1873_solutions

# X 1 2 and that the equation 88 x n 1 2 xn holds

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Unformatted text preview: et of real numbers that has exactly n  1 members. Choose a member 89 a of the set S. Since the set S a has exactly n members, it follows from the assertion P n that S a has a largest member that we shall call b. The larger of the two numbers a and b is clearly the largest member of the set S. Since P 1 is true and since the condition P n  P n1 holds for every positive integer n it follows from the principle of mathematical induction that P n is true for every positive integer n. 7. Given that S Z  , prove that the following three conditions are equivalent: a. The set S is infinite. b. The set S has no largest member. c. The set S is unbounded above. It it probably easier to explain why the denials of these three statements are equivalent to one another. We shall show that the following three satements are equivalent: a. The set S is finite. b. The set S has a largest member. c. The set S is bounded above. We already know that every finite set of numbers must have a largest member. Therefore condition a implies condition b. If a set of numbers has a largest member then that largest member is clearly an upper bound of the set. Therefore condition b implies condition c. Finally, we know that every nonempty set of integers that is bounded above must have a largest member. If the largest member of a set of postive integers is n then that set cannot have more than n members and must therefore be finite. Therefore condition c implies condition a. 8. Prove that for every positive integer n, if n horses run in a race and no two horses tie then there are exactly n! possible outcomes. For each positive integer n we define P n to be the assertion that if n norses run in a race and no two horses tie then there are exactly n! possible outcomes. The assertion P 1 is obviously true. Now suppose that n is any positive integer for which P n is true. To show that the assertion P n1 is also true, suppose that n  1 horses run in a race tha that no two horses tie. Choose one of the horses whom we shall call Dobbin and, for the moment let’s focus our attention on the other n horses. Those other horses present us with n! possible outcomes. For each of those outcomes, if we add in Dobbin then there are n  1 positions in which Dobbin can place among the other horses. Therefore the number of outcomes, including Dobbin is n  1 n!  n  1 !. Since P 1 is true and since the condition P n  P n1 holds for every positive integer n it follows from the principle of mathematical induction that P n is true for every positive integer n. 9. Given nonnegative integers n and r satisfying r n, the binomial coefficient n! n . r n r !r! n r is defined by the equation a. Prove that if n and r are nonnegative integers and r n then n n  n1 . r r r1 Hint: No special techniques are needed for this proof. Just combine the expressions on the left side. b. Prove that if n and r are nonnegative integers and r n then every set which has n members must have exactly n subsets that have r...
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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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