1873_solutions

A little more than the one in exercise 2 but it is

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Unformatted text preview: every number  0 there exists a number   0 such that for every number x and every number t satisfying |t x |   we have |t 2 x 2 |  . Solution: This statement is stronger than the statement in Exercise 4 but it is too strong to be true. The denial of this statement says that there exists a number  0 such that for every number   0 it is possible to find two numbers t and x such that |t x |   and |t 2 x 2 | . We shall prove that this denial is true. In order to prove the existence of a number as described in this assertion, we shall give an example. We define  1. Now we want to prove that something happens for every number   0 and so we continue: Suppose that   0. 25 We are now seeking two number t and x such that |t x |   and |t 2 x 2 | numbers we make the observation that if x  0 then 2 2 x 1 x 2  2x 2 1 2 x x We choose a positive number x such that 1  . x Then we define t  x 1 x and we observe that |t x |  1   x and |t 2 x 2 | 2  1. 7. For every number x |t x |   we have 1. To help ourselves find such 0, 1 there exists a number   0 such that for every number t 1 t 1 x 0, 1 satisfying  1. Hint: This statement is true. Write out a proof! Suppose that x 0, 1 . We shall prove the existence of the required number  by giving an example of one. We begin by observing that for each t 0, 1 we have 1  |x t | 1 x tx t In order to make the latter expression less than 1 we must prevent the denominator from being too small. In fact, looking at the following figure x 3x x 0 2 2 x x we see that whenever t is close enough to x we must have t  2 . In fact, whenever |t x |  2 we x have t  2 and for such numbers t we have 1  |x t | 1 |x t |. 1 x tx t x2 The latter number will be lesss than 1 as long as |x t |  x 2 . These observations tell us how to write our proof: We define  to be the smaller of the two numbers x x 2 and 2 . Then whenever t 0, 1 and |x t |   we have 1  |x t | 1 |x t |  1 x 2  1. 1 x tx t x2 x2 8. There exists a number   0 such that for every number x 0, 1 and every number t |t x |   we have 1  1. 1 x t 0, 1 satisfying Hint: This statement is false. Say why! Suppose that   0. We shall find two numbers t and x in the interval 0, 1 such that |x such that 1 1 1. x t For this purpose we define x   and t   1 26 t |   and  1 0 We see at once that |x  t |   and that 1  1. 1 x t This approach is actually too slick. A more motivated approach is to take x   and to look for a number t between 0 and x for which 1 1  1. x t The latter equation tells us that 1 1 1 t   and we see at once that the value of t that we want is 1 . 9. For every number p  0 there exists a number   0 such that for every number x t p, 1 satisfying |t x |   we have 1  1. 1 x t p, 1 and every number Hint: This statement is true. Write out a proof! In the event that p  1 the interval p, 1 is empty and every positive number  has the desired properties. We now assume that 0  p 1. The key to the proof we are s...
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