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Unformatted text preview: 3 8 Chapter 1 The Foundations: Logic and Proofs SECTION 1.7 Proof Methods and Strategy 11. The preamble to the solutions for Section 1.6 applies here as well, so you might want to reread it at this time.
In addition, the section near the back of this Guide, entitled “A Guide to Proof— 1l’Vriting, ”' provides an excellent
tutorial, with many additional examples. Don’t forget to take advantage of the many additional resources on
the website for this text, as well. If you are majoring in mathematics, then proofs are the bread and butter of your field. Most likely you
will take a. course devoted entirely to learning how to read and write proofs, using one of the many textbooks
available on this subject. For a review of many of them (as well as reviews of hundreds of mathematics books),
see this site provided by the Mathematical Association ofAmerica: http://wuw.maa.org/revieus/. . \Ve give an exhaustive pr00f~ just check the entire domain. For n : 1 we have 12 + 1 : 2 2 2 = 21. For n. = 2 we have 22+1= 5 2 4 : 22. For n : 3 we have 32+1: 10 .2 8 2 '23. For 'n. : 4 we have
42 + 1 = 17 2 16 : 24. Notice that for n. 2 5, the inequality is no longer true. Following the hint, we consider the two cases determined by the relative sizes of and 3). First suppose that
a 2 1;. Then by definition max(:r:,y) = a: and min(:r:,y) : '9. Therefore in this case max($,y) + min($,y) =
a" +y, exactly as desired. For the second (and final) case, suppose that 55 < 7;. Then. max(.7;,y) : y and
min(.r,y) : gr. Therefore in this case ma.x(33,y) + min(.r, y) : y + x = :L' + 9*, again the desired conclusion.
Hence in all cases, the equality holds. There are several cases to consider. If :t and y are both nonnegative, then rrl+ ly : x+y = Similarly,
if both are negative, then + = (—x) + (—y) = 7(33 + y) = )1: + y, since a: + y is negative in this case.
The complication (and strict inequality) comes if one of the variables is nonuegative and the other is negative.
By the symmetry of the roles of a: and 3; here (strictly speaking, by the commutativity of addition), we can
assume without loss of generality that it is x that is nonnegative and y that is negative. 80 we have a: 2 0
and y < 0. Now there are two subcases to consider within this case, depending on the relative sizes of the nonnegative
numbers x and —y. First suppose that .1: 2 Hy. Then a: + y 2 0. Therefore + = a‘ + y, and this
quantity is a nonnegative number smaller than .7: (since y is negative). On the other hand + : :17 + is a positive number bigger than Therefore we have + < :t < + , as desired. Finally, consider the possibility that a: < 7y. Then 33 + yl : #(a: + y) : + is a positive
number less than or equal to —y (since —x is nonpositive). On the other hand + Iyl = lxl + (739) is a
positive number greater than or equal to —y. Therefore we have cc + S Hg 3 + y, as desired. We want to find consecutive squares that are far apart. If n. is large enough, then (11+ 1)2 will be much
bigger than 712, and that will do it. Let’s take 12. = 100. Then 1002 : 10000 and 1012 = 10201, so the 201
consecutive numbers 10001, 10002, .. ., 10200 are not perfect Squares. The ﬁrst 100 of these will satisfy the
requirements of this exercise. Our proof was constructive, since we actually exhibited the numbers. . We try some small numbers and discover that 8 = 23 and 9 2 32. In fact, this is the only solution, but the proof of this fact is not trivial. One way to solve this is the following nonconstructive proof. Let 3r: 2 2 (rational) and y = J2 (irrational).
If 3:9 : 2ﬁ is irrational, we are done. If not, then let a: : 2‘5 and y : J2/4; 1' is rational by assumption,
and y is irrational (if it were rational, then \/2 would be rational). But now my 2 (ﬂip/5“ = $506)” 2
21/2 = J2, which is irrational, as desired. Section 1.7 13. 15. 17. 19. 21. 23. 25. Proof Methods and Strategy 39 a) This statement asserts the existence of :2: with a certain property. If we let y = :r, then we see that P(x)
is true. If y is anything other than (B, then is not true. Thus a: is the unique element that makes P
true. b) The ﬁrst clause here says that there is an element that makes P true. The second clause says that. whenever
two elements both make P true, they are in fact the same element. Together this says that P is satisﬁed by
exactly one element. c) This statement asserts the existence of an a; that makes P true and has the further property that whenever
we ﬁnd an element that makes P true, that element is a. hi other words. a; is the unique element that makes P
true. Note that this is essentially the same as the deﬁnition given in the text, except. that the ﬁnal conditional
statement has been replaced by its contrapositive. The equation la. —c 2 lb icl is equivalent to the disjunction of two equations: a. — c = b — c or a. 7c = ib+c.
The ﬁrst of these is equivalent to a 2 b, which contradicts the assumptions made in this problem, so the
original equation is equivalent to a — c = —b + c. By adding 6 + c to both sides and dividing by 2, we see
that this equation is equivalent to c 2 (e. + b)/2. Thus there is a unique solution. Furthermore, this c is an
integer, because the sum of the odd integers a. and b is even. We are being asked to solve in. : i 2) + (k + for is. Using the usual, reversible, rules of algebra, we see
that this equation is equivalent to k = (n — l)/2. In other words, this is the one and only value of k that
makes our equation true. Since n is odd, 1:. — 1 is even, so is is an integer. If :2; is itself an integer, then we can take n = a: and e = 0. No other solution is possible in this case, since
if the integer n is greater than 3:, then it is at least 32+ 1, which would make 6 2 1. If 1 is not an integer,
then round it up to the next integer, and call that integer n. We let 6 = n. — 3:. Clearly D S e < 1, this is the
only 6 that will work with this n, and n. cannot be any larger, since 6 is constrained to be less than 1. If = 5 and y = 8, then the harmonic mean is 258/(5+8) 2: 6.15, and the geometric mean is z 6.32.
If :1: : 10 and y = 100, then the harmonic mean is 2  10  100/(10 + 100) a: 18.18, and the geometric mean
is m 31.62. We conjecture that the harmonic mean of a: and y is always less than their geometric
mean if :5 and y are distinct positive real numbers (clearly if (1'? = y then both means are this common value).
So we want to verify the inequality icy/(33 + y) < \/a:_y. Multiplying both sides by (:L' + y) / (2 \/.?:_y) gives us
the equivalent inequality < (32 + y)/ 2, which is proved in Example 14. The key point here is that the parity (cddness or evenness) of the sum of the numbers written on the board
never changes. If 3' and k are both even or both odd, then their sum and their difference are both even, and
we are replacing the even sum 3' +14: by the even difference I j — k, leaving the parity of the total unchanged. If
j and is have different parities, then erasing them changes the parity of the total, but their difference j —k is
odd, so adding this difference restores the parity of the total. Therefore the integer we end up with at the end
of the process must have the same parity as 1 + ‘2 +    + It is easy to compute this sum. If we add the
ﬁrst and last terms we get 2n + 1; if we add the second and nextto—last terms we get 2 + (2n.7 1) = 271+ 1 ;
and so on. In all we get n sums of 2n + 1, so the total sum is n(2n + 1). If n is odd, this is the product of
two odd numbers and therefore is odd, as desired. Without loss of generality we can assume that n is nonnegative, since the fourth power of an integer and the
fourth power of its negative are the same. To get a handle on the last digit of n, we can divide n by 10,
obtaining a quotient k and remainder l, whence n 2 1016 +1, and i is an integer between 0 and 9, inclusive.
Then we compute n4 in each of these ten cases. We get the following values, where ‘?‘? is some integer that is 40 27'. 29. 31. 33. 35. 37. Chapter 1 The Foundations: Logic and Proofs a multiple of 10, whose exact value we do not care about (10k + 0)’1 : 10000a‘1 : 10000k“ + 0 (10k + 1)"1 = 10000;:4 + t»? . r3 + r?  A»? +?'?  I: + 1
(10;; + '2)"1 = 100001;“ + ‘3? . r3 + r?  a? + r?  r: +16
(10;; + 3)" : 10000;:4 + '9? . a3 + r?  a? +?? .1; + 81
(10k + 4)"‘ : 100001;“ + ‘2? . r3 + ‘2?  a2 + r? .1; + 2.56
(10:; + 5)"1 : 10(10qu + 1?? 4:3 + 2??  r»? + '3?  r: + 525
(10k + (3)4 = 1000051 + '3? . k3 + 1??  k2 + '1’“? . k +1296
(10;; + n" : 1000051 + ‘2?  k3 + ’3‘?  a? + r? . k + 2401
(10k + 8)4 2 10000161 + 1?? . 1&3 +1??ch +12?  k + 4096
(10k + £3)“ = 100(1de + ‘3?  t3 + 7? . k2 + “r? . k + 6561 Since each coefficient indicated by ?'? is a multiple of 10, the corresponding term has no eﬁ'ect on the ones
digit of the answer. Therefore the ones digits are 0, 1, 6, 1, 6, 5, 6, 1, 6, 1, respectively, so it is always a 0, 1, 5,01‘6. Because n3 > 100 for all n > 4, we need only note that n 2 1, n. 2 ‘2, n 2 3, and n. = 4 do not satisfy
n2 + r13 = 100. Since 54 2
This means that each of 374 and y4 is at most 4‘1 2 256, so their sum is at most 512 and cannot. be 625. 625, for there to be positive integer solutions to this equation both a: and ,1; must be less than 5. We give a proof by contraposition. Assume that it is not the case that a g or b 3 or c g Then
it must be true that a > and b > 6/5 and c > Multiplying these inequalities 01' positive numbers
together we obtain (the < 2 n, which implies the negation of our hypothesis that n 2 0116. The idea is to find a small irrational number to add to the smaller of the two given rational numbers. Because
we know that \/§ is irrational, we can use a small multiple of x/E. Here is our proof: By finding a common
denominator, we can assume that the given rational numbers are a/b and c/b, where b is a positive integer
and a and c are integers with a. < c. In particular, (a + 1)/b <_I c/b. Thus :3 2 (a + éx/iﬂb is between the
two given rational numbers, because 0 < ﬂ < 2. Furthermore, a: is irrational, because if ct were rational,
then 2(brr: — a.) = ﬂ would be as well, in violation of Example 10 in Section 1.6. a) Without loss of generality, we may assume that the .1: sequence is already sorted into nondecreasing order,
since we can relabel the indices. There are only a finite number of possible orderings for the y sequence, so if we
can show that we can increase the sum (or at least keep it the same) whenever we ﬁnd 1;, and yj that are out of
order (i.e., ’t' < j but y, > 93) by switching them, then we will have shown that the sum is largest when the y
sequence is in nondecreasing order. Indeed, if we perform the swap, then we have added airy} +:Ejy,g to the sum
stilt: 1173'in _ ($1 35i)(yi 1L1); and subtracted $1y5+33jy3 . The net effect, then, is to have added 93,113: +13, y,
which is nonnegative by our ordering assumptions. b) This is similar to part (21). Again we assume that the a; sequence is already sorted into nondecreasing
order. If the 1; sequence is not in nonincreasing order, then y, < y, for some i. < j . By swapping y, and y,
We increase the Sum by 51:ij + Ciji — 335?}; — :L'j‘yj 2 (ftj — — yj), which is nonpositive by our ordering
assumptions. In each case we just have to keep applying the function ,1“ until we reach 1, where 2 31' + i if at is odd
and 2 35/2 if .1: is even. Section 1. 7 39. 41. 43. 45. 47. Proof Methods and Strategy 41 51) f(6) = 3, _f(3) : 10. f(10) : 5, :16. f(16) :87 = 4, H4) 2 2, .ft2):1. We abbreviate this
t067>37> lU—>5—=167>8—Mi—>2—>1. b) 7—122—>11—»34—>17—>52>267) 13H4U—>QUHlUii5—>16—>8—=4~+2—>l
c)17H527i26ﬁ~>13H4OH207>10—>5—>167)8H4w>2—>1 d) 21—>6¢1—>32—>16—>8—>4—>2—)] “’0 give a constructive proof. Without loss of generality, we can assume that the upper left and upper right
corners of the board are removed. We can place three dominoes horizontally to till the remainng portion of
the ﬁrst row, and we can place four dominoes horizontally in each of the other seven rows to ﬁll them. The number of squares in a rectangular board is the product of the number of squares in each row and the
number of squares in each column. \Ve are given that this number is even, so there is either an even number
of squares in each row or an even number of squares in each column. In the former case. we can tilo the board
in the obvious way by placing the dominoes horizontally, and in the latter case, we can tile the board in the obvious way by placing the dominoes vertically. we follow the suggested labeling scheme. Clearly we can rotate the board if I'iecessary to make the removed
squares be 1 and 16. Square 2 must be covered by a domino. If that domino is placed to c0ver squares 2
and 6. then the following domino placements are forced in succession: 5—9, 13—14, and 1.0711. at which point
there is no way to cover square 15. Otl‘ierwise, square 2 must be covered by a domino placed at. 2—3. Then the
following domino plaeements are forced: 4—8, 11—12, 6—7, 59, and 1014, and again there is no way to cover square 15.
Remove the two black squares adjacent to one of the white corners. and remove two white squares other
than that corner. Then no domino can cover that white corner. because neither of the squares adjacent to it remains. a) It is not. hard to find the ﬁve patterns: agree. b) It is clear that the pattern labeled 1 and the pattern labeled ‘2 will tile the checkerboard. It is harder to
ﬁnd the tiling for patterns 3 and 4, but. a little experimentation shows that it is possible. 42 Chapter 1 The Foundations: Logic and Proofs It remains to argue that pattern 5 cannot tile the checkerboard. Label the squares from 1 to 64, one row
at a time from the top. from left to right in each row. Thus square 1 is the upper left corner, and square 64 is
the lower right. Suppose we did have a tiling. By symmetry and without loss of generality, we may suppose
that the tile is positioned in the upper left corner. covering Squares 1, 2. 10, and 11. This forces a tile to be
adjacent to it on the right, covering squares 3, 4, 12, and 13. Continue in this manner and we are forced to
have a tile covering squares 6., 7, 15, and 16. This makes it impossible to cover square 8. Thus no tiling is
possible. ...
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This note was uploaded on 02/18/2012 for the course MATH 640:244 taught by Professor Ming during the Spring '09 term at Rutgers.
 Spring '09
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