Question

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

Answer 1

Answer:

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