Solutions for week 2
1. Let Girl denote the statement the student is a girl, and Good denote
the statement the student is good at mathematics. Then (i) reduces to the
statement Girl Good. So its negation is the statement Girl
Good.
There must therefore be
Solutions for week 7
21. Long division of 7 into 2.00 gives 2/7 = 0.285714.
Let x = 1.7782345. Then 10000x = 17782.3452345 from which it follows that
9999x = 17782.345 1.778 = 17780.567 = 17780567/1000. We conclude that
x = 17780567/9999000.
22. We ha
Solutions for week 4
11. Suppose that (x, y) (A B) (C D). Then (x, y) A B or
(x, y) C D.
In the rst case (x, y) A B, which implies that x A and y B, and
therefore that x A C and y B D. Hence (x, y) (A C) (B D).
In the second case (x, y) C D, which implies
Solutions for week 5
16. (i) Suppose that x, y R. Then
y = 2x + 3 2x = y 3 x = (y 3)/2.
1
Hence f1 (y) = (y 3)/2.
(ii) Suppose that x, y R. Then
y = x3 1 x3 = y + 1 x =
1
Hence f2 (y) =
3
3
y + 1.
y + 1.
[Note 1: In general, the function h : R R dened by
Solutions for week 3
6. In this case the statement P (n) to be proved for all integers n 0 is that
n
i=0
ai = (1 an+1 )/(1 a).
Thus P (k) states that k ai = (1 ak+1 )/(1 a), and P (k + 1) states that
i=0
k+1 i
a = (1 a(k+1)+1 )/(1 a) or, simplifying, k+1
*7. Polynomials
Definition For a set F , a polynomial over F with variable x is of the form
an xn + an1 xn1 + an2 xn2 + . + a1 x + a0 ,
where an , an1 , ., a1 , a0 F . The ai , 0 i n are the coefficients of the
polynomial. If xn is the largest power of x
Appendix 2-ii
Appendix 2-ii
1) Definition Alternative definition of GCD. Let a and b be integers, at
least one of which is non-zero. Then the greatest common divisor is the
unique positive integer d such that
i) d|a and d|b , i.e. d is a common divisor,
i
Appendix 7
Appendix 7
Recall example from notes:
Example Let A = cfw_1, 2, 3, 4, 5, 6 and
1 2 3 4 5 6
=
.
5 6 3 1 4 2
It is easy to check that 6 = 1A . Choose 1 A and repeatedly apply to 1
six times to get
1 5 4 1 5 4 1.
If written as (1, 5, 4, 1, 5, 4),
7.2 Permutations continued
Definition
The positive powers n of a permutation are defined inductively by
setting 1 = and k+1 = k for all k N.
n
The negative powers of a permutation are defined by n = (1 )
for all n N, i.e. taking positive powers (just de
Problems for week 7
21. Find the innite recurring decimal expansion of the rational number 2/7,
and the rational number equal to the recurring innite decimal 1.7782345.
22. Given bijections f : Z+ A and g : Z+ B, explain how to combine them
into a bijec
Problems for week 5
16 . Find inverses for the following functions:
(i) f1 : R R given by f1 (x) = 2x + 3;
(ii) f2 : R R given by f2 (x) = x3 1;
(iii) f3 : R+ R given by f3 (x) = ln x.
Sketch the graph of f2 , and describe the elements of Gf2 R R.
17. Sup
SETS, NUMBERS, AND FUNCTIONS
MATH10101
http:/www.maths.man.ac.uk/nige/1101ssi.html
NIGEL RAY
School of Mathematics, University of Manchester
Oxford Road, Manchester M13 9PL, England
www.maths.manchester.ac.uk/nige/
September 21, 2015
Background
The rst 21
Problems for week 2
1. Consider the following statement.
(i) All girls are good at mathematics.
Which of the following statements is the negation of the above statement?
(ii) All girls are bad at mathematics.
(iii) All girls are not good at mathematics.
(
Problems for week 3
6. Prove by induction on n that, for any real number a = 1 and for integers
n 0,
n
ai =
i=0
1 an+1
.
1a
7. How many elements are there in each of the following sets?
(i) , (ii) cfw_, (iii) cfw_, , (iv) cfw_, .
8. The following are stan
Problems for week 4
11 . For sets A, B, C and D prove that
(A B) (C D) (A C) (B D).
Give an example in which equality holds, and an example in which it fails; at
which point does your proof fail to show (A B) (C D) (A C) (B D)?
12. Dene functions f and g
Appendix 6-i
Appendix 6-i (following section 6.1)
Contents
1. Unique factorization and gcd and lcm.
2. Sieve of Eratosthenes
3. Prime Number Theorem
4. Use of unique factorization
5. What is (100) and (1000)?
6. If gcd (r, n) = 1 and gcd (a, n) = 1 then g
Appendix 3-ii
Appendix 3-ii (at the end of Chapter 3)
Example Solve
2x 3 mod 5 and 3x 4 mod 7.
Solution. First, solve each individual congruence. The easiest way is to find
the inverse of the coefficients of x.
Note that 3 2 1 mod 5 so, on multiplying bot
Multiplication tables throughout the course.
Looking back through the course we have seen the following tables.
[2]10
[4]10
[6]10
[8]10
cfw_[2]10 ,[4]10 ,[6]10 ,[8]10 ,10
[2]10 [4]10 [6]10 [8]10
[4]10 [8]10 [2]10 [6]10
[8]10 [6]10 [4]10 [2]10
[2]10 [4]10
Solutions for week 2
1. Let Girl denote the statement the student is a girl, and Good denote
the statement the student is good at mathematics. Then (i) reduces to the
statement Girl Good. So its negation is the statement Girl ; Good.
There must therefore
Solutions for week 5
16. (i) Suppose that x, y R. Then
y = 2x + 3 2x = y 3 x = (y 3)/2.
Hence f11 (x) = (x 3)/2.
(ii) Suppose that x, y R. Then
y = x3 1 x3 = y + 1 x =
Hence f21 (x) =
p
3
y + 1.
3
x + 1.
[Note: In general, the function h : R R defined by
Solutions for week 7
5.
Then 10000x = 17782.345234
5 from which it follows
21. Let x = 1.778234
that 9999x = 17782.345 1.778 = 17780.567 = 17780567/1000. We conclude
that x = 17780567/9999000.
22. We have already proven the first part in class, by list
Solutions for week 4
11. Suppose that (x, y) (A B) (C D). Then (x, y) A B or
(x, y) C D.
In the first case (x, y) A B, which implies that x A and y B, and
therefore that x A C and y B D. Hence (x, y) (A C) (B D).
In the second case (x, y) C D, which impli
Solutions for week 3
6.
P In this case the statement P (n) to be proved for all integers n 0 is that
n
i=0
ai = (1 an+1 )/(1 a).
P
Thus P (k) states that ki=0 ai = (1 ak+1 )/(1 a), and P (k + 1) states that
Pk+1 i
P
(k+1)+1
i
k+2
)/(1 a) or, simplifying,
MATH10101 Exam 2013-14 Solutions & Feedback
A3. i). State Fermats Little Theorem.
ii) Prove that
230123 664436 + 123230 436664 0 mod 13
Solution i. If p is prime and a Z for which gcd (p, a) = 1 (i.e. p - a) then
ap1 1 mod p. (Also allow ap a mod p for al
MATH10101
Three hours
The number of marks on this paper is 75.
THE UNIVERSITY OF MANCHESTER
SETS, NUMBERS AND FUNCTIONS
23 January 2014
09.45 12.45
Answer ALL FIVE questions in Section A (30 marks in all) and THREE questions in Section B
(45 marks in all)