Description

Ahmed Semih Ozmekik No: 171044039
1. (5 points.) Prove that B ⊂ A where
A = {(a,b) | |a + b| < 21} and B = {(a,b) | |a − 1| < 10 and |b − 1| < 9}.
A = {(a,b) | −21 < a + b < 21} B = {(a,b) | −9 < a < 11 and −8 < b < 10}
Rearrange B :
B = {(a,b) | (−9 < a < 11) + (−8 < b < 10)}
B = {(a,b) | (−17 < a + b < 21)}
A has every pair of B has.(Not vice versa. ∴ B * A) ∴ B ⊂ A
2. (10 points.) Write the expressions which represent the sets, given the Venn diagrams in Figure 1.

Figure 1:
1) (A (B ∪ C)) ∪ (D ∪ E)
2) (B C) ∪ (F ∩ E) ∪ D
3) (E ∩ (G ∪ F)) ∪ (H ∩ (G ∪ F)) (G ∩ F) ∪ (B (A ∪ C)) ∪ (C (B ∪ E ∪ D))
3. (10 points.) Let R be the relation on R2, defined by:
(a1,b1) R (a2,b2) if and only if either a1 < a2 or both a1 = a2 and b1 ≤ b2.
Prove that R is a partial order on R2.
R is Reflexive: for all (a,b) ∈ R2, we have [(a,b),(a,b)] ∈ R Since a = a and b ≤ b.
R is Transitive: Whenever [(a1,b1),(a2,b2)] ∈ R and [(a2,b2),(a3,b3)] ∈ R, we have [(a1,b1),(a3,b3)] ∈ R. Since, if a1 < a2 and a2 < a3, therefore a1 < a3 holds; or if a1 = a2 and b1 ≤ b2, a2 = a3 and b2 ≤ b3, therefore a1 = a3 and b1 ≤ b3 holds.
R is Antisymmetric: Whenever (a1,b1) 6= (a2,b2) and [(a1,b1),(a2,b2)] ∈ R, we have [(a2,b2),(a1,b1)] ∈/ R. Since, if a1 < a2, then a2 ≮ a1, so there is no such element. Or, when a1 = a2 and b1 ≤ b2, if a2 = a1 and b2 ≤ b1. Therefore a2 = a1 and b2 = b1,
(a1,b1) = (a2,b2). Which conflicts with our assumption, therefore [(a2,b2),(a1,b1)] ∈/ R.
4. (15 points.) Let a partial order relation R on X = {n ∈ Z : 2 ≤ n ≤ 12}, defined by: xRy if and only if either (x is prime and x < y) or (x divides y).
Draw a Hasse diagram for R, and identify the least element and the maximal elements.
12

Maximal Elements: 12,9,10,8 Least Element: 2
5. (10 points.) Examine all Boolean lattices with cardinalities 1−5. Explain these lattice structures in detail.
6. (10 points.) Determine whether each of the following functions is injective, surjective, both or neither.
(a) f : Z → Z,f(n) = n + (−1)n
(b) f : R → R,f(n) = 2x
(c) f : R → R+ ∪ {0},f(n) = x + |x|
(a) Let x,y,k,l ∈ Z
Every image has preimage. It is surjective.
Say x = 2k and y = 2l. x + 1 = y + 1 ∴ x = y.
Say x = 2k + 1 and y = 2l + 1. x − 1 = y − 1 ∴ x = y. It is injective.
(b) Let x,y ∈ R
There is no image x such that f(x) = 0. It is not surjective.
2x = 2y ∴ x = y It is injective.
(c) Let x,y ∈ R
Every image has a preimage. It is surjective.
If x,y < 0 then, f(x) = f(y) = 0. It is not injective.
7. (15 points.) Let f : X → Y be a function.
Show that function f is bijective if and only if f(X −Z) = Y −f(Z), for every subset Z of X.
Let X = X + Z f(X + Z − Z) = f(X) + f(Z) − f(Z)
f(X) = f(X)
∴ f is bijective.
8. (15 points.) Let S denote the set of real 2 × 2 matrices of the form
where a and b are not both zero. Show that S is a group under the operation of matrix multiplication.
To be able to be a group, one must satisfy these things:
(a) Must be closed under the operation .
(b) Must be Associativitive.
(c) Must have identity element.
(d) Must have inverse element.
(a) It is closed, under operation of matrix multiplication. a,b ∈ R. Thus, a2+b2 ∈ R.

(d) An element exists in that group such that, a,b ∈ R.
Thus, a2 + b2 ∈ R and .
is the inverse element.)
9. (10 points.) Define the ring Zn. Show that Zn is a field if and only if n is a prime number.
A ring is a set S with two defined binary operators satisfying the following conditions:
(a) Additive associativity.
(b) Additive commutativity.
(c) Additive identity.
(d) Additive inverse.
(e) Left and right distributivity.
(f) Multiplicative associativity.
For a ring to be a field, also four operations must be defined. n=5 case:(n is a prime number)
1 2 3 4
1 1 2 3 4
2 2 4 1 3
3 3 1 4 2
4 4 3 2 1
n=6 case:(n is not a prime number)
1 2 3 4 5
1 1 2 3 4 5
2 2 4 0 2 4
3 3 0 4 0 3
4 4 3 0 4 2
5 5 4 3 2 1
Some of the element has no inverse element.
∴ This is not a field. Besides, having 0 in the table causing this ring not to become field anymore, since this is against the identity element 1.