1873_solutions

# Following exercises decide whether the statement is

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Unformatted text preview: hat N is a positive integer.  0 and an integer n nx . 1  n2x2 and we shall do so by giving an example of such a number . Now we want to prove that there exists a number N such that We define  To prove that there exists an integer n N 1  N2 1 2 1 2 2 . N such that nx 1  n2x2 we make the observation that Nx 1  N2x2 . 2. For every number x 0, 1 and every number  0 there exists a number   0 such that for every number t 0, 1 satisfying |t x |   we have |t 2 x 2 |  . Solution: We shall prove that this statement is true. Since we want to prove that something happens for every number x 0, 1 , we begin as follows: Suppose that x 0, 1 . Now we want to prove that something happens for every number Suppose that  0 and so we continue:  0. We are now seeking a number   0 such that the inequality |t 2 x 2 |  will hold whenever t 0, 1 we have |t x |  . We observe first that whenever t 2|t x |. |t 2 x 2 |  |t x ||t  x | |t x | |t |  |x | With this inequality in mind we define   /2. We see at once that whenever t 0, 1 and |t t2 | x2 | 0, 1 and x |   we have 2|t x|  2 2 . 3. For every number  0 and every number x 0, 1 there exists a number   0 such that for every number t 0, 1 satisfying |t x |   we have |t 2 x 2 |  . 24 Hint: The statement that appears here is identical to the one in Exercise 2. 4. For every number  0 there exists a number   0 such that for every number x t 0, 1 satisfying |t x |   we have |t 2 x 2 |  . 0, 1 and every number Solution: We shall prove that this statement is true. Since we want to prove that something happens for every number Suppose that  0 we begin:  0. We are now seeking a number   0 such that the intequality |t 2 x 2 |  will hold for all numbers t and x in the interval 0, 1 satisfying the condition |t x |  . To give an example of such a number  we continue: Define   /2. We see at once that whenever t 0, 1 and x t2 x2 | | 0, 1 and |t x|  2 2|t x |   we have 2 . 5. For every number  0 and every number x there exists a number   0 such that for every number t satisfying |t x |   we have |t 2 x 2 |  . Solution: The statement made in this exercise says a little more than the one in Exercise 2 but it is still true. Since we want to prove that something happens for every number  0 we begin: Suppose that  0. Now we want to prove that something happens for every number x and so we continue: Suppose that x is a real number. We are now seeking a number   0 such that for every number t satisfying |t x |   we have |t 2 x 2 |  . To help us find an example of such a number  we observe first that if |t x |  1 x x 1 t x1 then, since |t |  |t x  x| |t x |  |x |  1  |x | we have |t 2 x 2 |  |t x ||t  x | |t x | |t |  |x | |t x | 1  2|x | With this inequality in mind we define  to be the smaller of the two numbers 1 and / 1  2|x | . We observe that whenever a number t satisfies the inequality |t x |   we have 1  2|x |  . |t 2 x 2 | |t x | 1  2|x |  1  2|x | 6. For...
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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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