Spivak Calculus, Chapter 1 Exercises

1. Prove the following:

i. If ax = a for some number a 6= 0, then x = 1.

Since a 6= 0, there exists a multiplicative inverse a−1, such that

a−1ax = aa−1

x = 1

ii. x2 − y2 = (x− y)(x+ y).

x2 − y2 = x2 + xy− xy− y2

= x(x+ y)+ (−y)(x+ y)

= [x+ (−y)](x+ y)

= (x− y)(x+ y)

iii. If x2 = y2, then x = y or x =−y.

|x| =√

x2

1