# Law 19983 on

Classified in Mathematics

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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 = 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 *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 = 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**