MT2002 Analysis

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## Exercises 3

1. Use the axioms for a field to prove that 0 . a = 0 for any element a of a field.

Prove that -1 . -1 = 1 and thereby justify the old rhyme:

Minus times minus is equal to plus
The reasons for this we will not discuss.

Use the axioms for an ordered field to prove that 1 > 0 in any ordered field.

Suppose that a, b, c are elements of an ordered field. Prove the following.
if a > 0 then 0 > -a,
if a > 0 and 0 > b then 0 > a . b,
if a > b and 0 > c then b . c > a . c,
if a > 0 then a-1 > 0, if 0 > a then 0 > a-1.

2. Let F be the set {0, 1, 2, 3, 4, 5, 6} on which addition and multiplication are defined modulo 7. Show that F is a field under these operations.

Define an ordering 6 > 5 > 4 > 3 > 2 > 1 > 0 on F and show that F is not an ordered field under this ordering.
Prove that F is not an ordered field under any ordering.

3. Write down the least upper bound (lub) and greatest lower bound (glb) of each of the following sets.
In which cases are the lub and glb elements of the set?

(a) {xQ | x3 < 2}
(b) {xR - Q | x2 ≤ 2}
(c) {1 , 1/2 , 1/3 , 1/4 , 1/5 , ...}
(d) {xR | x2n+1 = 2 for some nN}
(e) {m/nQ | 0 ≤ m < n}
(f) {m/nQ | 0 < m < n with both m, n odd}
(g) real numbers in (0, 1) whose decimal expansions do not contain the digit 9,
(h) real numbers in (0, 1) whose decimal expansions contain only odd digits.

4. Show that the real numbers in (0, 1) whose first decimal digit is ≠ 9 form an interval of length 9/10.
Prove that the real numbers in (0, 1) whose first and second decimal digits are ≠ 9 form an union of nine intervals of total length 81/100.
Show that the real numbers with no 9 in the first three digits form a union of intervals of total length (9/10)3.

Hence prove that the set of Question 3(g) may be enclosed in a union of intervals of arbitrarily small total length. (Such sets are said to have measure zero.)

SOLUTIONS TO WHOLE SET
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JOC September 2001