Mathematical Thinking: Problem-Solving and Proofs (2nd Edition)

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Unformatted text preview: 1.7. Let x = 1 ,y =- 1. Then- 1 /x =- 1 6 < 1 =- 1 /y . The statement is true if we add the condition that y > 0. For if x > y > 0, then x/y > y/y = 1, so 1 /y > 1 /x . (Note that this depends crucially on the fact that x and y are positive.) But this yields 0 > 1 /x- 1 /y , and this implies that- 1 /x >- 1 /y , which is what we want. 1.21. To get from a ( x + y )( x- y ) + b ( x- y ) = 0 to a ( x + y ) + b = 0, we must divide by x- y , but this is only valid if x 6 = y , which is contrary to the later supposition that x = y . 1.45. (a) We must check that f ( x ) is well-defined for 2 < x < 4. If x > 2, then we have x- 1 > 1 0, so | x- 1 | = x- 1. If x > 2, then | x | = x , so | x |- 1 = x- 1. Thus f is well-defined on (2 , 4). Further, we see that the domain of f is R since any real number is either less than 4 or greater than 2. Hence the rules determine a function from R to R . (b) f is not well-defined at 0 since |- 1 | = |- 1 | = 1 6 =- 1 = | | - 1. Hence the rules do not determine a function from...
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HW2solns - 1.7. Let x = 1 ,y =- 1. Then- 1 /x =- 1 6...

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