Law 19983 on

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 = 2k + 1, y = 2l. Therefore x + y = 2k + 1 + 2l = 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 = 2k. Therefore n = 2k − 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 = 2k but n + 1 = 2l. Therefore 2k + 1 = 2l (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 n2 = (2k)2 = 2(2k2), which is even. If n is odd, then n2 = (2k + 1)2 = 2(2k2 + 2k) + 1, which is odd.

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

If m = 2k and n = 2l, then mn = 4kl. 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.            Prove or disprove: A − (B ∩ C) = (A − B) ∪ (A − C).

If A − (B ∩ C) = Ø

A − (B ∩ C) 6= Ø and let x ∈ A − (B ∩ C).

The x ∈ A and x 6∈ B ∩ C which implies that x ∈ A and (x 6∈ B OR x 6∈ C).

Case 1: x 6∈ B If x 6∈ B, then x ∈ A − B ,

x ∈ (A − B) ∪ (A − C).

Case 2: x 6∈ C If x 6∈ C, then x ∈ A − C which means x ∈ (A − B) ∪ (A − C).

x ∈ (A − B) ∪ (A − C). Hence A − (B ∩ C) ⊆ (A − B) ∪ (A − C)

 (A − B) ∪ (A − C) ⊆ A − (B ∩ C)

2.Prove that     

1- x∈ A ∩ B ↔ x∈ [A ∩ B]                  Complement

2- x∈ A ∩ B ↔  ⅂ [x∈ A ∩ B]               ∉

3- ↔ ⅂ [x∈ A v  x∈ B]                            Union

4- ↔  ⅂ x∈ A ∧ ⅂ x∈ B                           Demorgan Law

5- ↔ x∉ A ∧  x∉  B                                    ∉     

6- ↔ x∈ A ∧  x∈ B                                  Complement

7- ↔ x∈ A ∪ B                                           Intersection

x∈ A ∩ B ↔ x∈ A ∪  B