3100practiceproblems1 - PRACTICE PROBLEMS FOR FIRST MATH...

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PRACTICE PROBLEMS FOR FIRST MATH 3100 MIDTERM General Comments : These are practice problems. You should regard each indi- vidual problem as being a plausible exam problem. Certainly there are too many problems here to make one hour-long exam (roughly twice as many). The exam will be closed-book and calculators will not be permitted. I would urge you not to use calculators or software when doing the practice problems – that would not be such good practice. When in doubt, you should justify your answers by correct logical reasoning (brief explanations are often sufficient). 1) Let a,b R . a) Suppose a and b are both positive. Show that a < b ⇐⇒ a 2 < b 2 . b) Show that a < b ⇐⇒ a 3 < b 3 . c) Suggest a generalization of the first two parts to n th powers, for an arbitrary positive integer n . You need not prove it. First Solution: a) Suppose a,b > 0. We want to show that a b < 0 ⇐⇒ a 2 b 2 < 0. In general, given nonzero numbers A and B , to show that A and B have the same sign it is enough to show that A B is positive. As a 2 b 2 = ( a b )( a + b ), we have a 2 b 2 a b = a + b, which is indeed positive since a and b are both positive. b) Now suppose a and b are arbitrary but both zero (if a = b = 0, neither a b nor a 3 b 3 is negative). As above, we have a 3 b 3 a b = ( a b )( a 2 + ab + b 2 ) a b = a 2 + ab + b 2 , so it’s enough to show that a 2 + ab + b 2 > 0. This is clear if a and b have the same sign: all the terms are non-negative and at least one is positive. If a and b have opposite sign, then a 2 + ab + b 2 = | a | 2 − | ab | + | b | 2 ≥ | a | 2 2 | ab | + | b | 2 = ( | a | − | b | ) 2 0 , and is strictly positive unless b = a , in which case a 2 + ab + b 2 = a 2 > 0. c) The desired generalization is that if f ( x ) = x n and n is even, then for all a,b > 0, a < b ⇐⇒ a n < b n , whereas if n is odd, then for all a,b R , a < b ⇐⇒ a n < b n . Remark: To prove this by the current factorization method involves the identity a n b n = ( a b )( a n 1 + a n 2 b + ... + ab n 2 + b n 1 . When a and b are both positive, it is clear that a n 1 + a n 2 b + ... + ab n 2 + b n 1 > 0 , c Pete L. Clark, 2011. 1
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2 PRACTICE PROBLEMS FOR FIRST MATH 3100 MIDTERM and this is enough to show the result when n is even. However, when n is odd we need to establish the above inequality for all a,b which are not both zero, and off the top of my head this does not seem so straightforward. It is worth pointing out however that the result is actually equivalent to this inequality, so the alternate solution below gives an indirect proof of it. Second Solution:
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3100practiceproblems1 - PRACTICE PROBLEMS FOR FIRST MATH...

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