Download Solutions to assignment 6 File

MTH4110/MTH4210
Mathematical Structures
Solutions to assignment 6
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 (∗) Here is a definition of the factorial function, which maps the natural numbers
to the natural numbers. (Despite the curious notation, this really is a function!)
(i) 1! = 1 .
(ii) If n > 1, then n! = n × (n − 1)! .
(a) Prove by induction that this really does define a function; that is, prove that,
for any natural number n, the rule given enables us to calculate n! .
(b) Prove by induction that, for n ∈ N,
2n−1 ≤ n! ≤ nn .
(a) Let S be the set of natural numbers n for which we have defined the function n!.
Then the two parts of the definition show that
(i) 1 ∈ S;
(ii) if n ∈ S, then n + 1 ∈ S.
By the Principle of Induction, S contains all natural numbers.
So n! is defined for all n ∈ N. This also gives us a means to calculate it:
n! = n · (n − 1)! = n · (n − 1) · (n − 2)! = · · ·
= n · (n − 1) · (n − 2) · · · 2 · 1! = n · (n − 1) · (n − 2) · · · 1,
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the product of the natural numbers from 1 to n (inclusive).
(b) Let P(n) be the statement 2n−1 ≤ n! ≤ nn .
For n = 1, we have 2n−1 = 1, n! = 1, and nn = 1; so the three numbers are all
equal and so P(1) is true.
Suppose that n > 1 and that P(n − 1) is true; that is,
2n−2 ≤ (n − 1)! ≤ (n − 1)n−1 .
We have to prove P(n). Since it consists of two inequalities, we take the two
parts separately.
2n−1 = 2 · 2n−2
≤ n · (n − 1)! (n > 1; induction hypothesis)
= n!
(definition of n!)
Note that in the second line, we said that 2 ≤ n and 2n−2 ≤ (n − 1)!, and gave
a reason for each statement. You could separate this onto two separate lines if
you prefer.
Now also
n! =
≤
≤
=
n · (n − 1)!
(definition)
n−1
n · (n − 1)
(induction hypothesis)
n−1
n·n
(n − 1 ≤ n)
nn
Thus P(n) is proved, and the induction step is complete.
So by the Principle of Induction, P(n) holds for all n.
2 (†) How many zeros occur at the end of the number 100!?
The question asks for the power of 10 which divides 100!. Now 10 = 2 × 5, and
if and 5b are the powers of 2 and 5 which divide a number N, then the number of
zeros with which N ends is the smaller of a and b. (For example, 6000 = 24 · 31 · 53
and min(4, 3) = 3.)
Now 100! is the product of the natural numbers from 1 to 100. Half of these,
namely 2, 4, . . . , 100, are even, and so contribute 50 factors 2. But the 25 multiples of
4 contribute an extra factor of 2, and the 12 multiples of 8 an extra factor, similarly the
6 multiples of 16, the 3 multiples of 32, and the single multiple of 64. So the power of
2 dividing 100! is 250+25+12+6+3+1 = 297 .
2a
2
Similarly, there are 20 multiples of 5 which each contribute 5 to the product, and
4 multiples of 25 which contribute an extra 5. So the power of 5 is 520+4 = 524 .
So 100! ends with 24 zeros.
You can easily ask the computer to check this. In fact,
100! = 93326215443944152681699238856266700490715968264381621468592
96389521759999322991560894146397615651828625369792082722375
8251185210916864000000000000000000000000.
But I didn’t ask this question to see whether you could solve it by computer.
3 (†) Here is an extract from the Introduction to Probability lecture notes.
We say that the events A and B are independent if
P(A ∩ B) = P(A) × P(B).
You may assume that events are independent in the following situations:
(i) they are clearly physically unrelated (eg depend on different coin tosses),
(ii) you calculate their probabilities and find that P(A ∩ B) = P(A)P(B),
(iii) the question tells you that the events are independent!
(a) What is this? Is it a definition, a theorem, a proof, an example, or just some
chit-chat?
(b) Your boss asks you to summarise it in two or three brief bullet points, without
using any symbols. What would you say?
(c) Why is the P in the formula written in a different typeface from the other
symbols?
(a) The first part is a definition of the term “independent”. The second part is what
might be called “chit-chat”, or more formally, some useful information about the definition. There is an unusual feature here. Normally when we have a definition in a piece
of mathematics, that definition says everything about what the word means. But here
we are given more: statement (i) says that, if two events are physically unrelated, as
for example two events depending on different coin tosses, then they are independent:
this is either a mathematical theorem, or a statement about the real world, depending
on how you view it, and goes beyond the definition.
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(b) This is hard because the extract is already very condensed, and it is hard to
condense it further. Here is a possible answer:
• Two events are said to be independent if the probability of them occurring together is the product of the individual probabilities of the events.
• Two events which are physically unrelated are independent.
• Don’t assume independence without good reason!
(c) The typeface used for P is blackboard bold, which is used for concepts which
are very general and occur everywhere, such as N for the natural numbers. Probability
is another such concept.
4 (†) Let m and n be positive integers. Recall that the greatest common divisor of
m and n, written gcd(m, n), is the largest positive integer which divides both m and
n. Now define the least common multiple of m and n, written lcm(m, n), to be the
smallest positive integer which is divisible by both m and n.
(a) Prove that gcd(m, n) · lcm(m, n) = mn.
(b) Prove that gcd(m, n) ≤ lcm(m, n), with equality if and only if m = n.
With an unfamiliar question, it is a good idea to do an example first to check what
goes on. If m = 12 and n = 18, then gcd(12, 18) = 6 and lcm(12, 18) = 36; and
6 · 36 = 12 · 18. This is not a proof, but it will help you to think about a proof.
(a) Suppose that gcd(m, n) = d. Then d divides both m and n; say m = ud and
n = vd for some positive integers u and v. We immediately see that the number uvd is
a multiple of both m = ud and n = vd.
We also see that u and v have greatest common divisor 1; for if e | u and e | v, then
de | ud and de | vd, so the greatest common divisor of m and n is at least de, which
contradicts the fact that this gcd is d unless e = 1.
Could the least common multiple of m and n be smaller than uvd? Let N be this
l.c.m. Then certainly d divides N, and both u and v must divide N/d. Since u and v
have no common factor, this means that uv divides N/d, so duv | N. This shows that
duv really is the least common multiple.
But now
gcd(m, n) · lcm(m, n) = d · duv = du · dv = mn.
(b) Let d = gcd(m, n) and N = lcm(m, n). Then d | m and m | N, so d | N, whence
d ≤ N. Could equality hold? If d = N, then our argument shows that d = m = N, and
similarly d = n = N.
OFB 10/11/16
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