₁ − d) is a line, and every point (n, aₙ) lies on it.(4), P2

(7)aₙ₊₁ − aₙ = (aₙ + d) − aₙ = d and (n+1) − n = 1, so the gap is always (1, d).D1, D2, A2, A3, A5, A4

∎

Remark

Can A1 be removed from this proof?

Only one line in the body cites A1. Erase that line and see what remains.