MATH08064 Fundamentals of Pure Mathematics Solutions and comments 2021
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MATH08064
Fundamentals of Pure Mathematics
Solutions and comments
1. Let (xn )neN be a convergent sequence of real numbers with xn
Prove that the series
n1+北{
n=1
----→ L, where L > 0. n→+o
converges.
[10 marks]
Solution: Since xn → L > 0, there exists a positive integer N , such that, for all indices n with n > N ,
xn > .
For all such n we then have
1 < 1
The series n n之辛(1)L/2 converges because 1+L/2 > 1 (p-test). Therefore the series n n之辛(1)北{
converges (Comparison Test).
2.
(a) Let a, b e R, a < b. Let f : [a, b] → R be continuous in [a, b] and twice differentiable in
(a, b). Let A and B be the points with coordinates (a, f (a)) and (b, f (b)) respectively.
If the line segment with endpoints A and B intersects the graph of f at a point P with P A, B (see figure 1), prove that there exists a real number c in the interval
(a, b) such that f\\ (c) = 0. [10 marks]
Figure 1: Plot for Question 2a.
(b) Let a = -4, b = 1 and f (x) = l北(1)l . The points A(-4, ) and B(1, 1) are on the graph
of f and the line segment with endpoints A and B intersects the graph at a third point P (see figure 2). However, there is no point c in the interval (-4, 1) such that f\\ (c) = 0. (You are not asked to prove that AB intersects the graph, nor that f\\ doesn’t vanish).
Explain why this doesn’t violate the result of part (a). [5 marks]
Figure 2: Plot for Question 2b.
Solution: (a) Let (x0 , f (x0 )) with a < x0 < b be the coordinates of P . Applying the Mean Value Theorem (MVT) to f in each of the intervals [a, x0] and [x0 , b] we see that there exist c1 e (a, x0 ) and c2 e (x0 , b) such that f\ (c1 ) and f\ (c2 ) are equal to the slope of the line segment AB . In particular, f\ (c1 ) = f\ (c2 ). The function f\ satisfies the hypotheses of the MVT (alternatively, Rolle’s thm) in [c1 , c2]. Indeed, it is continuous in [c1 , c2] because it is differentiable all over (a, b), and differentiable in (c1 , c2 ). It follows that there exists c e (c1 , c2 ) such that f\\ (c) = 0. (b) We can’t apply part (a) to the interval [-4, 1] because f is not defined at 0, and we can’t apply it to [-4, 0) n (0, 1] either because it is not an interval. |
3. (a) Prove, by verifying the ε-δ property (Ross 17.2), that f (x) = is continuous
(b) Let f : [-1, 1] → [-1, 1] be continuous. Recall that by the Intermediate Value Theorem, there exists at least one x e [-1, 1] such that f (x) = x. (You are not asked to prove this.) (i) Prove that if in addition to continuity we have )f (a) - f (b)) < )a - b) for all a, b e [-1, 1], a b then there exists a unique x e [-1, 1] such that f (x) = x. (ii) In the setting of part (b)(i) either prove that f is monotone or give a carefully justified example of such a function f which is not monotone. [15 marks] |
f (x) = = , 2(北)
x e [-1, 0]
x e (0, 1] .
Then f is not monotone as f is strictly decreasing for x < 0 and strictly increasing for x > 0. Applying the reverse triangle inequality we have for x y
)f (x) - f (y)) = │ - │ = ))x) - )y)) < )x - y) < )x - y) .
4. Let G = GL(2, R) be the group of 2 x 2 invertible matrices under the operation of matrix multiplication, and consider the subset H c G of all matrices of the form (北(1) 1(0) ),
with x e R.
(a) Show that H is a subgroup of G. [5 marks]
(b) What familiar group is H isomorphic to? Show that H is indeed isomorphic to that
group. [5 marks]
Solution:
(a) We check that the set H satisfies the three conditions of the test for a subgroup.
Firstly, the set H is not empty because it contains the identity matrix (0(1) 1(0) ).
Secondly, the set H is closed under matrix multiplication because for any (北(1) 1(0) ) , ╱ y(1) 1(0) 、e H their product is also an element of H:
(北(1) 1(0) ) ╱ y(1) 1(0) 、= ╱ 北y1(0) 、e H.
Thirdly, the set H is closed under taking inverses because for any (北(1) 1(0) ) e H its inverse is also an element of H:
( 北(1) 1(0) )_1 = ( 1_北1(0) ) e H.
(b) We claim that the subgroup H is isomorphic to the group (R, +). To prove this, we will show that f : R → H defined via f (x) = (北(1) 1(0) ) is a bijective group homomorphism.
Firstly: f is a group homomorphism because
f (x + y) = ╱ 北y 1(0) 、= (北(1) 1(0) ) ╱ y(1) 1(0) 、= f (x) . f (y).
Secondly: f is injective because
f (x) = f (y) == (北(1) 1(0) ) = ╱ y(1) 1(0) 、 == x = y
and it is surjective because for any element (北(1) 1(0) ) of H we have a corresponding x e R and
f (x) = (北(1) 1(0) ) .
5. In each of the following cases state whether the given statement is TRUE or FALSE and provide a justification, including specific counterexamples where appropriate.
(a) The function f : S3 → Z3 defined by f (τ ) = o(τ ) - 1 is a group homomorphism.
Here, o(τ ) represents the order of the permutation τ .
(b) The groups Z8 x Z10 and Z40 x Z2 are isomorphic.
[5 marks]
[5 marks]
(c) All proper subgroups of D13 are cyclic. (Recall that a proper subgroup is a subgroup that is not equal to the whole group.) [5 marks]
Solution:
(a) False, because it is not always the case that f (σ o τ ) = f (σ) + f (τ ). For a specific
counterexample, note that f ((1 2) o (1 2)) = f (e) = 1 - 1 = 0 but f ((1 2)) + f ((1 2)) = (2 - 1) + (2 - 1) = 1 + 1 = 2 (mod 3).
(b) True, because both groups are isormorphic to Z8 x Z5 x Z2 and therefore isomorphic
to each other. Firstly, because 5 x 2 = 10 and gcd(5,2)=1 we have that Z5 x Z2 is a cyclic group of order ten and therefore isomorphic to Z10 . Thus Z8 x Z10 is isomorphic to Z8 x Z5 x Z2 . Similarly, because 8 x 5 = 40 and gcd(8,5)=1 we have that Z8 x Z5 is a cyclic group of order forty and therefore isomorphic to Z40 . Thus Z40 x Z2 is isomorphic to Z8 x Z5 x Z2 .
(c) True. The group D13 has order 26, so by Lagrange’s Theorem, any proper subgroup H < D13 must have order 1 or 2 or 13.
If )H) = 1, then H = {e} =〈< e〉which is cyclic.
If )H) = 2, let h e H such that h e. By a corollary of Lagrange’s theorem, the order of h must divide )H) and since h e the only possibility is that o(h) = 2. This
implies that )〈h〉) = o(h) = 2 and since )H) = 2, the inclusion〈h〉c H must in fact be an equality. In particular this means that H =〈h〉is a cyclic group.
If )H) = 13 we argue completely analogously to conclude that also in this situation H is a cyclic subgroup of D13 .
6. Consider the following graph:
Let G denote its group of symmetries. Justify all your answers to the questions below.
(a) How many symmetries does the graph have? [5 marks]
(b) Identify the symmetry group G (in terms of groups you have encountered in FPM).
[4 marks]
(c) Let X denote the set of nine edges of the graph.
Is the action of G on X faithful?
Is the action of G on X transitive?
[3 marks] [3 marks]
(d) Suppose we want to colour the edges of the graph with n colours. We consider two colourings the same if they are related by a symmetry of the graph.
How many different colourings of the edges are there? [10 marks]
Express your answer as a polynomial in n.
Solution:
(a) There are six symmetries.
2
Here is one way to count them. Number the vertices as follows: 7
6 4
5
(We also label the central edge ‘a’ for use further below.)
Vertices 1, 3, 5 have valency two, while 2, 4, 6, 7 have valency three. Since symmetries preserve valency of vertices, a symmetry can only send vertex 1 to one of {1, 3, 5}.
(Each of these choices is possible, as it is realized by a rotation of the graph.) After choosing where 1 is sent, it remains to choose where its neighbors 2 and 6 are sent. Either they stay in the same relative position (whence the symmetry is a rotation), or they are swapped (whence the symmetry is a rotation followed by a reflection). Altogether, there are 3 x 2 = 6 choices.
Alternatively, the symmetry group of the outer hexagon is
D6 = {e, g, g2 , g3 , g4 , g5 , h, gh, g2 h, g3 h, g4 h, g5 h} ,
where g is a rotation by 60 degrees and h is a reflection about a vertical axis. The symmetry group of the graph contains precisely the elements of D6 that preserve (i.e., are also symmetries of) the central “star” . These are rotations by 120 degrees and reflections:
G = {e, g2 , g4 , h, g2 h, g4 h}
Therefore, )G) = 6.
2022-08-05