Is false say why suppose that 0 we shall find two

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Unformatted text preview: eeking is the fact that if t and x are any numbers in the interval p, 1 then 1  |x t | 1 |x t | 1 x tx t p2 The latter expression will be less than 1 when |x t |  p 2 . With this observation we define   p 2 and we have found a number  with the required properties. 10. a. If either 0     or     2 then arctan tan /2  arctan tan /2   2 . We have two cases to consider: Case 1: Suppose that 0    . In this case we have 0     and also 2 0     . 2 2 Therefore arctan tan /2  arctan tan /2          . 2 2 2 Case 2: Suppose that     2. Since     2 2 we see that arctan tan     2 2 and since 3      2 2 2 and so arctan tan /2       2 Therefore arctan tan /2  arctan tan /2             . 2 2 2 27 b. Ask Scientific Notebook to solve the equation arctan tan /2  arctan tan /2   2 . Are you satisfied with the answer that it gives? 11. If x is any rational number then lim lim cos n!x m nÝmÝ  1. p We suppose that x is rational and we express x in the form q where p and q are integers and q  0. Whenever an integer n is greater than q, the number n! x is an even integer and for such integers n we have lim cos n!x m  m Ý 1  1 lim mÝ Since lim m Ý cos n!x m  1 whenever n is sufficiently large we have lim lim cos n!x m  1. lim lim cos n!x m  0. nÝmÝ 12. If x is any irrational number then nÝmÝ We suppose that x is irrational. Given any integer n we know from the fact that n! x is not an integer that | cos n!x |  1 and therefore that lim cos n!x m  0. mÝ Therefore lim lim cos n!x nÝmÝ m  0. 4 Elements of Set Theory Exercises on Set Notation 1. Given objects a, b and y and given that a, b  a, y , prove that b  y. Solution: Since y a, y and a, b  a, y , we know that y a, b . We know, therefore, that either y  a or y  b. However, if y  a then the equation a, b  a, y becomes a, b  a and we deduce from the fact that b a that b  a. So in this case too we have y  b. 2. Prove that if a, b, x and y are any given objects and if a, b  x, y then either a  x and b  y, or a  y and b  x. Solution: Since a a, b and a, b  x, y we know that a x, y . Therefore either a  x or a  y. In the event that a  x we have a, b  a, y and it follows from Exercise 1 that b  y. Similarly, if a  y then b  x. 3. Prove that if a, b, x and y are any given objects and if a , a, b  then a  x and b  y. 28 x , x, y , The preceding exercises guarantee that if a , a, b  x , x, y , then either a  x or a, b  x, y from which we can see at once that a  x and b  y. 4. Describe the set p . Solution: Since the only subset of the set 5. Describe the set p p is  itself we have p . . Hint: You should be able to prove that p p  , . 6. Given that A  a, b, c, d , list all of the members of the set p A . Solution: You should be able to show that p a, b, c, d is , a , b , c , d , a, b , a, c , a, d , b, c , b, d , c, d , a, b, c , a, b, d , a, c, d , b, c, d , a, b, c, d 7. Given tha...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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