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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).

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?  [Assume] What do you demonstrate must be true to complete the proof?  [Prove]

The function f : ℤ+ ⟶ ℕ defined by f(x) = ⌊ x – 1 ⌋ is a bijection. [Note the use of the floor function in the definition of function f.]

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 [1] [2] [3] [4] [5] 0 0 1 [6] [7] [8] [9] [10] 0 1 0 [11] [12] [13] [14] [15] 0 1 1 [16] [17] [18] [19] [20] 1 0 0 [21] [22] [23] [24] [25] 1 0 1 [26] [27] [28] [29] [30] 1 1 0 [31] [32] [33] [34] [35] 1 1 1 [36] [37] [38] [39] [40]  

Which of these given arguments uses the fallacy of circular reasoning (or begging the question)?  [Answer] 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. 

Prove the following statement using a proof by cases.   [Hint: there are 3 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?   [Assume]  What do you demonstrate must be true to complete the proof?   [Prove]  

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?

If A = { a, b, c } and B = { b, { c }}, then | ???? (A ⋃ B) | = 16.