OK, Beamer.

Rishabh Singh

April 2020

P-Set 1 Question 2

Q2Show that

A = {x ∈ R | cos x = 1}

is a subset of

B = {x ∈ R | sin x = 0}.

Proof. Let’s say x ∈ A. To prove A ⊆ B, we need to prove x ∈ B.Since x ∈ A, cos x = 1. It is known that

cos2 x + sin2 x = 1This means that

sin x =√

1− cos2 x

=√

1− 1

= 0This means that x ∈ B, which means A ⊆ B.

P-Set 1 Question 2

Q2, continuedShow that

A = {x ∈ R | cos x = 1}

is a subset of

B = {x ∈ R | sin x = 0}.

Is A a proper subset of B? Give reasons.

Proof. A ⊂ B if A 6= B. We need to prove that B 6⊆ A. We canprove this by providing a counter-example. Let x ∈ B.

sin2x + cos2 x = 1

cosx =√

1− sin2

= ±1

Since cos x = −1, x 6∈ A, A is a proper subset of B.

P-Set 1 Question 11

Q11In a class of 40 people studying music: 2 play violin, piano andrecorder, 7 play at least violin and piano, 6 play at least piano andrecorder, 5 play at least recorder and violin, 17 play at least violin,19 play at least piano, and 14 play at least recorder. How manyplay none of these instruments?

Answer

P-Set 1 Question 13

Q13Let A and B be general sets. Determine the containment relation(⊆, =, none) that holds between

a) P(A ∪ B) and P(A) ∪ P(B)

Proof. Let x ∈ P(A) ∪ P(B). This means that x ⊆ A or x ⊆ B.This means that x ⊆ A ∪ B, which implies x ∈ P(A ∪ B). Thisdirectly implies that P(A) ∪ P(B) ⊆ P(A ∪ B).

However, this doesn’t apply the other way around. If x ⊆ A ∪ B,that doesn’t mean x ⊆ A or x ⊆ B. (Take A = {1, 2}, B = {3, 4},x = {2, 3})

So P(A) ∪ P(B) ⊆ P(A ∪ B).

P-Set 1 Question 17

Q17Simplify

[A ∩ (A ∩ Bc)] ∪ [(A ∩ B) ∪ (B ∩ Ac)]

= [(A ∩ A) ∩ Bc ] ∪ [(A ∩ B) ∪ (B ∩ Ac)] (Associative Law)

= [A ∩ Bc ] ∪ [(A ∩ B) ∪ (B ∩ Ac)] (Idempotent Law)

= [A ∩ Bc ] ∪ [(B ∩ A) ∪ (B ∩ Ac)] (Commutative Law)

= [A ∩ Bc ] ∪ [B ∩ (A ∪ Ac)] (Distributive Law)

= [A ∩ Bc ] ∪ [B ∩ U ] (Negation Law)

= [A ∩ Bc ] ∪ B (Identity Law)

= B ∪ [A ∩ Bc ] (Commutative Law)

= (B ∪ A) ∩ (B ∪ Bc) (Distributive Law)

= (B ∪ A) ∩ U (Union with cmpt)

= B ∪ A (Identity Law)

P-Set 1 Question 17

Q17, continuedSimplify

[A ∩ (A ∩ Bc)] ∪ [(A ∩ B) ∪ (B ∩ Ac)]

and hence simplify

[A ∪ (A ∪ Bc)] ∩ [(A ∪ B) ∩ (B ∪ Ac)]

Proof. Observe that [A ∪ (A ∪ Bc)] ∩ [(A ∪ B) ∩ (B ∪ Ac)] is justthe dual of

[A ∩ (A ∩ Bc)] ∪ [(A ∩ B) ∪ (B ∩ Ac)].

This means that the simplified expressions of both laws are alsoduals of each other. Since

[A ∩ (A ∩ Bc)] ∪ [(A ∩ B) ∪ (B ∩ Ac)] = B ∪ A,

this directly implies that

[A ∪ (A ∪ Bc)] ∩ [(A ∪ B) ∩ (B ∪ Ac)] = B ∩ A.

P-Set 1 Question 19

Q19Use the laws of set algebra to prove for any R,P,Q

(R − P)− Q = R − (P ∪ Q)

Using the laws of set algebra,

(R − P)− Q = (R ∩ Pc) ∩ Qc (Difference Law)

= R ∩ (Pc ∩ Qc) (Associative Law)

= R ∩ (P ∪ Q)c (De Morgan’s Law)

= R − (P ∪ Q) (Difference Law)

P-Set 1 Question 20(c)

Q20(c)

Define the symmetric difference A⊕ B of two sets A and B to be

A⊕ B = (A− B) ∪ (B − A)

(c) Explain why A⊕ B can be written as (A ∪ B)− (A ∩ B)

A⊕ B = (A− B) ∪ (B − A)

= (A ∩ Bc) ∪ (B ∩ Ac) (Difference Law)

= [(A ∩ Bc) ∪ B] ∩ [(A ∩ Bc) ∪ Ac ] (Distributive Law)

= [B ∪ (A ∩ Bc)] ∩ [Ac ∪ (A ∩ Bc)] (Commutative Law)

= [(B ∪ A) ∩ (B ∪ Bc)] ∩ [(Ac ∪ A) ∩ (Ac ∪ Bc)] (DsLw)

= [(B ∪ A) ∩ U ] ∩ [U ∩ (Ac ∪ Bc)] (Negation Law)

= (B ∪ A) ∩ (Ac ∪ Bc) (Identity Law)

= (A ∪ B) ∩ (A ∩ B)c (De Morgan’s Law)

= (A ∪ B)− (A ∩ B) (Difference Law)

P-Set 1 Question 21

Q21Prove that for all sets A, B and C :

A× (B ∪ C ) = (A× B) ∪ (A× C )

Proof. Let x = (p, q) ∈ A× (B ∪ C ). This means p ∈ A andq ∈ B ∪ C . This means that q ∈ B or q ∈ C . This means thateither p ∈ A and q ∈ B or p ∈ A and q ∈ C .That is, (p, q) = x ∈ A× B or (p, q) = x ∈ A× C . So,x ∈ (A×B)∪ (A× C ). Hence, A× (B ∪ C ) ⊆ (A×B)∪ (A× C ).Similarly, (A× B) ∪ (A× C ) ⊆ A× (B ∪ C ). (Prove in exam)

Since both sides are subsets of each other, this means thatA× (B ∪ C ) = (A× B) ∪ (A× C )

P-Set 1 Question 22

Q22Let Ak = {n ∈ N | k ≤ n ≤ k2 + 5} for k = 1, 2, 3, . . . . Find

(a)4⋃

n=1

Ak

(b)90⋂

n=10

Ak

a)4⋃

n=1

Ak = {1, . . . , 6} ∪ {2, . . . , 9} ∪ {3, . . . , 14} ∪ {4, . . . , 21}

= {1, 2, . . . , 20, 21}

b)90⋂

n=10

Ak = {10, . . . , 105} ∩ · · · ∩ {90, . . . , 105, . . . , 8105}.

= {90, 91, . . . , 104, 105}

P-Set 1 Question 26

Q26Prove that if n is an integer, then

n −⌊

1

3n

⌋−⌊

2

3n

⌋equals either one or zero.

Let n = 3k for some k ∈ Z.

n −⌊

1

3n

⌋−⌊

2

3n

⌋= 3k − bkc − b2kc = 0

Let n = 3k ± 1 for some k ∈ Z,

3k±1−⌊k ± 1

3

⌋−⌊

2k ± 2

3

⌋=

{3k + 1− k − 2k = 1

3k − 1− (k − 1)− (2k − 1) = 1

P-Set 1 Question 28

Q28Let S = {x ∈ N | 0 ≤ x ≤ 11} and define f : S → S , wheref (x) = 5n + 2 mod 12. Is f one-to-one? Is it onto?

From the image above, we are able to see that not only is every yunique for all x ∈ S , every single y ∈ S is also ∈ f . Which meansthat f is bijective.

Cool Function Proof

Exercise. Prove if f : X → Y and g : Y → Z are invertible, thenso is g ◦ f : X → Z , and the inverse of g ◦ f is f −1 ◦ g−1.

Proof. For a function to be invertible, it must be bijective:one-to-one and onto. This, f : X → Y and g : Y → Z areone-to-one and onto. Now, considering g ◦ f ,

(g ◦ f )(x1) = (g ◦ f )(x2)

=⇒ f (x1) = f (x2) (g is one-to-one)

=⇒ x1 = x2 (f is one-to-one)

Hence, g ◦ f is one-to-one

Since f and g are onto, for every z ∈ Z , there is some y ∈ Y suchthat g(y) = z , and for every y ∈ Y there is one x ∈ X such thatf (x) = y . Hence, for every z ∈ Z , there is some x ∈ X such thatthere is some f (x) ∈ Y such that g ◦ f (x)z . Hence g ◦ f is onto.

Cool Function Proof Continued

Proof, continued. Since g ◦ f is both one-to-one and onto, itmust be bijective. Hence g ◦ f is invertible.

Now, let the inverse of g ◦ f be h : Z → X . Then it is evident that,

(g ◦ f ) ◦ h = ιZ (inverse functions)

(g−1 ◦ g) ◦ (f ◦ h) = g−1 ◦ ιZ (taking g−1)

ιY ◦ (f ◦ h) = g−1 ◦ ιZ (inverse functions)

f ◦ h = g−1 (identity function)

(f −1 ◦ f ) ◦ h = f −1 ◦ g−1 (taking f −1)

ιX ◦ h = f −1 ◦ g−1 (inverse functions)

h = f −1 ◦ g−1. (identity function)

Hence, (g ◦ f )−1 = f −1 ◦ g−1.

P-Set 1 Question 44

Q44Using the fact that(

1− 1

k2

)=

(k − 1)(k + 1)

k2

find an expression forN∏

k=2

(1− 1

k2

)in terms of N

Proof. Writing out the product,

N∏k=1

(1− 1

k2

)=

((1)(3)

22

)((2)(4)

32

)((3)(5)

42

)· · ·(

(N − 1)(N + 1)

N2

)We observe that every term gets canceled out, except for the termsin the first and last terms in the sequence, of which only partiallycancel out. Thus, the resultant expression is (N + 1)/2N

P-Set 2 Question 11

Q11Let a, b 6= 0 be integers. Let S be the set of integers defined by

S = {ax + by | x , y ∈ Z}

and let d0 be the smallest integer in the set S .Prove the following:

(a) If s ∈ S , then d0 is a divisor of s

(b) d0 is a divisor of both a and b.

(c) If d is a divisor of both a and b, then d is a divisor of d0

(d) d0 = gcd(a, b), and hence there exist x , y ∈ Z such thatax + by = gcd(a, b)

11(a)

(a) If s ∈ S , then d0 is a divisor of s.Proof. We can express s in the form

s = d0q + r (r < d0)

Let d0 = ax0 + by0 and s = ax1 + by1 for some x0, y0, x1, y1 ∈ Z.Then

s = d0q + r

ax1 + by1 = (ax0 + by0)q + r

a(x1 − qx0) + b(y1 − qy0) = r

ax2 + by2 = r (x2, y2 ∈ R)

which means r ∈ S . But r < d0 and d0 is the smallest member ofS . This proposed a contradiction - in that r doesn’t exist - r = 0.This means d0 | s.

11(b)

(b) d0 is a divisor of both a and b

Proof. For any x , y ∈ Z the number ax + by ∈ S .Let x = 0, y = 1, then ax + by = a · 0 + b · 1 = b. This meansthat b ∈ S .Let x = 1, y = 0, then ax + by = a · 1 + b · 0 = a. This means thata ∈ S .And since in part a, we proved that d | s given s ∈ S , sincea, b ∈ S , it follows that d | a, b.

11(c)

(c) If d is a divisor of both a and b, then d is a divisor of d0.

Proof. Let a = dm and b = dn. As established before,

d0 = ax0 + by0.

So, we observe that

d0 = ax0 + by0

= (dm)x0 + (dn)y0

= d(mx0) + d(ny0)

= d(mx0 + ny0)

Thus if d | a, b then d must also be a divisor of d0.

11(d)

(d) d0 = gcd(a, b), and hence there exist x , y ∈ Z such thatax + by = gcd(a, b)

Proof. We proved that d0 | a, b in parts a and b. Also, consideringthat any d that is a common factor of a and b satifies d | d0, thismeans that d0 is the largest number that is a common factor ofboth a and b or, gcd(a, b) = d0.

Since d0 ∈ S , which is defined as

S = {ax + by | a, b 6= 0, a, b,m, n ∈ Z}

this means that there is some x , y ∈ Z such that

ax + by = d0 = gcd(a, b)

Using the Bezout’s Property

Q12

a) Prove that if a, b,m ∈ Z such that gcd(a, b) = 1 and a | mand b | m, then ab | m.

b) Prove that if a and b are coprime integers and a | bc then a | c

a) Proof. By the Bezout Property, there exist x , y ∈ Z such thatax + by = 1. Multiplying this by m yields

amx + bmy = m.

Since a | m and b | m, we can say m = aq and m = bp forp, q ∈ Z. Substituting this yields:

a(bp)x + b(aq)y = m.

ab(px + qy) = m

and since px + qy ∈ Z. this means ab | m.

Using the Bezout’s Property

Q12

a) Prove that if a, b,m ∈ Z such that gcd(a, b) = 1 and a | mand b | m, then ab | m.

b) Prove that if a and b are coprime integers and a | bc then a | c

b) Proof. Since a and b are coprime, we can say ax + by = 1 forsome x , y ∈ Z. Multiplying this by c yields

acx + bcy = c

Also, since a | bc, it is true that an = bc, n ∈ Z. Substituting...

acx + any = c

a(cx + ny) = c

. Since cx + ny ∈ Z, this implies a | c

Using the Bezout Property

7g

Prove the following for a, b, c , d ∈ Z, k,m ∈ Z+:

a-f) Not important

g) If a ≡ b (mod m) then gcd(a,m) = gcd(b,m).

Proof. Since a ≡ b (mod m), the Division Theorem states that

a = b + md (d ∈ Z)

Now, considering the expression bx + my , we substitute to find

bx1 + my1 = (a−md)x + my

= ax −mdx + my

= ax + m(y − dx),

which is an expression of the form ax + my . Since the twoexpressions for the Bezout Property for a,m and b,m areequivalent, this implies that gcd(a,m) = gcd(b,m).