Question

Suppose a, b e Z. Show that if a for some integer c E Z, then a or b is even. Hint: Let P, A, and B be respectively the statements P {a2 + b2-c2}, A-fa is even), and B b is even]. In this problem you have to show that P (AVB). Use contradiction, i.e., prove that if the negation of this statement is true, then you come to a contradiction. Use that so that you have to assume that P, ~ A, and B are true, and come to a contradiction.

Answer 1

"A^2 + b^2 = c^2 on contrary assume that both a and b are odd As a and b are odd rightarrow = 2m + 1, b = 2n + 1 for some m, n elementof z Therefore a^2 + b^2 = (2m + 1)^2 + (2n + 1)^2 = 4m^2 + 4m + 1 + 4n^2 + 4n + 1 = 4(m^2 + n^2 + m + n) + 2 c is an integer rightarrow c is even or odd case i) c is even rightarrow 2/c i.e 2 divides c rightarrow 4/c^2 i.e 4 divides c^2 But a^2 + b^2 = c^2 rightarrow 4/c^2 rightarrow 4/a^2 + b^2 but 4 times a^2 + b^2 as 4