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25 Other Methods of Proof
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We give here a number of examples that illustrate proof techniques other than direct proof, proof by contradiction, and induction
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EXAMPLE 29 Show that if there are 23 people in a room then the odds are better than even that two of them have the same birthday Solution: The best strategy is to calculate the odds that no two people have the same birthday, and then to take complements Let us label the people p1 , p2 , , p23 Then, assuming that none of the p j have the same birthday, we see that p1 can have his birthday on any of the 365 days in the year, p2 can then have his birthday on any of the remaining 364 days, p3 can have his birthday on any of the remaining 363 days, and so forth So the number of different ways that these 23 people can all have different birthdays is 365 364 363 345 344 343
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On the other hand, the number of ways that birthdays could be distributed (with no restrictions) among 23 people is 365 365 365 365 = 36523
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Thus the probability that these 23 people all have different birthdays is p= 365 364 363 343 36523
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A quick calculation with a calculator shows that p 04927 < 05 That is the desired result
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EXAMPLE 210 Jill is dealt a poker hand of 5 cards from a standard deck of 52 What is the probability that she holds four of a kind Solution: If the hand holds 4 aces, then the fth card is any one of the other 48 cards If the hand holds 4 kings, then the fth card is any one of the other 48 cards And so forth So there are a total of 13 48 = 624 possible hands with four of a kind The total number of possible 5-card hands is 52 = 2598960 5 Therefore the probability of holding 4 of a kind is p= 624 = 000024 2598960
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252 OTHER ARGUMENTS
EXAMPLE 211 Let us show that there exist irrational numbers a and b such that a b is rational
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Solution: Let = 2 and = 2 If is rational then we are done, using a = and b = If is irrational, then observe that
2
= [
= 2 = [ 2]2 = 2
Thus, with a = and b = that a b = 2 is rational
2 we have found two irrational numbers a, b such
EXAMPLE 212 Show that if there are six people in a room then either three of them know each other or three of them do not know each other (Here three people know each other if each of the three pairs has met Three people do not know each other if each of the three pairs has not met) Solution: The tedious way to do this problem is to write out all possible acquaintance assignments for 6 people We now describe a more ef cient, and more satisfying, strategy Call one of the people Bob There are ve others Either Bob knows three of them, or he does not know three of them Say that Bob knows three of the others If any two of those three are acquainted, then those two and Bob form a mutually acquainted threesome If no two of those three know each other, then those three are a mutually unacquainted threesome Now suppose that Bob does not know three of the others If any two of those three are unacquainted, then those two and Bob form an unacquainted threesome If all pairs among the three are instead acquainted, then those three form a mutually acquainted threesome We have covered all possibilities, and in every instance come up either with a mutually acquainted threesome or a mutually unacquainted threesome That ends the proof It may be worth knowing that ve people is insuf cient to guarantee either a mutually acquainted threesome or a mutually unacquainted threesome We leave it to the reader to provide a suitable counterexample It is quite dif cult to determine the minimal number of people to solve the problem when threesome is replaced by foursome When foursome is replaced by ve people, the problem is considered to be grossly intractable This problem is a simple example from the mathematical subject known as Ramsey theory
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