Let Ak  = { x ∈ ℝ | k-1 ≤  x  ≤ k }, for each positive integer k.  What is , for some arbitrary integer n ≥ 0?

The function f : ℤ+ ⟶ {0, 1} defined by f(x) = ????A(????), the characteristic function of set A, where set A = {2n|n ∈ ℤ}, is injective (one-to-one).

Given arbitrary sets A, B, and C, complete the given membership table to verify whether the two sets, A – (B ⋂ C) and (A – B) ⋂ (A – C), are equal. Tip:  For any cells of the table that display beyond the right border of the question box, use the TAB →| key to move from cell to cell, rather than ‘clicking’ in a cell to make an entry. A B C (B ⋂ C) (A – B) (A – C) A – (B ⋂ C) (A – B) ⋂ (A – C) 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1  

Consider proving the following statement using a direct proof. “If a and b are real numbers with b > a, then b > (a+b)/2.” What do you assume as true to begin the proof?  What do you demonstrate must be true to complete the proof? 

Which of these given arguments uses the fallacy of circular reasoning (or begging the question)?  Argument A:    Proving:  For every real number x, x < x + 1. Let x be an arbitrary real number. We know that 0 < 1. Adding x to both sides, gives x + 0 < x + 1.    And that gives the equivalent inequality x < x + 1. So for every real number x, x < x + 1. Argument B:   Proving:  For integers x and y, if xy is a multiple of 5, then x is a multiple of 5 and y is a multiple of 5. Let x and y be integers with xy a multiple of 5. x is a multiple of 5 means x = 5k, for some integer k. Similarly, y is a multiple of 5 means y = 5j for some integer j. Substituting for x and y, we get xy = (5k)(5j) = 5(5kj).  Since 5kj is an integer, the product xy, which equals 5(5kj), is a multiple of 5. So xy is a multiple of 5, when x is a multiple of 5 and y is a multiple of 5. Argument C:    Proving: For every positive real number x, x + 1/x ≥ 2. Let x be a positive real number with x + 1/x ≥ 2. Multiplying both sides by x, we have x2 + 1 ≥ 2x. So by algebra, we get x2 - 2x + 1 ≥ 0, or (x-1)2 ≥ 0. Since it is true that the square of any real number is positive, (x-1)2 ≥ 0 confirms that x + 1/x ≥ 2, for every positive real number x. Argument D:  Proving:  For all integers m and n, if m and n are odd, then (m+n) is odd. Let m and n be integers. We know that when m and n are even, then (m+n) is even. So if m and n are odd, (m+n) is odd. 

The function f : ℤ+ ⟶ ℕ defined by f(x) = ⌊ x – 1 ⌋ is a bijection.

Prove the following statement using a proof by cases.   “For all non-negative integers n ≤ 2, n2 ≤ 2n.” Use good proof technique.   Grading rubric:1 pt. State any givens and assumptions. 3 pt. Clearly identify the cases and prove each case.1 pt. State the final conclusion at the end of the proof. Note:  To avoid the need for typing superscript exponents, you may use the expression ‘n-squared’ or ‘n^2’ to represent n2.  Also the ≤ symbol can be written as

Consider proving the following statement using a proof by contradiction. “The sum of an irrational number and rational number is irrational.” What do you assume as true to begin the proof?     What do you demonstrate must be true to complete the proof?    

For all sets A and B, ∅ ⊆ (A ⋂ B).

Let Ak  = { x ∈ ℝ | k-1 ≤  x  ≤ k }, for each positive integer k.   What is , where n is an arbitrary integer ≥ 2?