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**1. Give a direct proof of the following: “If x is an odd integer and y is an even integer, then x + y is odd”.**

Suppose *x* = 2*k* + 1, *y* = 2*l*. Therefore *x* + *y* = 2*k* + 1 + 2*l* = 2(*k* + *l*) + 1, which is **odd**.

**2. Consider the following theorem: If n is an even integer, then n + 1 is odd. Give a proof by contraposition of this theorem.**

Suppose *n* + 1 **is even**. Therefore *n* + 1 = 2*k*. Therefore *n* = 2*k* − 1 = 2(*k* − 1) + 1, which is odd

**3. Consider the following theorem: If n is an even integer, then n + 1 is odd. Give a proof by contradiction of this theorem.**

Suppose *n* = 2*k* but *n* + 1 = 2*l*. Therefore 2*k* + 1 = 2*l* (even = odd), which is a contradiction

**4. Prove the following theorem: n is even if and only if n2 is even.**

If *n* is even, then *n*2 = (2*k*)2 = 2(2*k*2), which is even. If *n* is odd, then *n*2 = (2*k* + 1)2 = 2(2*k*2 + 2*k*) + 1, which is odd.

**5. Prove: if m and n are even integers, then mn is a multiple of 4.**

If *m* = 2*k* and *n* = 2*l*, then *mn* = 4*kl*. Hence *mn* is a multiple of 4.**6. Prove:** |xy| = |x||y|, where x and y are real numbers. (recall that |a| is the absolute value of a, equals (a) if a>0 and equals (–a) if a<0>0>

**1. Give a direct proof of the following: “If x is an odd integer and y is an even integer, then x + y is odd”.**

Suppose *x* = 2*k* + 1, *y* = 2*l*. Therefore *x* + *y* = 2*k* + 1 + 2*l* = 2(*k* + *l*) + 1, which is odd.

**2. Consider the following theorem: If n is an even integer, then n + 1 is odd. Give a proof by contraposition of this theorem.**

Suppose *n* + 1 is even. Therefore *n* + 1 = 2*k*. Therefore *n* = 2*k* − 1 = 2(*k* − 1) + 1, which is odd

**3. Consider the following theorem: If n is an even integer, then n + 1 is odd. Give a proof by contradiction of this theorem.**

Suppose *n* = 2*k* but *n* + 1 = 2*l*. Therefore 2*k* + 1 = 2*l* (even = odd), which is a contradiction

**4. Prove the following theorem: n is even if and only if n2 is even.**

If *n* is even, then *n*2 = (2*k*)2 = 2(2*k*2), which is even. If *n* is odd, then *n*2 = (2*k* + 1)2 = 2(2*k*2 + 2*k*) + 1, which is odd.

**5. Prove: if m and n are even integers, then mn is a multiple of 4.**

If *m* = 2*k* and *n* = 2*l*, then *mn* = 4*kl*. Hence *mn* is a multiple of 4.**6. Prove:** |xy| = |x||y|, where x and y are real numbers. (recall that |a| is the absolute value of a, equals (a) if a>0 and equals (–a) if a<0>0>