Answers:

1. If a = {2, 4, 6} and b = {2, 4, 6}, the Method of Exhaustion requires that we test each possibility to determine the truth of the statement.

For a = 2 and b = 2      2 + 2 = 4 = 2(2)      2 - 2 = 0 = 2(0)

For a = 2 and b = 4      2 + 4 = 6 = 2(3)      2 - 4 = -2 = 2(-1)

For a = 2 and b = 6      2 + 6 = 8 = 2(4)      2 - 6 = -4 = 2(-2)

For a = 4 and b = 2      4 + 2 = 6 = 2(3)      4 - 2 = 2 = 2(1)

For a = 4 and b = 4      4 + 4 = 8 = 2(4)      4 - 4 = 0 = 2(0)

For a = 4 and b = 6      4 + 6 = 10 = 2(5)    4 - 6 = -2 = 2(-1)

For a = 6 and b = 2      6 + 2 = 8 = 2(4)      6 - 2 = 4 = 2(2)

For a = 6 and b = 4      6 + 4 = 10 = 2(5)     6 - 4 = 2 = 2(1)

For a = 6 and b = 6      6 + 6 = 12 = 2(6)     6 - 6 = 0 = 2(0)

Since a 2 can be factored out in each result, the sum and the difference are divisible by 2.

2. Let a = 2m and b = 2n for some integer m and n.

Sum of a and b = 2m + 2n = 2(m + n).      Difference of a and b = 2m - 2n = 2(m-n).

Since m and n are even integers, m + n and m - n are also even.    Hence, 2(m + n) and 2(m - n) are multiples of 2.

Therefore, for some even integers a and b, their sum and difference are divisible by 2.

3. In this case, we have to prove two things:

(a) if x = 1, then x³ - 3x² + 4x -2 = 0. and (b) if x³ - 3x² + 4x -2 = 0, then x = 1.

First, suppose x = 1. Then x³ - 3x² + 4x -2 = 1 - 3 + 4 - 2 = 0. This proves part (a), if x = 1, then

x³ - 3x² + 4x -2 = 0.

Second, assume that x³ - 3x² + 4x -2 = 0. Since x³ - 3x² + 4x -2 = (x -1) (x² -2x + 2), either

x -1= 0 or x² -2x + 2 = 0. If x² -2x + 2 = 0, the quadratic formula yields x = 1 + i. This contradicts

the hypothesis that x is real. Therefore x² -2x + 20. Hence x -1= 0 and therefore x = 1. This

proves part (b), if x³ - 3x² + 4x -2 = 0, then x = 1.