Problem Set 3
5 points

The problems are taken from the text. Keep in mind starred problems have solutions in the back!

Chapter 1.4
Exercises #5(all), 8(all), 11(all)

Chapter 1.5
Exercises 2 (a,b,c), 3 (b,c,g)


Comments explaining the answer are in RED. The answers are in BLUE.
Chapter 1.5
2. For this problem, just for clarity about WHAT is being said, I will introduce logical notation. Let
P = "A and B are invertible matrices" and Q="The product AB is invertible." The statement of the theorem mentioned is "If P, then Q"

(a) The contraposition is "If ~Q, then ~P". Note that ~P is the statement, "Not both A and B are invertible matrices," or "Either A is not invertible, or B is not invertible. The outline of the proof by contraposition is:
Assume that AB is not invertible. Then ......... Then either A is not invertible or B is not invertible.
(b) The converse is "If Q then P". So the outline of the proof of the converse is
If AB is an intervertible matrix, then...   Then both A and B are invertible matrices.
(c) To prove the theorem by contradiction, we need to assume the theorem does NOT hold, and find a contradiction.
Suppose there exist invertible matrices A and B whose product AB is not invertible. Then..... a proposition R holds. Then.... ~R. This is a contradition. Therefore AB is invertible.

3(g) The contraposition is the statement  "if x is not even, then 8 divdes x^2-1."
Suppose x is odd, then x = 2k+1 for some integer k. Therefore x^2-1 = (2k+1)^2-1 = 4k^2 +4k +1-1 = 4(k^2+k). On the other hand, k^2+k is itself even, since k even implies k = 2m for some integer m, and thus k^2+k = 4m^2+2m = 2(2m^2+m) is even, and k odd implies k = 2m+1 so that k^2+k = (4m^2+4m+1)+1 = 2(2m^2+2m+1) is even. Therefore k^2+k = 2n for some integer n, and x^2-1 = 4(k^2+k)=4(2n) = 8n is divisible by 8.