Research Methods in Mathematics Homework 4 solutions

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1 Research Methods in Mathematics Homework 4 solutions T. PERUTZ (1) Solution. (a) Since x 2 = 2, we have (p/q) 2 = 2, so p 2 = 2q 2. By definition, an integer is even if it is twice another integer. Since it is twice q 2, the integer p 2 is even. (b) The square of an odd number is again odd (because any odd integer can be written as (2n + 1) with n an integer, and (2n + 1) 2 = 2(2n 2 + 2n) + 1). We know by (a) that p 2 is even. Therefore p cannot be odd, and so it must be even. (c) Since p is even, we can write p = 2r for some integer r. Then p 2 = (2r) 2 = 4r 2. So 4r 2 = 2q 2, and therefore 2r 2 = q 2, which shows that q 2 is even. (d) Since q 2 is even by (c), q must be even by the same argument as in (b). (e) We want to prove that there is no rational x such that x 2 = 2. To do this, we suppose on the contrary that there is such an x, and show that this leads to an absurdity (a self-contradictory statement). From that we deduce that no such x exists. Now, every non-zero rational number x can be written as p/q where p and q have no common factor > 1. (For the purposes of this question, it s reasonable to take this for granted.) So let s choose p and q so that they have no common factor. We prove in (b) that p is necessarily even. We proved in (d) that q is even. So both are divisible by 2, contradicting our hypothesis that they have no common factor > 1. This is the contradiction. For completeness sake, let s prove that a rational number p/q 0 can be rewritten as p /q where p and q have no common factor > 1. Consider max( p, q ), that is, p or q, whichever is larger. Let P(m) be the statement: every non-zero rational p/q with max( p, q ) m is equal to some p /q where p and q have no common factor > 1. Proof by induction: for P(1), note that if max( p, q ) 1 then it must equal 1 since p/q 0. But then then p/q = ±1/1, and the numbers ±1 and ±1 have no common factor > 1. For the inductive step, assume P(m), and consider P(m + 1). So take a p/q where max( p, q ) m + 1. If p/q have no common factor > 1, there s nothing to prove. If they have a common factor d > 1 then write p = rd and q = sd. Then p/q = r/s, and max( r, s ) m. By P(m), we can write r/s = p /q where p and q have no common factor > 1. This proves P(m + 1) and completes the induction.

2 2 T. Perutz (2) (a) Refer to your class notes. (b) We wish to find ɛ > 0 such that (x ɛ) 2 > 2. If x 0 then ɛ = 1 will work (since then (x ɛ) 2 > x 2 > 2). So from now on we assume that x > 0. The condition (x ɛ) 2 > 2 is equivalent to Equivalently again, x 2 2ɛx + ɛ 2 > 2. 2ɛx ɛ 2 < x 2 2. This is our target. We know that x 2 > 2 (so x 2 2 > 0). Whatever ɛ we choose, we ll have 2ɛx ɛ 2 < 2ɛx. So it will be sufficient to choose ɛ > 0 such that 2ɛx < x 2 2. So let s take Then ɛ = x2 2. 3x 2ɛx = 2x x2 2 = 2 3x 3 (x2 2) < x 2 2, as required. Notice that this problem is slightly simpler than (a), because the sign of the ɛ 2 term worked in our favor here. (3) Solution. (a) The set of integers-squared has no upper bound. Indeed, if x is an integer then x 2 x. So an upper bound U would be at least as large as any integer. In particular, U n for every natural number n. But there is no such rational number. Indeed, such a U would evidently have to be positive. But if U is a positive rational number then the integer part of U + 1 is a natural number strictly larger than U. (b) The set of reciprocals of natural numbers is bounded above by 1. Indeed, every natural number n satisfies n 1. It follows that every reciprocal 1/n satisfies 1/n 1 (because if 1/n > 1 then, by one of the properties of the ordered field Q, we have n(1/n) > n, i.e. 1 > n). (c) The inequality (1 + x 2 ) 1 3 is equivalent to x 2 (here we use the fact that 1 + x 2 > 0), hence to 3x 2 2. But every rational x satisfies 3x 2 0, hence 3x 2 2. Hence this set is Q, which has no upper bound. (d) This inequality rearranges to x 2, hence to 2 x 2. An upper bound U is U = 100. Indeed, if x > 100 then x 2 > 100, so x does not satisfy the inequality. Now for lower bounds: (a) This set has lower bound 0, since the square of an integer is non-negative. (b) This set has has lower bound 0, since the reciprocal of a natural number is positive. (c) This set has no lower bound, since it is Q.

3 Research Methods in Mathematics Homework 4 solutions 3 (d) This set is bounded below by 100, since if x < 100 then x 2 > 100, so x does not satisfy the defining inequality. (4) Solution. Least upper bounds: (a) has no upper bound, hence no least upper bound. (b) has least upper bound 1. Indeed, 1 is an upper bound; and since 1 is in the set, every upper bound x satisfies x 1. (c) This set has no upper bound, hence no least upper bound. (d) This is the set of rationals x such that x 2 < 2. As shown in class and in q.2, any least upper bound u must satisfy u 2 = 2. But by q.1, no such rational number u exists. Hence there s no least upper bound. Greatest lower bounds: (a) has greatest lower bound 0. Indeed, every member in the set is 0, and 0 is in the set so any lower bound must be 0. (b) The greatest lower bound is 0. Indeed, 0 is a lower bound, and if l were another lower bound and l > 0, we could write l = p/q for natural numbers p and q. But then l = p q 1 q > 1 2q, so 1/(2q) is a member of the set smaller than l, contradicting the assumption that l is a lower bound. (c) This set had no lower bound, hence has no greatest lower bound. (d) This set has no greatest lower bound because if l were a greatest lower bound, we would have l 2 = 2 (same argument as for least upper bounds). There s no such rational l by question 2. (5) Solution. First, the diagram in Figure 1 illustrates the five ways to add four numbers. Each tree picture is read top to bottom. We start with four inputs (branches of the tree). Each Y point represent an addition. We end up with one output. There are 14 ways to add five numbers a, b, c, d, e. In diagrammatic form, they are as in Figure 2. The first five are of the form ( ) + e, where ( ) is some way of adding a, b, c and d. The next two take the form ( ) + (d + e) where now ( ) adds a, b and c; the two after that, (a + b) + ( ); and the last five, a + ( ). We have caught all the possibilities here because in any procedure for adding a, b, c, d and e, the last addition we perform will add the sum of the first k inputs (calculated in some fashion) to the sum of the last n k inputs (also calculated in some fashion). Here k could be 4, 3, 2 or 1. We just described what happens in each of these cases.

4 4 T. Perutz a b c d ((a+b)+c)+d Figure 1: Four inputs Extra credit. Suppose we are carrying out an addition a a n+1. The number C n+1 represents the number of ways of doing this. The first step is to use brackets to divide the summands into two groups: (a a j ) + (a j a n+1 ). There are C j ways to carry out the addition a a j, and for each of those, there are C n j ways to carry out the addition a j a n+1. Hence C n+1 = C 0 C n + C 1 C n C n C 0. We can also this express this in terms of the tree diagrams. Starting at the base of a tree, the tree branches into two parts, which we can think of as sub-trees. We look at all the possibilities for those sub-trees. You can see this in the picture for C 4, which shows the relation We have C 4 = C 5 = = 42.

5 Research Methods in Mathematics Homework 4 solutions 5 Figure 2: Five inputs