Which of these given arguments uses the fallacy of circular…
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.