ipm g5087 2011 past exams

APPENDIX X

Yearly Exams

0

IPM (G5087) Autumn 2011 Yearly Exams p.1

24.1. BSc/MMath EXAMINATIONS 2009MATHEMATICS: Introduction to Pure Mathematics

G5087

n-th June 2009 2.00 pm–3:30 pm

Attempt all questions.Time allowed: ONE and a HALF hours.Each question carries TWENTY marks. The numbers beside the questions

indicate the approximate marks that can be gained from the corresponding partsof the questions.

1. (a) Let E := {2n : n ∈ N0}, F := {n ∈ N : n is prime} and G := {k + 5 : k ∈ E}.Decide whether each of the following is true or false and back your answer withan explanation [8 marks]

(i) E ∩ F = ∅, (ii) E ∪G = ∅, (iii) E ∩G = ∅, (iv) F ∩G = ∅.

(b) Let Ai be a given set for each i ∈ I , where I is a nonempty set of indexes.Define (in words or symbols) their intersection

⋂i=I Ai. [4 marks]

(c) Build a sequence of sets {Ai : i ∈ N} such that [4 marks]

∀N ∈ N :

N⋂i=1

Ai 6= ∅ and

∞⋂i=1

Ai = ∅. (1)

[4 marks]

(d) A sweet maker produces two types of chocolate: dark and milk, with the rule thatthe dark chocolate must be wrapped in red or blue paper. A workshop sample trayis composed of three wrapped chocolates: one yellow, one red and one blue.

(i) What is the minimum number of chocolates one must unwrap, in order toensure the rule holds true for this sample? Explain.

(ii) Two persons, of which one is a dark-only chocolate lover and one is a milk-only lover, are presented with the tray. Both persons know the rule andare asked to choose exactly one of the wrapped chocolates. Which one isguaranteed to satisfy her taste? Explain.

2. (a) Using induction, prove thatn∑

k=1

k2 =1

6n(n+ 1)(2n+ 1), ∀ n ∈ N. (1)

(b) Let a ∈ R and r ∈ Rr {1}; using induction, prove that

n−1∑k=0

ark = arn − 1

r − 1, ∀ n ∈ N.

3. (a) Fill in the boxes in the following: [3 marks]

Theorem (Euclidean division). For each m ∈ N0 and n ∈ N there exists a uniquepair (q, r) ∈ N2

0 such that

m = (i) + r and (ii) ≤ (iii) ≤ n− 1. (1)

In this problem we denote the rest of the division of m by n by

modnm := r for m,n, r as in the Euclidean Division Theorem’s statement. (2)

(b) Given a ∈ N0, b ∈ N define their highest common factor (also known as greatestcommon divisor) hcf (a, b). [3 marks]

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(c) Which of hcf (9, 12) and hcf (0, 12) is bigger? [2 marks]

Hint. All positive integers are factors (i.e., divisors) of 0.(d) Show that hcf (0,m) = m for m ∈ N. [2 marks]

(e) Prove that for any m ∈ N0 and n ∈ N we have [6 marks]

hcf (n,m) = hcf ( modnm ,n) . (3)

Hint. Put r := modnm, h := hcf (n,m) and g := hcf (r, n), and use parts (a) and(b) to show that h ≤ g and g ≤ h.

(f) Fix now a, b ∈ N0 with b > 0 and consider the sequence of numbers rk, k =−1, 0, 1, 2, . . . , defined recursively by [4 marks]

rk :=

b, if k = −1,

a, if k = 0,

modrk−1rk−2, if k ≥ 1 and rk−1 > 0

0, if k ≥ 1 and rk−1 = 0.

(4)

Show that there exists a K ∈ N such that rK > 0, rK+1 = 0 and that hcf (a, b) =rK .

4. For each of the following five functions find out if it is injective, surjective, bijective ornone. Back your claims with proofs or counterexamples, as appropriate. [4× 5 marks]

(a) f : Z→ Z with f(x) = x+ 3,(b) g : Qr {0} → Q with g(x) = 1/x,(c) h : N→ N with h(x) = x+ 3,

(d) k : Q→ Q+0 with k(x) = x2.

(e) j : P(A) → P(A) with j(X) =A r X and P(A), the power set ofA, for a given set A.

5. Let n, k ∈ N0 and k ≤ n.(a) Give an expression for

(nk

)involving n, k and the factorial. [4 marks]

(b) Show that the number of subsets of [1 : n] with exactly k elements are(nk

). (You

do not need to use induction, an “intuitive” argument is enough.) [8 marks]

(c) Show that the number of all the possible injective maps from [1 : k] into [1 : n] isgiven by k!

(nk

). [8 marks]

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Solution of Exam IPMG5087, Summer 2009

1. (a) Let E := {2n : n ∈ N0}, F := {n ∈ N : n is prime} and G := {k + 5 : k ∈ E}.Decide whether each of the following is true or false and back your answer with anexplanation [8 marks]

(i) E ∩ F = ∅, (ii) E ∪G = ∅, (iii) E ∩G = ∅, (iv) F ∩G = ∅.

(b) Let Ai be a given set for each i ∈ I , where I is a nonempty set of indexes. Define(in words or symbols) their intersection

⋂i=I Ai. [4 marks]

(c) Build a sequence of sets {Ai : i ∈ N} such that [4 marks]

∀N ∈ N :N⋂i=1

Ai 6= ∅ and∞⋂i=1

Ai = ∅. (1)

[4 marks]

(d) A sweet maker produces two types of chocolate: dark and milk, with the rule thatthe dark chocolate must be wrapped in red or blue paper. A workshop sample tray iscomposed of three wrapped chocolates: one yellow, one red and one blue.

(i) What is the minimum number of chocolates one must unwrap, in order to ensurethe rule holds true for this sample? Explain.

(ii) Two persons, of which one is a dark-only chocolate lover and one is a milk-onlylover, are presented with the tray. Both persons know the rule and are askedto choose exactly one of the wrapped chocolates. Which one is guaranteed tosatisfy her taste? Explain.

Solution.

(a) Note that E is just the set of non-negative even numbers.

(i) False because 2 is both prime (so in F ) and even (so in E). Thus 2 ∈ E ∩ F , which makesit nonempty.

(ii) False because both E and G are nonempty and so must be their union, E ∪ G, which hasboth as subsets.

(iii) True. If a number m ∈ E ∪G then n = 2n for some n ∈ N and m = 2l + 5 for some l ∈ Nand thus 2n = 2l+ 5, which implies that 5 = 2(n− l) which is false. By contraposition therecan be no number m ∈ N in E ∪G, but E ∪G ⊆ N so it is empty.

(iv) False. Both 7 ∈ F and 7 ∈ G are true, thus 7 ∈ F ∩G.

(b) The intersection of the collection {Ai : i ∈ I } is the set that consists of those elements thatare in each single Ai, for all i ∈ I . In symbols this is written as follows, let i∗ be any fixed elementin I (who is non-empty by assumption) then⋂

i∈I

Ai := {x ∈ Ai∗ : ∀ i ∈ I : x ∈ Ai} . (2)

As observed in class, the choice of which i∗ to work with makes no difference.

(c) A possible example is to take Ai to be the set of positive multiples of i, for i ∈ N. Then if

N ∈ N, the set⋂N

i=1Ai = A1 ∩ A2 ∩ · · · ∩ AN (call it BN for short) is that of numbers that aresimulatneously multiples of 1, 2, . . . , N . One such number is N ! = 1× 2× · · · ×N . So N ! ∈ BN isthus BN 6=.

On the other hand, let B∗ :=⋂

i∈NAi, we want to show that B∗ = ∅. Arguing by contradiction,suppose B∗ 6= ∅, so there is a number M ∈ N and M ∈ B∗ =

⋂n∈NAn. By taking n = M+1 in the

intersection, it follows that M ∈ AM+1, i.e., M is a multiple of M + 1, which implies M ≥M + 1which is absurd So B∗ is empty as claimed.

(d) Although not necessary, it may help you if you write the rule’s statement as follows

t(c) = dark ⇒ w(c) = blue or w(c) = red, (3)

where t is the “type” function and w is the wrapper color function.

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Note that the equivalent contrapositive statement, which will be quite useful, is that a non-blueand non-red wrapped chocolate must be (non-dark, hence) milk. Indeed this is the translation inEnglish of

w(c) 6= blue and w(c) 6= red ⇒ t(c) = milk ( 6= dark). (4)

(i) To ensure the rule, it is enough to unwrap the yellow chocolate. It is milk, then the rule istrue. If it is dark then the rule is false. The answer is thus one.

(ii) A milk-only lover can guarantee a choccy to their taste by choosing yellow. A dark-onlylover, who will clearly avoid the yellow, may still be disappointed as red or blue need notcontain dark.

4


2. (a) Using induction, prove that

n∑k=1

k2 =1

6n(n+ 1)(2n+ 1), ∀ n ∈ N. (1)

(b) Let a ∈ R and r ∈ Rr {1}; using induction, prove that

n−1∑k=0

ark = arn − 1

r − 1, ∀ n ∈ N.

Solution.

(a) Let us write L(n) :=∑n

k=1 k2 and R(n) := n(n+ 1)(2n+ 1)/6; we must prove L(n) = R(n)

for all n ∈ N. First we check the base case, n = 1,

L(1) =

1∑k=1

k2 = 12 = 1 and R(1) =1

61(1 + 1)(2× 1 + 1) =

6

6= 1 ⇒ L(1) = R(1). (2)

Next let us prove the Inductive Step. For this assume L(n− 1) = R(n− 1) (IH), and let us showL(n) = R(n). We have

L(n) =

n−1∑k=1

k2 + n2

(by definition of L) = L(n− 1) + n2

(by the Inductive Hypothesis (IH)) = R(n− 1) + n2

(by definition of R and algebra) =(n− 1)n(2(n− 1) + 1) + 6n2

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(more algebra) =n ((n− 1)(2n− 1) + 6n)

6.

(3)

To finish we must just prove that (n − 1)(2n − 1) + 6n = (n + 1)(2n + 1), but this is true aselementary algebra shows:

(n− 1)(2n− 1) + n = 2n2 − 2n− n+ 1 + 6n = 2n2 + 3n+ 1, (4)

(n+ 1)(2n+ 1) = 2n2 + 2n+ n+ 1 = 2n2 + 3n+ 1. (5)

Thus L(n) = R(n) and the proof is complete.

(b) The Base Case, n = 1, is checked as follows:

0∑k=0

ark = ar0 = a and arn − 1

r − 1= a

r − 1

r − 1= a, (6)

which implies that the two members are equal for n = 1.

To prove the Inductive Step, assume the Inductive Hypothesis for n, i.e.,

n−1∑k=0

ark = arn − 1

r − 1, (IH)

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we want to prove the equality with n+ 1 replacing n.

n∑k=0

ark =

n−1∑k=0

ark + arn (algebra)

= arn − 1

r − 1+ arn (using (IH))

= arn − 1 + rn+1 − rn

r − 1(algebra)

= arn+1 − 1

r − 1,

(7)

as desired.

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3. (a) Fill in the boxes in the following: [3 marks]

Theorem (Euclidean division). For each m ∈ N0 and n ∈ N there exists a uniquepair (q, r) ∈ N2

0 such that

m = (i) + r and (ii) ≤ (iii) ≤ n− 1. (1)

In this problem we denote the rest of the division of m by n by

modnm := r for m,n, r as in the Euclidean Division Theorem’s statement. (2)

(b) Given a ∈ N0, b ∈ N define their highest common factor (also known as greatestcommon divisor) hcf (a, b). [3 marks]

(c) Which of hcf (9, 12) and hcf (0, 12) is bigger? [2 marks]

Hint. All positive integers are factors (i.e., divisors) of 0.(d) Show that hcf (0,m) = m for m ∈ N. [2 marks]

(e) Prove that for any m ∈ N0 and n ∈ N we have [6 marks]

hcf (n,m) = hcf ( modnm ,n) . (3)

Hint. Put r := modnm, h := hcf (n,m) and g := hcf (r, n), and use parts (a) and (b)to show that h ≤ g and g ≤ h.

(f) Fix now a, b ∈ N0 with b > 0 and consider the sequence of numbers rk, k =−1, 0, 1, 2, . . . , defined recursively by [4 marks]

rk :=

b, if k = −1,

a, if k = 0,

modrk−1rk−2, if k ≥ 1 and rk−1 > 0

0, if k ≥ 1 and rk−1 = 0.

(4)

Show that there exists a K ∈ N such that rK > 0, rK+1 = 0 and that hcf (a, b) = rK .

Solution.

(a) The complete statement reads:

Theorem (Euclidean division). For each m ∈ N0 and n ∈ N there exists a unique pair(q, r) ∈ N2

0 such that

m = nq + r and 0 ≤ r ≤ n− 1. (5)

(b) The highest common factor of a and b is that number h ∈ N such that

h | a and h | b (6)

and

(k | a and k | b) ⇒ k ≤ h. (7)

We denote h = hcf (a, b).More concisely this can be written as

hcf (a, b) := max {k ∈ N : k | a and k | b} (8)

(c) We have hcf (9, 12) = 3 and hcf (0, 12) = 12. So hcf (9, 12) < hcf (0, 12).(d) All natural numbers divide 0. Hence 0 and n’s common factors are all the factors of n, and

the highest one is n. Thus hcf (0, n) = n.

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(e) Let us denote modnm =: r, h := hcf (n,m) and g = hcf (r, n). We want to show thath = g.We first show that g ≤ h. Since g = hcf (r, n) we have g | r and g | n, which implies thatr = r′g and n = n′g for some r′, n′ ∈ N. But by the Euclidean Division we have thatm = qn+ r for some q ∈ N, thus

m = qn′g + r′g =(qn′ + r′) g, (9)

which means that g | m. Hence, recalling that h = hcf (n,m) and g | m, we have g ≤ h.To conclude we still have to show h ≤ g. Since h = hcf (n,m) we have h | n and h | m, son = n′′h and m = m′′h. Now, by the Euclidean Division we have

0 ≤ r = m− qn =(m′′ − qn′′)h, (10)

so m′′ − qn′′ ∈ N0 and h divides r. But h divides also n, so it is a common factor of n and rand thus h ≤ g, as claimed.

(f) By the Euclidean Division (a) we have that for each k ∈ N0 either that rk 6= 0 and rk < rk−1

or rk−1 = 0 and rk = 0. Consider the set R := {rk : rk > 0}. Since r−1 = m > 0, thenr−1 ∈ R and the set R is thus non-empty. By the Well-Ordering Principle, it follows thatR ∈ N has a smallest element, i.e., there exists a K ∈ N such that rK = minR. Note alsothat rK+1 6∈ R (because rK+1 < rK and rK is the smallest element in R) so rK+1 = 0.Furthermore, by (e), we have that

hcf (rk−2, rk−1) = hcf (rk, rk−1) , for k ≥ 1 and rk−1 6= 0. (11)

In particular it follows that

hcf (a, b) = hcf (r1, a) = · · · = hcf (rK+1, rK) . (12)

But rK+1 = 0 and rK > 0, so by (d) we have that hcf (rK+1, rK) = hcf (0, rK) = rK . Hence

hcf (a, b) = rK . (13)

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4. For each of the following five functions find out if it is injective, surjective, bijectiveor none. Back your claims with proofs or counterexamples, as appropriate. [4× 5 marks]

(a) f : Z→ Z with f(x) = x+ 3,(b) g : Qr {0} → Q with g(x) = 1/x,(c) h : N→ N with h(x) = x+ 3,

(d) k : Q→ Q+0 with k(x) = x2.

(e) j : P(A)→P(A) with j(X) = ArXand P(A), the power set of A, for agiven set A.

Solution. Reminders

A function φ : Domφ→ Codφ is injective if and only if

φ(x) = φ(y) ⇒ x = y (1)

and it is surjective if and only if

∀ y ∈ Codφ : ∃ x ∈ Domφ : φ(x) = y. (2)

A function is bijective if and only if it is simultaneously injective and surjective.

(a) f is injective.

Indeed, let x1, x2 ∈ Z such that f(x1) = f(x2). If follows that

x1 + 3 = x2 + 3 (by definition of f)

so x1 = x2 (by subtracting 3 from each member).(3)

f is surjective. Indeed, suppose y ∈ Z, consider the element x := y − 3, then

f(x) = x+ 3 = (y − 3) + 3 = y. (4)

f is thus bijective.

(b) g is injective. Indeed, suppose x1, x2 ∈ Qr {0} with g(x1) = g(x2), it follows that

1

x1=

1

x2(by definition of g)

thus x1 = x2 (by inverting both members).(5)

g is not surjective. Indeed the element 0 belongs to the codomain Q, but there is no x ∈ Qr {0}such that 1/x = 0.

g is therefore not bijective.

(c) h is injective for the same arguments used for f and because of the cancellation law in N.

h is not surjective (unlike f). Indeed, the element 2 cannot be written as h(x) for any x ∈ N; theonly x that works is x = −1 which is not an element of N.

h is not bijective.

(d) k is not injective. Indeed, we have f(1) = 1 = f(−1), yet 1 6= −1.

k is not surjective. Indeed, the element 2 ∈ Q+0 has no counterimage in Q. The only elements that

work are ±√

2 which are known to be irrational numbers, therefore not in Q.

k is not bijective

(e) j is injective. Let X,Y ∈ P(A), which the same as X,Y ⊆ A and suppose j(X) = j(Y ).Then ArX = Ar Y . But, by definition of complementation, we know that

Z = Ar(Ar Z) ∀ Z ∈P(A) . (6)

It follows that

X = Ar(ArX) = Ar(Ar Y ) = Y. (7)

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j is surjective. Let Y ∈ Cod j = P(A), we want X ∈ Dom j = P(A) such that j(X) = Y , whichis equivalent to ArX = Y . Let X := Ar Y , it follows then, by (6), that

j(X) = ArX = Ar(Ar Y ) = Y. (8)

j is thus bijective.

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5. Let n, k ∈ N0 and k ≤ n.

(a) Give an expression for(nk

)involving n, k and the factorial. [4 marks]

(b) Show that the number of subsets of [1 : n] with exactly k elements are(nk

). (You do

not need to use induction, an “intuitive” argument is enough.) [8 marks]

(c) Show that the number of all the possible injective maps from [1 : k] into [1 : n] isgiven by k!

(nk

). [8 marks]

Solution.

(a) An expression is given by the formula(n

k

)=

n!

(n− k)!k!. (1)

(b) Let us use a counting argument, by building a generic k-element subsetK = {x1, x2, . . . , xk} ⊆[1 : n]. We have n elements to choose x1 from; then n−1 elements are left to choose x2 from; thenn− 2 to choose x3 from;. . . ; then n− k + 1 to choose xk from. In total we have

n× (n− 1)× (n− 2)× · · · × (n− k + 1) =n!

(n− k)!(2)

choices. Notice however that we have counted sets too many times; indeed, different ways ofchoosing can lead to the same set K, because sets do not change under permutations of theirelements (for example K = {x1, x2, . . . , xk} and J = {x2, x1, . . . , xk} represent the same set,although “picked” in a different order). We have therefore to divide the total by all the possibleways of ordering a set of k elements. But this is known to be k! (indeed, the first element can bechosen among k, the second among k− 1, etc.). Thus the total number of subsets K ⊆ [1 : n] withk elements is given by

n!

(n− k)!k!=

(n

k

). (3)

(An alternative, rigorous, proof can be given using induction. See the homework problems for sucha proof.)

(c) Let us count the number of injective maps f : [1 : k]→ [1 : n]:

? choose f(1) in {1, . . . , n}, i.e., in n different ways;

? choose f(2) in {1, . . . , n} r {f(1)}, i.e., in n − 1 different ways (f(1) has to be excludedbecause, by injectivity of f we must have f(1) 6= f(2));

? etc.

? choose f(k) in {1, . . . , n}r {f(1), . . . , f(k − 1)}, i.e., in n− k + 1 possible different ways.

In total the number of possible choices is

n× (n− 1)× (n− 2)× · · · × (n− k + 1) =n!

(n− k)!=

(n

k

)k!, (4)

as claimed.

(Also here the result can be proved by induction on n.)

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G5087

n-th June 2010 2.00 pm–3:30 pm



1. For each positive integer i ∈ N, consider the set

Di := {k ∈ N : k divides i} . (♣)

(a) Describe D9 and D15 by listing all their elements. [6 marks]

(b) Find max(D9 ∩D15). What is this number, with respect to 9 and 15. [6 marks]

(c) Simplify the expression [4 marks]⋂n∈N

D3n, (♦)

by listing all the elements of the resulting set, or otherwise.

(d) Prove your answer in ((b)) is correct. [4 marks]

2. Prove the following 2 identities by induction on n: [2× 10 marks]

n∑k=1

k =n(n+ 1)

2, ∀ n ∈ N, (a)

n∏k=2

(1− 1

k2

)=n+ 1

2n, ∀ n ∈ N and n ≥ 2. (b)

3. (a) Complete the boxes in the following [3 marks]

Definition. We say that p ∈ N is a prime number if and only if

n divides p ⇒ (i) or (ii)

and p ≥ (iii)(1)

(b) Let p be a prime number, and a, b ∈ N such that p divides ab but p does not [4 marks]

divide a. Based on a theorem from class, prove that

hcf (b, p) = p. (2)

(c) What if p were not prime in (b), would the result still hold? Back your answerwith an example. [3 marks]

4. For each one of the following functions decide if it is (i) injective, (ii) surjective, (iii)bijective and prove your answer, in each case. [4× 5 marks]

(a) g : Z→ Z with g(x) := 2x.(b) h : N→ N with h(x) := x+ 5.

(c) k : Z→ Z with k(x) := x+ 5.(d) b·c : Q→ Z with bxc := floor of x.

(The floor, bxc, of x ∈ R is by definition the largest possible k ∈ Z such that k ≤ x.)5. (a) Let n be the cardinality of a finite set A, and let k ∈ N0 and k ≤ n. Complete

the following definition [5 marks](n

k

):= number of subsets of (i) containing exactly (ii) elements. (♣)

12


(b) Give an explicit formula for(nk

)in terms of n and k. [5 marks]

(c) Among all digital expansions of exactly 7 digits, how many are there with only “1”and “2” appearing in them (e.g., 1111211 and 1212121)? Explain your answer. [5 marks]

(d) How many are there with exactly 4 figures “1” and 3 figures “2” (e.g., 1121122)?Explain your answer. [5 marks]

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1. For each positive integer i ∈ N, consider the set

Di := {k ∈ N : k divides i} . (♣)

(a) Describe D9 and D15 by listing all their elements. [6 marks]

(b) Find max(D9 ∩D15). What is this number, with respect to 9 and 15. [6 marks]

(c) Simplify the expression [4 marks]⋂n∈N

D3n, (♦)

by listing all the elements of the resulting set, or otherwise.

(d) Prove your answer in ((b)) is correct. [4 marks]

Solution.

(a) D9 = {1, 3, 9} and D15 = {1, 3, 5, 15}.(b) max(D9 ∩D15) = max {1, 3} = 3, this is the highest common factor of 9 and 15 denotedhcf (9, 15).

(c) We have ⋂n∈N

D3n = D3. (1)

(d) From the definition of intersection and that of the sets (♣) we have⋂n∈N

D3n := {k ∈ N} k divides 3n, for all n ∈ N. (2)

Now 1 and 3 both divide all numbers of the form 3n with n ∈ N, hence

{1, 3} ⊆⋂n∈N

D3n. (3)

To prove the reverse inclusion, we note that {1, 3} = D3 and that an intersection of a collection ofsets is subset of each one the collection’s sets, hence⋂

n∈ND3n ⊆ D3 = {1, 3} . (4)

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2. Prove the following 2 identities by induction on n: [2× 10 marks]

n∑k=1

k =n(n+ 1)

2, ∀ n ∈ N, (a)

n∏k=2

(1− 1

k2

)=n+ 1

2n, ∀ n ∈ N and n ≥ 2. (b)

Solution. Proof of (a). The base case is clearly satisfied

1∑k=1

k = 1 =1(1 + 1)

2. (3)

To prove, the inductive step, fix n and assume the inductive hypothesis holds true:

n−1∑k=1

=(n− 1)n

2. (4)

The we haven∑

k=1

=

n−1∑k=1

+n

(inductive hypothesis (4)) =(n− 1)n

2+ n

(basic algebra) =(n− 1 + 2)n

2=

(n+ 1)n

2,

(5)

as desired.

Proof of (b). The Base Case (for n = 2, because n = 1 does not make sense in this formula) is

2∏k=2

(1− 1

22

)= 1− 1

4=

3

4=

2 + 1

2× 2. (6)

To prove the inductive step, fix n ≥ 3 assume the inductive hypothesis

n−1∏k=2

(1− 1

k2

)=

n

2(n− 1). (7)

Then we haven∏

k=2

(1− 1

k2

)=

(n−1∏k=2

(1− 1

k2

)) (1− 1

n2

)=

n

2(n− 1)

(1− 1

n2

)=

n

2(n− 1)

n2 − 1

n2

=(n+ 1)(n− 1)

2(n− 1)n=

(n+ 1)

2n,

(8)

as claimed.

15


3. (a) Complete the boxes in the following [3 marks]

Definition. We say that p ∈ N is a prime number if and only if

n divides p ⇒ (i) or (ii)

and p ≥ (iii)(1)

(b) Let p be a prime number, and a, b ∈ N such that p divides ab but p does not divide [4 marks]

a. Based on a theorem from class, prove that

hcf (b, p) = p. (2)

(c) What if p were not prime in (b), would the result still hold? Back your answer withan example. [3 marks]

Solution.

(a) Definition. We say that p ∈ N is a prime number if and only if

n divides p ⇒ n = 1 or n = p

and p ≥ 2

(3)

(b) Since p is prime and p | ab, but p - a then p | b, by the Atomic Property of Primes (Theoremproved in class). Hence p is a common factor of p and b, but it is also the highest such facto, becausthere is no number bigger than p that divides p. (More precisely, we could say that p | hcf (p, b)and thus p ≤ hcf (p, b) ≤ p, which implies that p = b.)

(c) If p is not prime then (2) then the result does not hold. A counterexample is given by p = 6(not prime), a = 3 and b = 4, in which case we have

p = 6 - 3 = a

p = 6 | 12 = ab

yet hcf (p, b) = 2 6= 6 = p.

(4)

16


4. For each one of the following functions decide if it is (i) injective, (ii) surjective, (iii)bijective and prove your answer, in each case. [4× 5 marks]

(a) g : Z→ Z with g(x) := 2x.(b) h : N→ N with h(x) := x+ 5.

(c) k : Z→ Z with k(x) := x+ 5.(d) b·c : Q→ Z with bxc := floor of x.

(The floor, bxc, of x ∈ R is by definition the largest possible k ∈ Z such that k ≤ x.)

Solution.

(i) The function g is injective. Indeed, for any x, y ∈ Z such that g(x) = g(y), it follows that [3 marks]

2x = 2y ⇒ x = y. (1)

The function g is not surjective. For example, the element 3 ∈ Z in the codomain has nocounterimage for g in Z, because this would mean that 2x = 3 for some x ∈ Z, i.e., x = 3/2 ∈ Zwhich is clearly false. [2 marks]

(ii) The function h is injective. Indeed, for any x, y ∈ N with h(x) = h(y) it follows that [3 marks]

x+ 5 = y + 5 ⇒ x = y. (2)

The function h is not surjective. Indeed, the element 1 ∈ N has no counterimage for h in N

as this would imply that 1 = h(x) = x + 5 for some x ∈ N, leading to −4 = x ∈ N which isclearly false. [2 marks]

(iii) The function k is injective and this can be shown in the same way as for h. [2 marks]

The function k is surjective. Indeed, for any y ∈ Z, taking x = y − 5 ∈ Z it follows thatk(x) = x+ 5 = (y − 5) + 5 = y. I.e., any y ∈ Z has a counterimage for k in Z.

(iv) The function b·c is not injective. For example x = 1.1 and y = 1.2 which are clearly differentyet they satisfy bxc = 1 = byc. [3 marks]

The function b·c is surjective. Indeed for any y ∈ Z, since y ∈ Q, y is its own counterimage. [2 marks]

17


5. (a) Let n be the cardinality of a finite set A, and let k ∈ N0 and k ≤ n. Completethe following definition [5 marks](

n

k

):= number of subsets of (i) containing exactly (ii) elements. (♣)

(b) Give an explicit formula for(nk

)in terms of n and k. [5 marks]

(c) Among all digital expansions of exactly 7 digits, how many are there with only “1”and “2” appearing in them (e.g., 1111211 and 1212121)? Explain your answer. [5 marks]

(d) How many are there with exactly 4 figures “1” and 3 figures “2” (e.g., 1121122)?Explain your answer. [5 marks]

Solution.

(a) (n

k

):= number of subsets of A containing exactly k elements. (♣)

(b) (n

k

)=

n!

k!(n− k)!=n(n− 1) . . . (n− k + 1)!

k!. (1)

(c) All we need to find out is where we put the 1 figure (then the 2’s are automatically determinedby the vacant positions).

Since there are 7 possible positions to choose from, we need to count the number of all possiblesubsets of a set of 7 elements. That is 27 = 128.

(d) We have to find out in how many ways we can choose 4 elements out of 7. That is 7 choose4, i.e., (

7

4

)=

7× 6× 5

3× 2= 35. (2)

18



G5087

n-th June 2011 2.00 pm–3:30 pm



1. (a) Let A and B denote two sets. By using words or logical symbols, write thedefinition, of their union, A ∪B. [5 marks]

(b) Recall that N denotes the set of all positive integers. Let E := {5k : k ∈ N and k ≥ 2}and F := {3k : k ∈ N and k ≥ 2}. For each of the following statements, saywhether it is true or false:

(i) 5 ∈ E,(ii) 10 ∈ E,

(iii) 10 ∈ E ∩ F,(iv) 10 ∈ E ∪ F,

(v) x prime ⇒ x 6∈E ∪ F.

[5 marks]

(c) Consider the indexed collection of sets {Ai : i ∈ I }, with the index set I 6=∅. State, with the help of words and/or logical symbols, the definition of thecollection’s union,

⋃i∈I Ai. [5 marks]

(d) Let En := {x ∈ Q : 1/n ≤ x ≤ 1}, for n ∈ N. Find out which one of the followingequals

⋃n∈NEn, and prove your claim.

(i) {x ∈ Q : 0 < x ≤ 1} (ii) {x ∈ Q : 0 ≤ x ≤ 1} (iii) {0, 1}[5 marks]

19


2. (a) State the four properties that make (Z,+) an Abelian group. [4 marks]

(b) Suppose x1, x2 ∈ R. Show that if x1 ∈ Z and x1 + x2 ∈ Z then x2 must be in Z. [4 marks]

(c) Suppose x1, x2 ∈ R both statisfy, for some a, b, c ∈ R, the equality

ax2i + bxi + c = 0, for i = 1, 2, (1)

and that x1 6= x2. Show that x1 + x2 = −b/a. [4 marks]

(d) Let f : A→ B be a function, what does it mean for f to be injective? [4 marks]

(e) Show that the function

f : Z → Z

x 7→ 2x2 + x(2)

is injective. [4 marks]

Hint. Use (d), suppose f(x1) = f(x2) and recall (c).3. [(a) ]

Prove by induction thatn∑

k=1

k2 =1

6(2n+ 1)(n+ 1)n, ∀ n ∈ N, (1)

?? Prove by induction that

n3 ≤ 3n ∀ n ∈ N. (2)

Hint. For (2) it will be useful to remember that (a+ b)3 = a3 + 3a2b+ 3ab2 + b3,to show the cases n = 1, 2 “by hand”, and to put the base case of the inductionat n = 3.

20


4. Let m ∈ N0 and M := [1 : m] := {k ∈ N : 1 ≤ k ≤ m}.(a) Fix k ∈ N0 such that k ≤ m. Give the definition of the binomial coefficient

(mk

)as the cardinality of a subcollection of the power set P(M). [5 marks]

(b) What is the cardinality of P(M)? [5 marks]

(c) Using the definition in (a) and the answer in (b), explain why the following formulais true:

2m =m∑k=0

(m

k

). (1)

[5 marks]

(d) Expand (a + b)m using the Binomial Theorem, and use the result to show alge-braically that (1) is true. [5 marks]

5. (a) Illustrate the Euclidean Algorithm by finding hcf (287, 205). [5 marks]

(b) Find x, y ∈ Z that solve the equation

287x+ 205y = 164. (1)

(Non-integer solutions are not accepted.) [5 marks]

(c) Find another solution x′, y′ to (1). [5 marks]

(d) Can you do the same for

287x+ 205y = 165? (2)

Explain your answer [5 marks]

21



1. (a) Let A and B denote two sets. By using words or logical symbols, write thedefinition, of their union, A ∪B. [5 marks]

(b) Recall that N denotes the set of all positive integers. Let E := {5k : k ∈ N and k ≥ 2}and F := {3k : k ∈ N and k ≥ 2}. For each of the following statements, say whetherit is true or false:

(i) 5 ∈ E,(ii) 10 ∈ E,

(iii) 10 ∈ E ∩ F,(iv) 10 ∈ E ∪ F,

(v) x prime ⇒ x 6∈ E ∪F.

[5 marks]

(c) Consider the indexed collection of sets {Ai : i ∈ I }, with the index set I 6= ∅. State,with the help of words and/or logical symbols, the definition of the collection’s union,⋃

i∈I Ai. [5 marks]

(d) Let En := {x ∈ Q : 1/n ≤ x ≤ 1}, for n ∈ N. Find out which one of the followingequals

⋃n∈NEn, and prove your claim.

(i) {x ∈ Q : 0 < x ≤ 1} (ii) {x ∈ Q : 0 ≤ x ≤ 1} (iii) {0, 1}[5 marks]

Solution.

(a) The union of A with B, A ∪B, is the set

{x : x ∈ A or x ∈ B} . (1)

In words A ∪ B is the set whose elements are all those elements which are at least in one of A orB.

(b) (i) false, (ii) true, (iii) false, (iv) true, (v) true.

(c) Let i0 ∈ I (such and i0 exists given that I 6= ∅). The union of the collection {Ai : i ∈ I },⋃i∈I Ai, is defined as the set

{x : ∃ i ∈ I : x ∈ Ai} . (2)

In words,⋃

i∈I Ai is the set whose elements consist of the elements that are in some Ai for atleast one i ∈ I .

(d) The correct guess is (i):

U :=⋃n∈N

En = {x ∈ Q : 0 < x ≤ 1} =: E. (3)

To show the result this, let’s prove that U ⊆ E and E ⊆ U . Since En ⊆ E for each n, thenthe union U ⊆ E. To prove the reverse inequality, we have to show that if some x0 ∈ Q satisfies0 < x0 ≤ 1 then there exists n0 for which x0 ∈ En0

, which is equivalent to say that

1

n0≤ x0 ≤ 1. (4)

The second inequality is trivial; to prove the second one, recalling that x0 ∈ Q then x0 = a/b forsome a, b ∈ N (they may be chosen both postive). By the Archimedean property of N, there isalways n0 such that n0 ≥ b/a. Since both n0 and a/b are positive it follows that 1/n0 ≤ a/b.

22


2. (a) State the four properties that make (Z,+) an Abelian group. [4 marks]

(b) Suppose x1, x2 ∈ R. Show that if x1 ∈ Z and x1 + x2 ∈ Z then x2 must be in Z. [4 marks]

(c) Suppose x1, x2 ∈ R both statisfy, for some a, b, c ∈ R, the equality

ax2i + bxi + c = 0, for i = 1, 2, (1)

and that x1 6= x2. Show that x1 + x2 = −b/a. [4 marks]

(d) Let f : A→ B be a function, what does it mean for f to be injective? [4 marks]

(e) Show that the function

f : Z → Z

x 7→ 2x2 + x(2)

is injective. [4 marks]

Hint. Use (d), suppose f(x1) = f(x2) and recall (c).

Solution.

(a) (Z,+) is an Abelian group because

AG1. + is associative in Z, i.e., (x+ y) + z = x+ (y + z) for all x, y, z ∈ Z;

AG2. + has a neutral element in Z, namely 0 ∈ Z satisfies x+ 0 = 0 + x = x, for all x ∈ Z;

AG3. each element x in Z has an additive inverse, namely −x because x+ (−x) = 0;

AG4. + is commutative in Z.

(b) Since x1 ∈ Z we have −x1 ∈ Z. Also x1 + x2 ∈ Z and (Z,+), hence

x2 = (x1 + x2)︸︷︷︸∈Z

+ (−x1)︸︷︷︸∈Z

∈ Z. (3)

(c) Suppose x1 and x2 satisfy the quadratic equation (1), then

x1,2 =−b±

√b2 − 4ac

2aand thus x1 + x2 = 2× −b

2a= − b

a. (4)

(Alternatively, if x1 and x2 are the roots of the quadratic polynomial ax2 + bx+ c then

ax2 + bx+ c = a(x− x1)(x− x2) = ax2 − a(x1 + x2)x+ ax1x2 ∀ x ∈ R, (5)

hence, by the identity principle for polynomials, we must have

− a(x1 + x2) = b and thus x1 + x2 = − ba.) (6)

(Even more simple solution, suppose x1 and x2 satisfy

ax21 + bx1 + c = 0 = ax2

2 + bx2 + c (7)

then we have0 = a(x2

1 − x22) + b(x1 − x2) =(a(x1 + x2) + b) (x1 − x2). (8)

Since x1 − x2 6= 0 this implies that

a(x1 + x2) + b = 0, (9)

whence

x1 + x2 = − ba.) (10)

(d) The function f is called injective if and only if

f(x1) = f(x2) ⇒ x1 = x2. (11)

(e) The function is injective. Suppose, by contradiction, that there are x1, x2 ∈ Z, x1 6= x2, suchthat f(x1) = f(x2). Then for some c ∈ Z we have

2x2i + xi + c = 0 for both i = 1, 2. (12)

By (c) it follows that x1 + x2 = −1/2. On the other hand from (b) we have x1 + x2 ∈ Z, hence−1/2 ∈ Z which is a contradiction.

23


3. [(a) ]

Prove by induction thatn∑

k=1

k2 =1

6(2n+ 1)(n+ 1)n, ∀ n ∈ N, (1)

?? Prove by induction thatn3 ≤ 3n ∀ n ∈ N. (2)

Hint. For (2) it will be useful to remember that (a + b)3 = a3 + 3a2b + 3ab2 + b3,to show the cases n = 1, 2 “by hand”, and to put the base case of the induction atn = 3.

Solution.

(a) Inductive step: suppose (1) is true for n− 1, i.e.,

n−1∑k=1

k2 =1

6(2n− 1)n(n− 1), (3)

let us prove (1):

n∑k=1

k2 =

n−1∑k=1

k2 + n2

=1

6

((2n− 1)n(n− 1) + 6n2

)=n

6

(2n2 − 3n+ 1 + 6n

)=n

6

(2n2 + 3n+ 1

)=n

6(2n+ 1)(n+ 1),

(4)

as desired.

The base case is easily checked for n = 1, as∑1

k=1 k2 = 1 = 3× 2× 1/6.

(b) For n = 1 we have n3 = 13 = 1 and 3n = 31 = 3, thus (2) is true.

For n = 2 we have n3 = 23 = 8 and 3n = 32 = 9, thus (2) is also true.

For n = 3 we have n3 = 33 and 3n = 33, thus (2) is (barely) true and this proves the base case.

Let us prove the inductive step: let n ≥ 3, suppose that (2) is true for n we want to prove it forn+ 1. By expanding the cube and using n ≥ 3 we get

(n+ 1)3 = n3 + 3n2 + 3n+ 1

(using 3 ≤ n) ≤ n3 + nn2 + (3n+ 1)

(thanks to 3n + 1 ≤ n3) ≤ n3 + n3 + n3

(by basic algebra) ≤ 3n3

(by Inductive Hypothesis) = 3n+1,

(5)

as desired.

24


4. Let m ∈ N0 and M := [1 : m] := {k ∈ N : 1 ≤ k ≤ m}.(a) Fix k ∈ N0 such that k ≤ m. Give the definition of the binomial coefficient

(mk

)as

the cardinality of a subcollection of the power set P(M). [5 marks]

(b) What is the cardinality of P(M)? [5 marks]

(c) Using the definition in (a) and the answer in (b), explain why the following formulais true:

2m =

m∑k=0

(m

k

). (1)

[5 marks]

(d) Expand (a+b)m using the Binomial Theorem, and use the result to show algebraicallythat (1) is true. [5 marks]

Solution.

(a) The binomial coefficient m choose k is defined as(m

k

):= # {S ∈P(M) : #S = k} . (2)

In words,(mk

)is the number of all possible different choices of a set S consisting of exactly k

elements as a subset of M , which has m elements.

(b) #P(M) = 2#M = 2m

(c) The set P(M) can be partitioned into the sets

M (k) := {S ∈P(M) : #S = k} (3)

for k = 0, . . . ,m. We have P(M) =⋃m

k=0 M (k) and M (k) ∩M (j) = ∅, if k 6= j, which meansthat the collection is disjoint; thus the cardinality of the union is given by

#P(M) =

m∑k=0

#M (k) =

m∑k=0

(m

k

). (4)

On the other hand we have #P(M) = 2m, whence (1) follows.

(d) The Binomial Theorem says that

(a+ b)m =

m∑k=0

(n

k

)am−kbk. (5)

Taking a = 1 and b = 1 we obtain

2m = (1 + 1)m =

m∑k=0

(n

k

), (6)

which is (1).

25


5. (a) Illustrate the Euclidean Algorithm by finding hcf (287, 205). [5 marks]

(b) Find x, y ∈ Z that solve the equation

287x+ 205y = 164. (1)

(Non-integer solutions are not accepted.) [5 marks]

(c) Find another solution x′, y′ to (1). [5 marks]

(d) Can you do the same for287x+ 205y = 165? (2)

Explain your answer [5 marks]

Solution.

(a) The Euclidean algorithm gives:

287 = 1× 205 + 82

205 = 2× 82 + 41

82 = 2× 41,

hence hcf (287, 205) = 41.

(3)

(b) Noting that 164 is a multiple of 41 and using the relations in (3) above we may write

164 = 4× 41 = 4× (205− 2× 82) = 4× 205− 8× (287− 205) = 12× 205− 8× 287. (4)

So one possible solution of (1) is thus

x = −8 and y = 12. (5)

(c) There are (infinitely many) more solutions. In fact, noting that 205 = 5×41 and 287 = 7×41we infer that

7× 205− 5× 287 = 0. (6)

Combining this identity with (4) we get, for any k ∈ Z, that

164 = 164 + 0k = (−8− 5k)287 + (12− 7k)205, (7)

which means, for example, for k = 2, that also x′ = 2 and y′ = −2 is a solution. (For each otherchoice of k ∈ Z we get another solution.)

(d) Equation (2) is unsolvable for x, y ∈ Z. If there were x and y such that 287x + 205y = 165then since, 41 divides both 287 and 205, we would have that 41 divides 165. This would meanmod41 165 = 0, whereas by Euclidean division we know that mod41 165 = 1.

26

ipm g5087 2011 past exams

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