MATH08064 Fundamentals of Pure Mathematics
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MATH08064
Fundamentals of Pure Mathematics
PART A
Answer all six questions in this part.
They are worth 40 marks in total.
A1. For each of the following statements, state whether it is TRUE or FALSE. Give a brief proof or a counterexample.
(a) If A is a non-empty subset of 皿 and 1 ≤ a ≤ 2 for all a e A, then 1 ≤ infA ≤ supA ≤ 2.
[3 marks]
(b) If A is a non-empty bounded subset of 皿, and B is defined by B = {a4 : a e A}, then
supB = (supA)4 . [3 marks]
Solution:
True.
Proof: 1 is a lower bound of A and infA is the largest lower bound, therefore 1 ≤ infA. Similarly, 2 is an upper bound and supA is the smallest upper bound, therefore supA ≤ 2. Finally, infA ≤ supA is true for all non-empty bounded sets. Putting it all together we have 1 ≤ infA ≤ supA ≤ 2.
False.
Counterexample: Let A = {一1, 0}. Then B = {0, 1}, supA = 0, supB = 1 (supA)4 .
A2. Prove, by verifying the ← -吝 property of Ross, Theorem 17.2, that the function f : 皿 → 皿 given by f (x) = 2x3 +x +7 is continuous at the point x0 = 2.
[7 marks]
Solution Let ← > 0. Pick a positive number 吝 such that 吝 < 1 and 吝 < , for example 吝 = min ,1, 、. For all x e 皿 with lx 一 2l < 吝 we have lx 一 2l < 1, therefore 1 < x < 3, therefore
l2x2 +4x +9l = 22x +4x +9 ≤ 39,
therefore
lf (x) 一 f (2)l = l(2x3 +x +7) 一 25l = l2x3 +x 一 18l = l2x2 +4x +9llx 一 2l ≤ 39吝 < ← .
A3. Use Taylor’s Theorem to prove that if x e [0, 1] then
x 一 + 一 ≤ log(x + 1) ≤ x 一 + .
[Recall that we write log for the logarithm to the base e, loge. ] [7 marks]
Solution: If x = 0 then we have 0 ≤ 0 ≤ 0, so going forward we assume x > 0. The derivatives of log(x + 1) can be calculated as
log(n)(x + 1) = (一1)n+1(n 一 1)!
The Taylor series about zero is then
(一 k+1xk
(this is in the book and can be used without proof). Let Rn(x) be the nth remainder then by subtracting log(1 + x) across the inequality, and multiplying by 一1, we are asked to prove if x e (0, 1] then R4(x) ≤ 0 ≤ R5(x). By Taylor’s theorem there exists y1 ,y2e (0,x) such that
R4(x) = 一!(x4 and R5(x) = 一!(x5 .
Since x,y1 ,y2 > 0 we see R4(x) < 0 and R5(x) > 0 so that the inequalities hold.
A4. Give an example of each of the following. You do not need to explain why your example is correct.
(a) a group of order 12 that is abelian but not cyclic |
[1 mark] |
(b) a nonabelian group of order 2018 that contains an element of order 1009 |
|
|
[1 mark] |
(c) a group of order less than 10 that acts transitively on {1, 2, 3, 4, 5} |
[1 mark] |
(d) a nontrivial proper subgroup of 勿25 |
[1 mark] |
(e) a nonconstant homomorphism from 勿4 to 勿2 |
[1 mark] |
(f) a group with exactly one conjugacy class |
[1 mark] |
(g) a subgroup of D3 that is not normal |
[1 mark] |
Solution: |
|
(a) up to isomorphism, the only possibility is 勿2 × 勿6 |
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(b) the dihedral group D1009 |
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(c) up to isomorphism, the only possibility is 勿5 |
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(d) the only possibility is the cyclic subgroup〈5)≤ 勿25 |
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(e) there is only one; it sends any generator to 1 e 勿2 |
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(f) the trivial group {e} is the only possibility |
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(g) the only possibilities are the subgroups generated by a single reflection |
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A5. Suppose that G is a finite group, and that A,B ≤ G are subgroups. Let H := A n B be the intersection of A and B.
(a) Prove that H is a subgroup of G.
(b) Prove that if lAl and lBl are coprime, then H is trivial.
Solution:
[3 marks] [3 marks]
(a) We apply the subgroup test.
Firstly, H is nonempty: A and B are subgroups of G, so e e A and e e B. Therefore e e A n B = H.
Secondly, H is closed under products: if g, h e H, then g, h e A and g, h e B. There- fore gh e A and gh e B since A and B are subgroups. Therefore gh e A n B = H.
Finally, H is closed under inversion: if g e H, then g e A and g e B, so g一1 e A and g一1 e B since A and B are subgroups. Therefore g一1 e A n B = H.
(b) By Lagrange’s theorem, lHl divides both lAl and lBl. Since lAl and lBl are coprime, we must have lHl = 1, so H is trivial.
A6. In the following question, you may use any theorems from the course but you must clearly state which theorems you are using.
(a) Show that a group G of order 35 cannot act transitively on a set X of order 9.
[4 marks]
(b) Is 勿9 × 勿8 勿72? Justify your answer. [3 marks]
Solution:
(a) If G acts transitively on X then X is an (the only) orbit of the action. By the orbit- stabilizer theorem, lXl = 9 divides lGl = 35. This is impossible, so no such action can exist.
(b) 勿a × 勿b is cyclic if and only if gcd(a, b) = 1. Since 9 and 8 are relatively prime,
勿9 × 勿8 is cylic. We have l勿9× 勿8l = 9 × 8 = 72 = l勿72l. Two cylic groups of the same order are isomorphic, so 勿9 × 勿8 勿72 .
PART B
There are four questions in this part, each worth 20 marks.
The best three answers will count.
B1. (a) Prove from the ← -N definition of convergence (without using any Limit Theorems)
that the sequence (an)neN given by an = converges to 2. [6 marks]
(b) Find the limit of the sequence (bn)neN given by
bn = 3n2 ╱ 1 一 弋1 一 ← .
(You may use any Limit Theorems you find useful provided that you make clear what it is you are using). [7 marks]
(c) Let (an)neN be a sequence of real numbers such that 0 < an < 1 for all n.
(i) Prove that, if the series 左(}) an converges, then the series 左(}) 1a一a(n)n converges as
well. [4 marks]
(ii) Is the converse true? Give a proof or a counterexample. [3 marks]
Solution:
(a) Let ← > 0. Let N be a positive integer such that N > 10/←. For all n ≥ N we have
| 一 2 | = < ≤ < ← .
(b) We’ll show that 0 ≤ bn ≤ for all n. The squeeze theorem then implies bn → 0 as n → }.
The inequality bn ≥ 0 is obvious. The inequality bn ≤ can be proved as follows:
bn = 3n2 ╱ 1 一 弋1 一 ←
= 3n2 ╱ 1 一 │ 1 一 、╱ 1 + │ 1 一 、
1 + │ 1 一 n3(1)
= 3n2 1 一 ╱ 1 一 、
1 + │ 1 一 n3(1)
3
=
n ╱1 + │ 1 一 、
≤ 3
Another solution: By Taylor’s formula with remainder, ′ 1 一x = 1 一 + R(x), where we have the following estimate for the remainder R(x): there exists a positive constant C such that for all x e (一1/2, 1/2), lR(x)l ≤ Cx2 . It follows that for all n ≥ 2,
0 ≤ bn = 3n2 │ 1 一 │ 1 一 + R │ 、、、 = 一 3n2R │ 、 ≤ + .
By the squeeze theorem, bn → 0.
2022-05-06