Bridge to Abstract Mathematics: Mathematical Proof and Structures

5.4 PROOF BY MATHEMATICAL INDUCTION 189

Finally, we have

[sin 2mx + 2 sin x(cos 2mx cos x -'sin 2mx sin x)]/2 sin x

= [sin 2mx - 2 sin2x sin 2mx + 2 sin x cos x cos 2mx]/2 sin x

= [(I - 2 sin2 x) sin 2mx + (2 sin x cos x) cos 2mx]/2 sin x

= [cos 2x sin 2mx + sin 2% cos 2mx]/2 sin x

= [sin (2x + 2mx)]/2 sin x

= sin [2(m + 1)x]/2 sin x, as desired

(b) THEOREM If n E N and x is a positive real number then x2"-I > 0.

"Proof" Assume x2"-' > 0. Since x > 0, then x2" = x(x2"-') > 0, as desired.

(c) THEOREM If n E N, then I;=, rk-' = (1 - rn)/(l - r), r # 1.

Start of "Proof" The theorem is clearly true for n = 1, since r0 = 1 = (1 - r)/(l - r).
To verify condition (ii), assume the theorem true for m. To show the theorem
true for m + 1, we must prove that E=, rk = [(I - rm)/(l - r)] + 1....