to Answers Second Problem Set
Math E311 Spring 2008
1) First, critique the following proof by cases (i.e. is it a valid proof? are there holes in the logic? be sure to explain your answer carefully). Theorem: If x is any real number then x2 x. Proof: Assume x is any real number. Then we prove this by consideration of cases: Case 1: Assume x 0. Then x2 0, while x 0. Therefore x2 x. Case 2: Assume x 1. Then multiply both sides of this inequality by x to obtain 2 x x. Since both cases lead to the desired conclusion then the result is proved. Answer: No, this is not a valid proof there is a missing case which leads to a major problem (i.e. the theorem as stated is not true). The "proof" does not consider 0 < x < 1 (this would not have been an issue if the proof had been restricted to integers, but in this case it specifically references real numbers). When 0 < x < 1, then following the Case 2 lead, multiplying both sides of x < 1 by x (which is positive, so the inequality does not change direction), then in fact x2 < x exactly when 0 < x < 1. 2) And one more proof to critique (again, is it a valid proof? are there holes in the logic? be sure to explain your answer carefully). Theorem: If N is a prime integer then 2N 1 is prime. Proof: Proof by contradiction. Suppose N is not prime, i.e. N = A x B for integers A and B. Let M = 2A So 2N 1 = 2AB 1 = (2A)B 1 = MB 1 . Now MB 1 = (M 1) (1 + M + M2 + ... + MB-1), MB 1 is divisible by (M 1). Therefore 2N 1 (which equals MB 1) must have a factor and so 2N 1 is not prime. This completes the proof. Answer: Again, this is not a valid proof. First off, it is not even a proof by contradiction. To prove an "If P then Q" statement by contradiction, one would assume the negation of the implication and show that this leads to a contradiction. Since "If P then Q" is equivalent to "Not P or Q" then its negation is equivalent to "P and Not Q" Here, this would translate into assuming that N is a prime integer and then showing that assuming that 2N 1 is not prime leads to a contradiction. This is not possible to do as the original implication is itself false(!), as an easy counterexample shows: 211 1 = 2047 is not prime (equals 23 times 89), even though 11 is prime. Here, the "proof" actually (almost) proves not that "If P then Q" but "If Not P then Not Q," which is an entirely different implication (check the truth tables for these two implications to see that they are not propositionally equivalent). I say "almost" as even this implication is not quite proved there's a mild hole in the logic. When it's assumed that N is not prime and that N = A x B, it needs to be stated that it is assumed that N is a positive integer and that A and B are also positive and that neither one is equal to 1. In
the current version of the proof assume that N = 6. Then N = 1 x 6 (i.e. take A = 1 and B = 6). Eventually the proof then simply shows that 26 1 (which equals 63) is equal to 1 times 63, which doesn't show that 63 is not prime. 3) Next, create a (real!) proof - use any method you want to prove that for any pair of integers x and y, the expression 3(x+ 1) + (x + y)2 + 3y + 2 is never equal to a power of 2 (i.e. 2 raised to an integer). Answer: There were many creative approaches to this problem. Many people considered four separate cases for the various combinations of odd and even for x and y. In fact it's possible to avoid having to do four separate cases by the following. Rewriting the expression as: 3x+ 3 + (x + y)2 + 3y + 2 = 3x + 3y + (x + y)2 + 5 = (x + y)2 + 3(x + y) + 5 which equals M2 + 3M + 5, where M = (x + y) Now M2 + 3M + 5 = M(M+3) + 5. If M is even then M+3 is odd, so M(M+3) is even. If M is odd then M+3 is even, so again M(M+3) is even. In either case M2 + 3M + 5 = M(M+3) + 5 is even + odd which equals odd. Thus when n > 0 is a positive integer then 2n is even, and so since M2 + 3M + 5 is odd, then 3(x+ 1) + (x + y)2 + 3y + 2 cannot equal a positive power of 2. When n < 0 then 2n is not even an integer and so since 3x+ 3 + (x + y)2 + 3y + 2 is an integer for integer valued x and y then it is not possible for 3x+ 3 + (x + y)2 + 3y + 2 to be a (negative) integer power of 2. Finally, when n = 0, then 20 = 1, and we need to show that 3x+ 3 + (x + y)2 + 3y + 2 cannot equal 1 for integer valued x and y. Again rewrite this expression as M2 + 3M + 5 where M = x + y. Checking that M2 + 3M + 5 = 1 has no integer roots clears this last case (solving M2 + 3M + 4 = 0 leads to two complex solutions for M using the quadratic equation). This then completes the proof.
4) Consider the series 1/2+ 1/6 + 1/12 + ... + 1/n(n+1) (where n is a positive integer) (a) Calculate the sum of this series for several small values of n, and use these results to form a conjecture for the sum of this series in general. (b) Use the fact that 1/n(n+1) = 1/n 1/(n+1) to find a trick for deriving the sum of this series without using mathematical induction. Answer:
(a) It's relatively easy to see that one ends up with 1/2+ 1/6 + 1/12 + ... + 1/n(n+1) = n/ (n+1) in each case. (b) So now, using the observation given in the question, we see that 1/2+ 1/6 + 1/12 + ... + 1/n(n+1) = (1 1/2) + (1/2 1/3) + (1/3 1/4) + (1/4 1/5) + ... + [(1/n) - (1/(n+1))] This is where the "magic" occurs notice that all (almost!) the terms beautifully cancel each other out. For instance the "-1/2" term is followed immediately by a "+ 1/2" term and then the "-1/3" is followed by a "+1/3" term. The only terms that escape this are the first (equaling 1) and last (equaling 1/(n+1)) terms, and so the entire sum collapses to simply 1 (1/(n+1)) which equals n/(n+1). 5) And now a couple of pigeonhole principle questions. First, show that given any 11 integers from 1 to 20, two of the integers have an odd sum. Answer: Here there are 10 odd integers from 1 to 20, and 10 even integers. Thus any set of 11 integers will necessarily have at least one odd and one even integer (it's impossible to fit 11 "pigeons" into either 10 odd or 10 even "pigeonholes.") Thus, there are two integers whose sum is odd (odd plus even yielding an odd sum). 6) Next show that given any 11 integers from 1 to 20 then in the set of 11 integers there must be one integer that divides another integer in the set (for instance in the set {2, 3, 5, 7, 9, 11, 13, 17, 18, 19, 20} 18 is divisible by 3 (and 2) (and 20 is divisible by and 5 by 2 as well). Answer: A more involved one, and one that people tried a variety of approaches on. The trick is to find 10 pigeonholes for the 11 pigeons (the set of 11 integers). This is hard to do if one starts going down the route of trying out many different cases it's not clear that all possible combinations will have been checked. The easiest approach is to create sets of integers any two of which divide each other. First off, notice that the subset of 11 integers cannot include the number 1, for it divides all the other 19 numbers. Next, create a subset of even integers {2, 4, 8, 16} (note not {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}) In this {2, 4, 8, 16} set it is impossible to take any two members of this set without creating a pair of integers that divide each other. So, in the set of 11 integers there can be at most one integer from this {2, 4, 8, 16} set. Next, consider the following additional sets, starting with larger and larger (odd integers) simply those integers that haven't shown up in previous subsets yet: {3, 6, 12} then {5, 10, 20} then {7, 14} then {9, 18}
This makes five sets so far, again, no more than one integer can be taken from any of these five subsets without creating a pair of integers that divide each other. Finally there are the remaining 5 integers 11, 13, 15, 17, and 19. So consider the following 10 sets of integers: {3, 6, 12}, {5, 10, 20}, {7, 14}, {9, 18}, {11}, {13}, {15}, {17}, and {19}. Since there are only 10 subsets, then any set of 11 integers from 1 to 20 must contain 2 from one of the sets with 2 or more elements, in which case those two integers divide each other (the smaller dividing evenly into the larger, that is).
7) Finishing off the last problem I showed you in class (with a slight twist - subbing in 17 for 2001 for now) - show that there exists an integer U so that 17 x U is a number composed only of the digit "1" (e.g. 11,111,111,111). Note - you don't need to find U, just prove that it must exist. Big hint(s): use the pigeonhole principle where you consider the infinite set of numbers 1, 11, 111, 1111, 11111, 111111, ... and consider their remainders when you divide them by 17 (and so there are 17 pigeonholes - the 17 possible remainders 0, 1, 2, ..., 16). So at least two such numbers share the same remainder, so their difference is divisible by 17... now what does their difference look like? This same approach can be generalized to show that for any prime P there is an integer M such that P x M is a number composed only of the digit 1. Answer: Here there are an infinite number of pigeons the infinite set of integers that are composed only of 1's and a finite number of pigeons the 17 remainders {0, 1, 2, ..., 15, 16} that are possible upon division by 17. So two numbers composed of just 1's leave the same remainder with respect to 17 (an equivalent statement is to say that the two numbers leave the same remainder "modulo" 17). Next, we need some notation there are many ways to go about this, but for the sake of simplicity let's let *A* stand for a number with A 1's in a row (e.g. *3* will stand for 111). Then suppose *A* and *B* leave the same remainder modulo 17, i.e. *A* = 17M + R, and *B* = 17N + R for some integers M and N, and some integer R with 0 < R < 17 (note if R = 0 then we're done as both *A* and *B* are already divisible by 17). This means that *A* - *B* = 17M + R (17N + R) = 17(M-N) (assuming that *A* is the larger of the two numbers). So *A* - *B* = 17V for some integer V. Now comes the critical part... notice that when one subtracts a number composed of 1's from another number composed of 1's the result is a number that starts with 1's and ends with 0's. For instance 111111 1111 = 110000. Using this observation then it's easy to check that *A* - *B* equals a number starting with (A-B) 1's and ending with B 0's. This is the same as a number composed of (A-B) 1's times 10B (thin...