Unformatted text preview: Jennifer Klarfeld Homework #3 Section1.4‐1.5 Section 1.4 6. Let a and be be real numbers. Prove that: (a) ab=ab. Let a and b be real numbers. Case 1: Let a ≥ 0 and b ≥ 0. Then ab ≥ 0, and ab = ab = ab. Case 2: Let a < 0 and b < 0. Then ab ≥ 0, and ab = ab = (‐a)(‐b) = ab. Case 3: Let a ≥ 0 and b < 0. Then ab ≤ 0, and ab = ‐(ab) = ‐(ba) = (‐
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 . . . for ≠ 0. 1 Jennifer Klarfeld Homework #3 Section1.4‐1.5 7. Suppose a, b, c, and d are integers. Prove that: (h) if a divides b, then a divides bc. Suppose a divides b. Then b = ak, for some integer k. So bc = (ak)c = a(kc). Since k and c are both integers, kc is also an integer. Thus, a divides bc. (k) if a divides b and c divides d, then ac divides bd. Suppose that a divides b and c divides d. Then b = ak for some integer k and d = cj for some integer j. Therefore, bd = (ak)(cj) = a(kc)j = a(ck)j = ac(kj). Since k and j are both integers, kj is also an integer. Thus, ac divides bd. 2 Jennifer Klarfeld Homework #3 Section1.4‐1.5 Section 1.5 3. Let x, y, and z be integers. Write a proof by contraposition to show: (c) if is not divisible by 4, then is odd. Assume is even. Then = 2 for some integer . So 7 = 2 7 = 4 7 , which is divisible by 4. So if is even, then 7 is divisible by 4. Therefore, if 7 is not divisible by 4, then is odd. (d) if is even, then either or is even. → ( ∨ ) Assume is odd and is odd. Then = 2 + 1 and = 2 + 1 for some integers k and j. So = 2 + 1 2 + 1 = 4 + 2 + 2 + 1 = 2 2 + + + 1. Because 2 + + is an integer, is odd. Hence, if is odd and is odd, then is odd. Therefore, if is even, then either is even or is even. ~( ∨ )
= ~ ∧ ~ 3 Jennifer Klarfeld Homework #3 Section1.4‐1.5 5. A circle has center (2,4). (a) Prove that (‐1,5) and (5,1) are not both on the circle. Assume (‐1,5) and (5,1) are both on the circle. Then, since the center is (2,4), (‐1,5) and (5,1) are the same distance from (2,4). By the distance formula, we have, for (‐1,5), the distance from (2,4) is −1 − 2 7 + 5 − 4 7 = −3 7 + 1 7 = 10, and, for (5,1), the distance from (2,4) is 5 − 2 7 + 1 − 4 7 = 3 7 + −3 7 = 18. Since 10 ≠ 18, then (‐1,5) and (5,1) are not the same distance from (2,4). Therefore, (‐1,5) and (5,1) are not both on the circle. (b) Prove that if the radius of the circle is less than 5, then the circle does not intersect the line y=x‐6. Assume that the circle with center (2,4) has a radius that is less than 5, but that the circle does intersect with y=x‐6. If the radius is less than five, then there is a point on the circle, (x,y), such that − 2 7 + − 4 7 < 5. Since the circle intersects with y=x‐6, then we can substitute x=y+6 into the radical, giving +6 −2 7 + −4 7 < 5, 4 Jennifer Klarfeld Homework #3 Section1.4‐1.5 which can be simplified to 2 7 + 32 < 5. K
Hence, 2 7 + 32 ≥ 25, so 7 ≥ − , which is not possible since 7 ≥ 0. 7 Thus, we have a contradiction. So if the radius of the circle is less than 5, then the circle does not intersect the line y=x‐6. 5 Jennifer Klarfeld Homework #3 Section1.4‐1.5 6. (c) Suppose is a positive integer. Write a proof by contradiction to show that if is odd, then + is even. Assume that is odd, and + 1 is odd. Then + 1 = 2 + 1 for some integer . Subtracting 1 from both sides, we see that = 2, which means that is even. But this contradicts the hypothesis that is odd. Therefore, if is odd, then + 1 is even. 6 Jennifer Klarfeld Homework #3 Section1.4‐1.5 7. (a) Prove the following biconditional statement. divides if and only if divides . Assume that divides . Then = for some integer . By dividing both sides my , we see that = , and that divides . Now assume that divides . Then = for some integer . Then = () = (), which means that divides . Therefore, divides if and only if divides . 7 ...
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 Spring '16
 Math, divides, Jennifer Klarfeld

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