Download MTH4110/MTH4210 Mathematical Structures

MTH4110/MTH4210
Mathematical Structures
Solutions to Assignment 7
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 (∗) Let f : N → N be given by f (n) = n + 1. Convince yourself that f is not
surjective and explain what is wrong with each of the following answers taken from
last year’s mid-term test. Your task is not just to say what was meant, but to try to
make some helpful comment which will clear up the misunderstanding that led to
each statement.
(a) f is not surjective, since not every input has an output;
(b) f is not surjective, since f (1) has no input;
(c) f is not surjective, since it is not defined for f (0);
(d) f is not surjective, since not every f (n) ∈ N has an n ∈ N such that f (n) =
n + 1.
This exercise is based on last year’s mid-term test, which asked (amongst other
things) whether the function f given above was surjective or not. Almost everybody
spotted that the function was not surjective, but quite a number had problems explaining why. The wrong statements (a)-(d) above are actual answers I found in last year’s
test.
This exercise is not really a mathematical question; there is no right answer. But
thinking about it might help you with your own exams in future. When you have written an answer, read it through, and check for these kinds of inaccuracies and inclarities.
(a) The property that every input has an output is one of the defining characteristics
of a function, and not that of a surjective function. Or put differently, the rule n 7→ n+1
unambiguously associates the output n + 1 ∈ N with the input n ∈ N, so it is not true
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that not every input has an output, and, in particular, this cannot be the reason why f
is not surjective.
(b) At the heart of this is a common confusion, namely mixing up input and output
of a function. Using the definition of f , the assertion “ f (1) has no input” is the same
as “2 has no input”, which is false, since the number 2 in the codomain of f does
indeed have an input that produces it, namely 1. What was presumably meant was that
“ f is not surjective, since 1 has no input”, which is a mildly sloppy way of saying
f is not surjective since 1 is not in the range of f
which would have been correct.
(c) This is similar to the confusion in (b) above but ads another one into the mix.
What was presumably meant was that f is not surjective since f is not defined at 0, or
that 0 is not in the domain of f , the idea being that adding 0 to the domain of f would
have produced a surjective function. The question, however, was not how to turn f
into a surjective function, but to explain why f was not surjective in the first instance.
(d) This is again an instance of confusing input and output of a function. The
intention was presumably to write that f is not surjective, since not every m in the
codomain N has an n in the domain N, such that f (n) = m.
2 (†) The function F : R → R is defined by
x
if x ∈ Q,
F(x) =
2x if x ∈
/ Q.
(a) Is F really a function? Why?
(b) Is it injective? surjective?
√
(c) Calculate F(1.41), F(1.414), F(1.4142), F(1.41421), F( 2).
(d) Is F a continuous function?
(a) Yes, it is a function. Don’t be confused by the notation into thinking that there
are two different outputs for each input. When a real number x is input, the black box
decides whether x is rational or not; if it is rational, it is passed straight to the output,
but if it is irrational, then it is doubled first.
(b) Suppose that x1 6= x2 . Could we have F(x1 ) = F(x2 )? We look at three cases:
Case 1: x1 and x2 are both rational. Then
F(x1 ) = x1 6= x2 = F(x2 ).
Case 2: x1 and x2 are both irrational. Then
F(x1 ) = 2x1 6= 2x2 = F(x2 ).
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Case 3: One is rational and the other irrational; without loss of generality, x1 is
rational and x2 irrational. [What does the phrase “Without loss of generality”
mean?] Then F(x1 ) = x1 , which is rational; and F(x2 ) = 2x2 , which is irrational
(see below). So F(x1 ) 6= F(x2 ).
In all cases, F(x1 ) 6= F(x2 ). So F is injective.
Does every real number y occur as a value of F? If y is rational, then F(y) = y;
if y is irrational, then so is y/2 (again see below), so F(y/2) = 2 · y/2 = y. So F is
surjective.
We used twice the fact that, if 2x is irrational, then x is irrational. (In the second
case, put x = y/2.) It is easiest to prove this in the contrapositive form, which is logically the same: if x is rational, then 2x is rational. (We will discuss the contrapositive
in Chapter 9 of the notes.) Now suppose that x = m/n, where m and n are integers
with n 6= 0. Then 2x = 2m/n. So the result is proved.
(c) 1.41 = 141/100 is rational, so F(1.41) = 1.41.
√ = 1.414,
√
√ Similarly F(1.414)
F(1.4142) = 1.4142, F(1.41421) = 1.41421. But 2 is irrational, so F( 2) = 2 2.
√
(d) There are rational numbers which get arbitrarily close to 2, such as the truncations of the decimal expansion (as indicated in part (c)). For√any such
√ rational
number x, we have F(x) = x, which will be about 1.4 . . .. But F( 2) = 2 2, which
is about 2.8 . . .. So the function is not continuous.
This is an explanation, not a mathematical proof. Can you take the definition of
a continuous function (from the Calculus I module) and give a formal proof that our
function F is not continuous?
3 (†) The logarithm to base 10 of a number x, written log10 x, is the number y such
that 10y = x. Show that there do not exist natural numbers m and n such that
log10 2 =
m
.
n
Suppose that log10 2 is rational, say log10 2 = m/n, where m and n are integers.
As usual we may cancel common factors and assume that gcd(m, n) = 1. Now, by
definition, this means that
10m/n = 2 .
Raise both sides of the equation to the nth power:
10m = 2n .
Now we have a contradiction, since the left-hand side is divisible by the prime 5,
whereas the right-hand side is not (as it is a power of 2).
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4 (†) Non-mathematicians often have a lot of trouble with the assertion
0.999 . . . = 1.
(a) Is the assertion true or false?
(b) Someone taking the opposite view to yours accosts you at a party. What
argument would you use to convince them that you are correct?
(a) The assertion is true: using the formula for the sum of a geometric series,
0.999 . . . =
9
9
9/10
9
+
+
+··· =
= 1.
10 100 1000
1 − 1/10
[The sum of the series a + ar + ar2 + · · · is a/(1 − r) if |r| < 1. In our case a = 9/10
and r = 1/10.]
It is important to recall what this means. The sum of an infinite series is a limit; in
this case, the sum of any finite number of terms will fall slightly short of the limit, but
the difference gets smaller as the number of terms increases. You will see this in more
detail in Calculus.
(b) There are various methods you could try. Here are a couple.
(i) If 0.999 . . . < 1, then there should be a number lying between them, that is,
bigger than 0.999 . . . and smaller than 1. Can you show me such a number?
(ii) You agree that 0.333 . . . = 1/3, yes? (Divide 1 by 3, using long division, to
see this.) Now multiply both sides by 3.
5 (From the summer exam 2016)
(a) Let n be a natural number. Show that n3 is even if and only if n is even.
√
(b) Prove that 3 2 is irrational.
(a) We start with the ‘if’ clause. Suppose that n is even. Then n = 2k for some
k ∈ N. Then n3 = 8k3 , so n3 is even. For the converse, it suffices to show that if n is
odd, then n3 is odd. Suppose now that n is odd, that is, n = 2k +1 for some k ∈ N∪{0}.
Then
n3 = (2k + 1)3 = 8k3 + 12k2 + 6k + 1 = 2(4k3 + 6k2 + 3k) + 1 ,
so n3 is odd.
√
(b) Suppose to the contrary that 3 2 is rational. Then we can write
√
m
3
2= ,
n
4
where m, n ∈ Z with n 6= 0 and gcd(m, n) = 1. Raising both sides of the equation above
to the power three and multiplying through by n3 , we see that
2n3 = m3 ,
(1)
so m3 is even. By (a) it follows that m is even. Thus m = 2k for some k ∈ Z. Equation
(1) now implies that
n3 = 4m3 ,
so n3 is even. Again by (a) it follows that n is even. √Thus both m and n are even,
contradicting our assumption that gcd(a, b) = 1. Thus 3 2 cannot be rational.
OFB 18/11/16
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