Prove each statement. Proof by using cases

(a) If x is an integer, then x2 + 5x - 1 is odd.

(b) If x and y are real numbers, then max (x, y) + min (x, y) = x + y.

(c) If integers x and y have the same parity, then x + y is even.

The parity of a number tells whether the number is odd or even. If x and y have the same parity, they are either both even or both odd

see answer below

Step-by-step explanation:

for

a) for x²+5*x-1 to be odd, then x²+5*x should be even, therefore

if x is even, x=2*N → x²+5*x = (2*N) ²+5 * (2*N) = 4*N²+10*N = 2 * (2*N+5) = 2*m → then x²+5*x is even

if x is odd, x=2*N+1 → x²+5*x = (2*N+1) ²+5 * (2*N+1) = 4*N²+4*N+1 + 10*N+5 = 4*N² + 14*N+6 = 2 * (2*N² + 7*N+3) = 2*m → then x²+5*x is even

since always x²+5*x is even, then x²+5*x-1 is always odd

b) for max (x, y) + min (x, y) = x + y, then

if the max (x, y) = x → then min (x, y) = y → max (x, y) + min (x, y) = x + y

if the max (x, y) = y → then min (x, y) = x → max (x, y) + min (x, y) = y + x = x + y

then always max (x, y) + min (x, y) = x + y

c) If integers x and y have the same parity, then

if x is even and y is even, x=2*N₁ + y=2*N₂ → x+y = 2*N₁ + 2*N₂ = 2 * (N₁ + N₂) = 2*m → then x+y is even

if x is odd and y is odd, x=2*N₁+1 + y=2*N₂+1 → x+y = 2*N₁+1 + 2*N₂+1 = 2 * (N₁ + N₂+1) = 2*m → then x+y is even

then always x+y is even