MATH ASSIGNMENT 1 - SOLUTIONS September 11, 2006

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1 MATH ASSIGNMENT - September, 00. Using the Trichotomy Law prove that if a and b are real numbers then one and only one of the following is possible: a < b, a b, or a > b. Since a and b are real numbers then a b is a real number. By the Trichotomy Law we know that a b < 0, a b 0 or a b > 0. These immediately translate into a < b, a b or a > b.. We define the absolute value of a real number a by { a, a 0 a a, a 0 Prove the following: (a) a + b a + b. We will each of these by cases. The case where either ab 0 is not interesting, so we will leave it. We must have a < 0 or a > 0 and b < 0 or b > 0. Thus, we are left with cases to check: () a > 0 and b > 0, () a < 0 and b > 0, () a > 0 and b < 0, and () a < 0 and b < 0. In case () since both a and b are positive, a + b is positive and a a, b b, and a + b a + b. Therefore a + b a + b a + b and the statement is true. In case (), since a < 0, a a. To show that a + b a + b we must show that a + b a + b 0. Either a + b 0 or a + b 0. If a + b 0 If a + b 0 a + b a + b ( a) + b (a + b) a > 0 since a > 0 a + b a + b ( a) + b ( (a + b)) ( a) + b + a + b)) b > 0 since b > 0 Thus a + b a + b in this case. Case () is similar since the roles of a and b are reversed. Case () is similar to Case ().

2 (b) y y. Here we break the proof up into the same cases: () > 0, y > 0, () < 0, y > 0, () > 0, y < 0, and () < 0, y < 0. In Case () since > 0 and y > 0, then y > 0, and it easily follows that y y y. In Case () since < 0 and y > 0, then y < 0, and it easily follows that y (y) ( )y y. In Case () since > 0 and y < 0, then y < 0, and it easily follows that y (y) ( y) y. In Case () since < 0 and y < 0, then y > 0, and it follows that y y ( )( y) y. (c), if 0. Since 0, we know that is its multiplicative inverse, so. Solving gives us that. (d) y, if y 0. y Use the above again and the fact that y y : y y y y y. (e) y + y. Solution Method I: The eloquent solution uses the results of Part a to show this: y + ( y) + y + y, where the inequality comes from a. Solution Method II: You can do this one much like the first one. Break it into cases and do them one at a time. Cases: () > 0, y > 0, () < 0, y > 0, () > 0, y < 0, and () < 0, y < 0. We need to show in each case that + y y 0. In Case () we have to deal with two cases y 0 and y 0. If y 0, then y y and + y y + y ( y) y > 0. If y 0, Fall 00 MATH 0-090

3 then y ( y) y and + y y + y (y ) > 0. Thus, this is true in Case (). Case (): In this case and y y. Again, we have to consider two cases: y 0 and y 0. However, note that if < 0 and y > 0, it cannot happen that y 0. So, y 0, then y ( y) y and + y y + y (y ) 0 0. Thus, this is true in Case (). Case (): In this case and y y. Again, we have to consider two cases: y 0 and y 0. Again, as in Case () it is impossible for y 0. So, y 0, then y y and + y y y ( y) 0 0. Thus, this is true in Case (). For Case (), and y y. If y 0, then y y and + y y + ( y) ( y) > 0. If y 0, then y ( y) y and + y y +( y) (y ) y 0. Thus, this is true. (f) y y. Solution Method I: There is an eloquent solution here as well. y + y y + y y y Solution Method II: You can also break it into cases and do them one at a time. Cases: () > 0, y > 0, () < 0, y > 0, () > 0, y < 0, and () < 0, y < 0. We need to show in each case that y ( y ) y + y 0. In Case () we have to deal with two cases y 0 and y 0. If y 0, then y y and y + y y + y > 0. If y 0, then y ( y) y and y + y y +y (y ) > 0. Thus, this is true in Case (). Case (): In this case and y y. This time it is possible for y 0 but impossible for y 0. If y 0, then y ( y) y and y + y y + + y y > 0. Thus, this is true in Case (), Case (): In this case and y y. This time it is possible for y 0 but impossible for y 0. If y 0, then y y and y + y y y y 0. Thus, this is true in Case (). For Case (), and y y. If y 0, then y y and y + y y ( ) + ( y) ( y) > 0. If y 0, then y ( y) y and y + y y ( ) + ( y) 0 0. Thus, this is true. Fall 00 MATH 0-090

4 . The fact that a 0 for all real numbers a has tremendous implications. The most widely used of all inequalities is the Schwarz inequality: y + y + y + y Do ONE of the following: (a) Prove the Schwarz inequality by using y + y (how is this derived?) with i, y + y i y + y first for i and then for i. The first inequality comes from the fact that 0 ( y) y + y, so y + y. Thus, doing the algebra y + y + y y + y + y y + y ( + ) + ( ( ) ( + + y y + y y y + y ) ) y + y + y y + y + y Adding these y + y + y + y y + y + y + y y + y and y y + y + y y + y + + y + y + y + y + y + y (b) Prove the Schwarz inequality by first proving that ( + )(y + y ) ( y + y ) + ( y y ). First, ( + )(y + y ) y + y + y + y. Fall 00 MATH 0-090

5 Now, ( y + y ) + ( y y ) y + y y + y + y y y + y y + y + y + y ( + )(y + y) ( + )(y + y ) ( y + y ) + ( y y ) ( y + y ) Thus, ( + ) (y + y) y + y and we are done.. Prove the following formulæ by induction (a) n i i n(n + )(n + ) First, check it for n and we have, so it is true for n. Now, assume it is true for k. We must prove that it is true for k +. k+ i i k i + (k + ) i k(k + )(k + ) + (k + ) k(k + )(k + ) + (k + ) (k + )[k(k + ) + (k + )] (k + )[k + 7k + )] (k + )(k + )(k + ) which is what we needed, and we are done. (b) n ( n) First, check it for n and we have (), so it is true for n. Now, assume it is true for k. We must prove that it is true for k +. 5 Fall 00 MATH 0-090

6 + + + (k + ) k + (k + ) ( k) + (k + ) ( ) k(k + ) + (k + ) k (k + ) + (k + ) (k + ) (k + k + ) (k + ) (k + ) ( ) (k + )(k + ) ( (k + )) which is what we needed, and we are done. 5. Find a formula for (a) (i ) (n ) i k (k ) + k (n)(n + ) (k ) + k (k ) n n(n + ) + n n + n (n + n) n (b) (i ) (n ) i Solution Method I: k (k ) (k ) + (k) (n)(n + )(n + ) (n)(n + )(n + ) n(n + )(n ) k n(n + )(n + ) n n Fall 00 MATH 0-090

7 Solution Method II: (k ) (k k + ) k k + k k + n ( ) ( ) n(n + )(n + ) n(n + ) + n n(n + )(n + ) n n n(n + )(n ) n n. Use the given method to find: (a) n Following the eample from the homework sheet we note that (k + ) k k + k + k +. Then proceeding as the eample we would have: (n + ) n + n + n + n n (n + ) n(n + )(n + ) n(n + ) n n (n + ) n(n + )(n + ) n(n + ) n n n + n + n n n (n + ) (b) n 7 Fall 00 MATH 0-090

8 First, (k + ) 5 k 5 5k + 0k + 0k + 5k +. So, (n + ) n + 0 n + 0 n + 5 n (n + ) 5 0 n (n + ) n n 5 + 5n + 5n n n n(n + )(n + )(n + n ) n n(n + )(n + )(n + n ) 0 n + n n(n + )(n + ) n(n + ) 0 5 n (c) n(n + ) For this one you need to realize that each term is of the form can be rewritten as n(n + ) k (k + ) k k +. Thus, ( ) n + ( + ) + + n n + This is a classic eample of what is known as a telescoping sum. k (k + ) and this ( n ) n + 8 Fall 00 MATH 0-090