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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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 Fall '08
 STAFF
 Math, Calculus

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