S ∪ T = {x | (x ∈ S ) ∨ (x ∈ T )}
S ∩ T = {x | (x ∈ S) ∧ (x ∈ T)}
Set of all elements in S but not T.
S - T = {x | (x ∈ S) and ∉ T} = {x | (x ∈ S) ∧ (x ∉ T)}
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S = {x | x ∉ S}
S × T ={ (x,y) |x ∈ S,y ∈ T}
What common sets do the following sets refer to?
A {m ∈ \(\mathbb{Z}\):2|m} ∪ {2k + 1:k ∈ Z}
B {m ∈ \(\mathbb{Z}\):2|m} ∩ {2k + 1:k ∈ Z}
A This is the set of all integers
B This is an empty set
A set S is called a subset of a set T, written S ⊆ T, when the implication “if x ∈ S then x ∈ T” is true for every x in the universe of discourse.
To prove S ⊆ T , we prove the implication (x ∈ S) => (x ∈ T).
Prove:
{n ∈ \(\mathbb{N}\) : 5 | n + 2} ⊆ {2k + 1 : k ∈ \(\mathbb{N}\)}
when n = 8
8 ∈ {n ∈ \(\mathbb{N}\) : 5 | n + 2}
but 8 ∉ {2k + 1 : k ∈ \(\mathbb{N}\)}
If for any integers x and y, a | (bx + cy)
Then a | b and a | c
Proof:
Assume for any integer x,y a | (bx + cy)
then when x = 1, y = 0
a | (b(1) + c(0)), i.e. a | b
Use specific scenarios when it is true
then when x = 0, y = 1
a | (b(0) + c(1)), i.e. a | c
To prove S ⊆ T we prove the implication
x ∈ S and x ∈ T
Two sets S and T are equal i.e. S = T,
S and T have the same elements
[(x ∈ S) ∧ (x ∈ T) AND (x ∈ T) ∧ (x ∈ S)]
i.e. we prove
(1) S ⊆ T
(2) T ⊆ S