Premium Resources

PAPER TITLE: FURTHER PURE MATHEMATICS

EXAM DATE: THURSDAY 11, JUNE 2019

COURSE CODE: M303/D

Question 1

(a) Use the Euclidean algorithm to find hcf(217,483).

(b) Determine the least positive integer that satisfies all three of the following linear congruences.

x ≡ 3 (mod 4); 3x +4≡ 6 (mod 7); 7 x ≡ 6 (mod 11).

(c) Is the following statement true or false? If it is true, prove it. If it is false, justify your answer. Any number of the form 17m + 2, where m ≥ 0 is a non-negative integer, must have a prime divisor of this same form.

ANSWERS(Purchase full paper to get all the solutions):

1a)

hcf(217,483) using Euclidean algorithm.

483 = 2(217) + 49

217 = 4(49) + 21

49 = 2(21) + 7

21 = 3(7) + 0

Therefore, the hcf = 7

1b)

The least possible integers

For:

x ≡ 3 (mod 4)

the least possible integer that satisfies x ≡ 3 (mod 4)  is 3

for:

3x +4≡ 6 (mod 7)

3x ≡ 2(mod 7)

3x ≡ 9(mod 7)

x ≡ 3(mod 7)

the least possible integer that satisfies 3x+4 ≡ 6 (mod 7) is 3

for:

7 x ≡ 6 (mod 11)

11x-4x ≡ 6 (mod 11)

0-4x ≡ 6 (mod 11)

-4x≡ 28 (mod 11)

x ≡ -7 (mod 11)

x ≡ 4 (mod 11)

the least possible integer that satisfies 7 x ≡ 6 (mod 11) is 4

1c)

Yes, statement is true. “the Any number of the form 17m + 2, where m ≥ 0 is a non-negative integer, must have a prime divisor of this same form” and this can be proved using mathematical induction.

Proof:

First positive divisor of the form 17m +2 = 2, as 2 is a prime, then 2 is the prime divisor of 2 that is clearly of the form 17m +2.

Suppose that every integer greater than or equal to 2 but less than k =17n + 2 as the prime divisor of the form 17l +2.

Case1:

If k = 17n +2 is prime

→ k = 17n +2 is itself a prime divisor of the form 17l +2.

Case2:

If k = 17n +2 is not prime

→ k = 17n +2 is a composite

Then,

k = 17n +2  = ab for a,b∈Z. 1

since  the product of ab is of the form 17n +2: one of a and b is of the form . And the other of the form .

→ k = 17n +2   =

Where 1

By induction hypothesis,

Question 2

(a) Let G = Z₉ × Z₅.

(i) Write down the order of G.

(ii) Write down the order of (3,0).

(iii) Write down an element of order 5.

(iv) Is G cyclic? Justify your answer.

(b) Consider the following two groups, G and H, both of which have order 360.

G = Z₂ × Z₁₂ × Z₁₅

H = Z₃₀ × Z₂ × Z₆

1. Write each of G and H as a direct product of cyclic groups of prime power order (and hence deduce that G and H are not isomorphic groups).
2. Is either of G or H isomorphic to Z₃₆₀? Briefly justify your answer.

(iii) Is either of G or H isomorphic to D_ic90? Briefly justify your answer.

Question 3

(a) Evaluate the Legendre symbol (127/167). (Note that 127 and 167 are primes.)

(b) Determine whether or not the quadratic congruence 3x² +4x +5≡ 0 (mod 17) has solutions.

(c) Use the rational root test to show that f(x)=x³ + 12x² −5x +3 is irreducible over Q.

Question 4

Let

A = {(a,b) ∈ R² :12 + b² ≤ 3},B={(a,b) ∈ R² : a>0}

and

C = A−B.

(a) Sketch on separate diagrams, or describe in words, each of A, B and C taking care to indicate which points lie in the sets.

(b) Determine whether (−1,0) is a closure point of C for the Euclidean metric d⁽²⁾

(c) Write down in set notation the closure, interior and boundary of C for the Euclidean metric d⁽²⁾ on the plane.

(d) Determine whether the closure of C for the Euclidean metric on the plane is compact

Question 5

Let α = 8 − 7; you may assume that

[Q(8) : Q]=[Q(7) : Q] = 2, 8 ∉ Q(7) and 7 ∉ Q(8).

(a) Express α³ in the form a+b for suitable a,b ∈ Q.

(b) With the help of (a), or otherwise, show that  ∈ Q(α).

(c) Hence show that Q(α) = Q().

(d) Determine the degree [Q(α):Q].

(e) By considering α² find the minimal polynomial for α over Q.

Question 6

 Define a metric, d, on X = R² −{(0,b):b ∈ R } by

d(x, y) =|x₁ −y₁|+2 |x₂ −y₂| where x =( x₁, x₂) and y = ( y₁, y₂) are points in R². (You do not have to show that d is a metric.)

(a) Write down d((−1,−1),(1,1)).

(b) Write down a d-disconnection of X.

(c) (i) Show that for (a, b) ∈ X, d((1/n,0),(a, b)) ≥ 2|b|.

     (ii) Hence, or otherwise, show that (1/n,0) is not a d-convergent sequence in X.

     (iii) Hence, or otherwise, show that X is not d-complete.

      (iv) Deduce that X is not d-sequentially compact.

Question 7

(a) Let D₅₀ be the dihedral group of order 100:

            D₅₀ ={r,s|r⁵⁰ = s² = e, sr = r⁴⁹s}

(i) Show that r²⁵sr²⁵s = e.

(ii) Hence (or otherwise) deduce that r²⁵ ∈ Z(D₅₀) where Z(D₅₀) is the centre of D₅₀

(b) Let G be an arbitrary group of order 100 = 4×25.

(i) Write down the order of the Sylow 2-subgroups and the order of the Sylow 5-subgroups of G.

(ii) Calculate the possible numbers of Sylow 2-subgroups of G.

(iii) Prove that there is only one Sylow 5-subgroup of G and hence deduce that this subgroup is normal.

(c) Now let G be a group with order p²q² where p and q are distinct, odd primes, with p>q. Let m_p be the number of Sylow p-subgroups and m_q be the number of Sylow q-subgroups.

(i) Show that if m_p ≠  1 then p divides q² −1.

(ii) Hence (or otherwise) deduce that G has a unique Sylow p-subgroup.

(iii) Hence deduce that G cannot be simple.

 

Question 8

Let R = Z[] = {a + b-: a, b ∈ Z}, and let

N(a + b-) = a² +6b² be a norm on R, which you may assume is multiplicative.

(a) Show that the only elements r ∈ R with N(r) = 1 are r = ±1, and that there are no elements with N(r) = 2 or N(r) = 3.

(b) Show that each of 2, 3 and  is irreducible in R.

(c) Hence, or otherwise, show that R is not a unique factorization domain.

Although the ring R = Z[] ={a + b√−6:a,b ∈ Z} is not a unique factorization domain, some reducible elements of R may still be uniquely factorizable. Indeed, show that 7 has a unique factorization in R as follows.

Suppose that 7 = u v where u = a + b, v = c + d and u, v are not units of R.

(d) By taking norms in the equation 7 = u v show that the norms of both u and v must be equal to 7.

(e) Show that there are exactly four elements u = a + b in R such that N(u) = 7.

(f) Hence show that 7 has a unique factorization in R.

(g) Using the factorization in (f) or otherwise, show that the ideal I ={2,1} of R is equal to R, that is, I = R. [Hint: Establish that 1 ∈ I.

Question 9

(a) Let

            A = {f ∈ C[0,1] : |f(x)−f(y)|≤exp(|x−y|)|x−y| for each x, y ∈ [0,1]}

and

B ={f ∈ C[0,1] : |f(x)−2019|≤1 for 0≤ x ≤ 1}.

1. Show that A is d_max-closed.
2. Deduce that A∩B is d_max-closed.
3. Hence or otherwise show that A∩B is d_max-sequentially compact. (You may assume that A is an equicontinuous set of functions.

(b) Define a distance function d: [0 ,∞)×[0,∞)→ R as follows:

     

(i) Write down d(0,1).

(ii) Prove that d is a metric on [0,∞) by showing that

(1) d satisfies (M₁).

(2) d satisfies (M₂).

(3) d satisfies (M₃).

(Purchase full paper by adding to cart):

Last updated: Sep 02, 2021 03:07 PM

add to cart ← Back