klarfeld-sec-1.4-1.5

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern