klarfeld-sec-1.4-1.5 - Jennifer Klarfeld Homework#3...

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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|=|a||b|. Let a and b be real numbers. Case 1: Let a ≥ 0 and b ≥ 0. Then ab ≥ 0, and |ab| = ab = |a||b|. Case 2: Let a < 0 and b < 0. Then ab ≥ 0, and |ab| = ab = (-­‐a)(-­‐b) = |a||b|. Case 3: Let a ≥ 0 and b < 0. Then ab ≤ 0, and |ab| = -­‐(ab) = -­‐(ba) = (-­‐ b)(a) = |b||a| = |a||b|. A similar argument follows for a < 0 and b ≥ 0. Thus, in all cases, |ab|=|a||b|. (c) = , for b≠0. Assume b ≠ 0. , , , , , , , , - - Case 1: Let ≥ 0 and > 0. Then ≥ 0, and Case 2: Let ≥ 0 and < 0. Then ≤ 0, and Case 3: Let < 0 and > 0. Then < 0, and Case 4: Let < 0 and < 0. Then > 0, and Thus, in all cases, , - = , - = , - = = − = − = , - = , , . , 1, 1- = = = , 11, , - = = , , - . . . 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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