Math 2510: Real Analysis I – Homework Solutions §7
Fall 2008
§11 Ordered Fields
11.3b (−x) · y = −(xy) and (−x) · (−y) = xy.
For the first statement:
(−x) · y = ((−1) · x) · y
= (−1)(x · y)
= −(x · y)
Theorem 11.1(c)
M3
Theorem 11.1(c)
For the second statement notice first that if a + b = 0, then b is the additive inverse of a and
a is the additive inverse of b. Therefore, a is the additive inverse of b and so a is the additive
inverse of the additive inverse of a. This shows that a = −(−a) for any real number a. In
particular, −(−(x · y)) = xy.
Now:
(−x) · (−y) = −(x · (−y))
= −(x · ((−1) · y))
= −((−1) · (xy))
= −(−(xy))
by the first part of the problem
Theorem 11.1(c)
M2 and M3
Theorem 11.1(c)
= xy
11.3d If x · z = y · z and z 6= 0, then x = y.
Supose that x · z = y · z and that z 6= 0. Then z1 is defined and z · z1 = 1. Multiplying both
sides of the equation x · z = y · z by z1 we get (x · z) · z1 = (y · z) · z1 . Using M3 we see that
x · (z · z1 ) = y · (z · z1 ). Therefore, x · 1 = y · 1 and so by M4, x = y.
11.3h If 0 < x < 1, then x2 < 1.
In class.
11.4 Prove: If x ≥ 0 and x ≤ for all > 0, then x = 0.
In class.
11.6a Prove: ||x| − |y|| ≤ |x − y|.
By the Triangle inequality, |z + y| ≤ |z| + |y| for every z and y. Take z = x − y. Then
|x| = |x − y + y| ≤ |x − y| + |y|. Subtracting |y| from both sides we see that |x| − |y| ≤ |x − y|
for all x and y. Switching x and y gives that |y| − |x| ≤ |y − x| = |x − y|. Multiplying this last
inequality by −1 we get, −|x − y| ≤ |x| − |y|. This shows that −|x − y| ≤ |x| − |y| ≤ |x − y|
and so ||x| − |y|| ≤ |x − y|.