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3 February, 14:21

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

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  1. 3 February, 14:28
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    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
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