Download Solutions to assignment 5

MTH4110/MTH4210
Mathematical Structures
Solutions to assignment 5
Autumn 2016
The solutions to homework problems don’t just give you the answer; they do two
other things. It is your job to disentangle these!
(a) They give some pointers on how to attack an unfamiliar question.
(b) They give specimens of how to write out solutions to maths problems.
1 (∗) Which of the following relations R on sets X are (i) reflexive, (ii) symmetric,
(iii) transitive? Give proofs or counterexamples.
(a) X = Q, a R b if and only if ab > 0.
(b) X = P(R), a R b if and only if a ⊂ b.
(c) X = Z, a R b if and only if a − b is even.
(d) X = N, a R b if and only if a + b is a multiple of 5.
(e) X is the set of countries of Europe, a R b if and only if a and b have a common
border.
In the preceding question, in each case where R is an equivalence relation, describe
the equivalence classes of R.
Remember that to prove that one of these properties holds, we need a general
argument; but to show it does not hold, a single example where it fails is enough.
(a)
(i) Not reflexive, since 02 is not strictly positive.
(ii) Symmetric, since if ab > 0, then ba > 0 as well.
(iii) Transitive. First we observe that ab > 0 if and only if both a and b are
non-zero and have the same sign. Suppose now that ab > 0 and bc > 0.
We need to show that ac > 0. This, however, is the case, since, if b > 0
then a > 0 and c > 0,so ac > 0, while if b < 0 then a < 0 and c < 0, so
again ac > 0.
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(b)
(i) Not reflexive, since, for example, R is not a proper subset of R.
(ii) Not symmetric, since, for example, 0/ is a proper subset of R, but R is not
a proper subset of 0.
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(iii) Transitive, since if a is a proper subset of b and b is a proper subset of c,
then a is a proper subset of c.
(c)
(i) Reflexive, since a − a is even for every integer a.
(ii) Symmetric, since if a − b is even then so is b − a.
(iii) Transitive. In order to see this, note that if a − b is even and b − c is even
then a − c = (a − b) + (b − c) is a sum of two even integers, so must itself
be even.
(d)
(i) Not reflexive, since, for example, 2 + 2 is not a multiple of 5.
(ii) Symmetric, since if a + b is a multiple of 5, then so is b + a.
(iii) Not transitive, since, for example, 2 + 3 is a multiple of 5, and so is 3 + 7,
but 2 + 7 is not.
(e)
(i) Does every country share a border with itself? This is a question of definition; I don’t think there is an obvious answer.
(ii) Symmetric, since if a and b share a border, so clearly do b and a.
(iii) Not transitive, since, for example, France and Spain share a border, so do
Spain and Portugal, but France and Portugal do not.
The only case where R is an equivalence relation is (c). Noting that we have a R b if
either both a and b are even or if both a and b are odd, it follows that the equivalence
class containing an even integer is the set of all even integers, while the equivalence
class containing an odd integer is the set of all odd integers.
2 (†) Consider the following argument:
False proposition If a relation is symmetric and transitive then it is reflexive.
Proof Let R be a symmetric and transitive relation. Take elements x, y satisfying
x R y. Then y R x (since R is symmetric), and so x R x (since R is transitive; put z = x
in the transitive law). So R is reflexive.
(a) Say what is wrong with this argument.
(b) Give a counterexample to the false proposition.
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The mistake is quite hard to spot. Here is a small example. Let R be the relation
on the set {a, b, c, d}, in which every pair of a, b, c are related but d is not related to
anything else (even itself). The relation is symmetric and transitive, but not reflexive
since d is not related to itself.
A simpler but more spectacular example consists of a set in which no pair of elements is related. Each of the symmetric and transitive laws start off with at least one
pair satisfying the relation; so if no pairs satisfy R, then the symmetric and transitive
laws hold “vacuously”. (For example, a failure of the symmetric law would mean that
there are elements x and y with x R y but not y R x.
The mistake occurs when we say “Take elements x, y satisfying x R y.” There may
be no such pair.
Remark The argument actually shows that, if R is symmetric and transitive, and if
x is related to any element at all, then it is related to itself.
3 (†) Prove that 53n + 2n+1 is a multiple of 3 for every natural number n.
As this is a statement about natural numbers we should stand a good chance proving it by induction. For a natural number n, let P(n) be the statement that 53n + 2n+1
is a multiple of 3.
We start by observing that P(1) is true, since 53 + 22 = 129 = 3 · 43 is indeed a
multiple of 3.
Next we show that P(n) implies P(n + 1). To this end, suppose that P(n) is true for
some natural number n, that is, 53n + 2n+1 is a multiple of 3. Thus, for some natural
number k, we have
53n + 2n+1 = 3k
(1)
Now
53(n+1) + 2(n+1)+1 = 53 · 53n + 2 · 2n+1
= 53 · 53n + 2(3k − 53n )
3n
= 123 · 5 + 6k
(using exponent laws)
(using (1))
(simplifying)
3n
= 3(41 · 5 + 2k)
(factorising)
is again a multiple of 3, so P(n + 1) holds.
Thus, by the Principle of Induction, the statement P(n) is true for all natural numbers n.
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4 (From the summer exam 2014)
(a) Suppose that the relation R on the set N of natural numbers is defined by x R y
if and only if x + y is even. Is R reflexive? Is it symmetric? Is it transitive?
(b) You are given that the relation S on the set {1, 2, 3} is an equivalence relation
and has equivalence classes {1, 2} and {3}. Write down all the pairs (a, b)
for which a S b holds.
(a) R is reflexive, since for every x ∈ N the number x + x = 2x is even, so x R x
holds for every x ∈ N. Furthermore, R is symmetric, since if x R y holds for some
x, y ∈ N, then x + y is even, so y + x = x + y is even as well, so y R x holds. Finally, R
is also transitive. In order to see this, suppose that x R y and y R z hold. Then there are
integers k and l, such that
x + y = 2k ,
y + z = 2l .
(2)
(3)
Using the above equations we see that x + z = (2k − y) + (2l − y) = 2(k + l − y), so
x + z is even, that is, x R z holds.
(b) The relation a S b holds for the following pairs (a, b):
(1, 1), (1, 2), (2, 2), (2, 1), (3, 3) .
In order to see this, note that since S is reflexive, me know that 1 S 1, 2 S 2, and 3 S 3
hold. Finally, since 1 and 2 belong to the same equivalence class, we know that 1 S 2
and 2 S 1 must hold as well.
OFB 26/10/16
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