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
NAME:
Number:
Group:
MATH10101: WEEK SIX TEST 2015 (45 minutes)
Answer both questions; they are worth 15% of your final 10101 mark.
Write on both sides if necessary: the use of calculators is forbidde
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
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 sh
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 grap
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
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 negat
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
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 +
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
*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 coefficie
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 math
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.manchest
Problems for week 3
6. Prove by induction on n that, for any real number a 6= 1 and for integers
n 0,
n
X
ai =
i=0
1 an+1
.
1a
7. How many elements are there in each of the following sets?
(i) , (ii)
Solutions to the Additional Questions
1. Use the result from the notes that states that
n
r
=
n!
.
r! (n r)!
The left hand side of the proposed result equals
n1
r1
=
n
r+1
n+1
r
(n 1)!
n!
(n + 1)!
,
(
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 s
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 t
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 =
7 Permutations Very little of this section comes from PJE.
7.1 Definition A permutation (p.147) of a set A is a bijection : A A.
Notation If A = cfw_a, b, c, . and is a permutation on A we can express
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 negat
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 +
Solutions for week 4
11. If (x, y) (A B) (C D), then (x, y) A B or (x, y) C D.
In the first case, x A and y B, which implies that x A C and
y B D. Hence (x, y) (A C) (B D). In the second case, x C and
Problems for week 2
1. Consider the following statement:
(i) All girls are good at mathematics.
Which of the following statements is its negation?
(ii) All girls are bad at mathematics.
(iii) All girl
Problems for week 4
?
11 . For any sets A, B, C and D prove that
(A B) (C D) (A C) (B D).
Explain why your logic cannot be adapted to prove that
(A B) (C D) = (A C) (B D)
[in other words, find an impl
Problems for week 5
?
16 . Find inverses for the following functions, giving reasons:
(i) f1 : R R given by y = f1 (x) = 2x + 3;
(ii) f2 : R R given by y = f2 (x) = x3 1;
(iii) f3 : R+ R given by y =
Solutions for week 5
16. (i) Suppose that x, y R. Then
y = 2x + 3 2x = y 3 x = (y 3)/2.
Hence f11 (y) = (y 3)/2.
(ii) Suppose that x, y R. Then
y = x3 1 x3 = y + 1 x =
Hence f21 (y) =
3
p
3
y + 1.
y +