“Overthinking It: The Unnecessary Application of Advanced Mathematics to Popular Culture”
Sean Nixon, Geneseo
Wed. Nov. 2, 2:30 PM, Newton 203
Extra credit for going and giving me a paragraph or so re connections you make to the ideas
GROW STEM Speaker
“Everything You Need to Know About Succeeding in STEM (But Were Afraid to Ask)”
Phillip A. Ortiz, SUNY Empire State STEM Learning Network
Wednesday, Nov. 2, 4:00 PM, Newton 204
Questions?
Proofs of Set Relationships
Section 5.2
Prove { 4n | n ∈ Z } ⊆ { 2n | n ∈ Z }
To show S ⊆ T
“Choose an element” method
Pick arbitrary/generic element of S
Prove that element also in T
Let x be in { 4n | n ∈ Z }, so x = 2(2m) for integer m so (2m) is an integer meaning that
x ∈ { 2n | n ∈ Z }
In fact, prove { 4n | n ∈ Z } ⊂ { 2n | n ∈ Z } ⊂ Z
Show there’s an element of T not in S
2 ∈ { 2n | n ∈ Z } but 2 ∉ { 4n | n ∈ Z }
{ 2n | n ∈ Z } ⊂ Z b/c 1 ∈ Z but 1 ∉ { 2n | n ∈ Z }
But prove { 4n | n ∈ Q } = Q
Prove each set is a subset of the other
Prove x ∈ S iff x ∈ T
First direction: let x ∈ { 4n | n ∈ Q }, so x = 4n and n = a/b for integers a, b, b ≠ 0 so x = 4a/b
which is a ratio of 2 integers (b≠0) and so 4a/b is in Q
Second direction: let x ∈ Q so x = a/b for integers a, b, b ≠ 0 x = 4 (a/4b)
= 4n for n = a/4b, thus x ∈ { 4n | n ∈ Q } b/c a and 4b are both integers,
4b ≠ 0
Prove that { x ∈ Z | x ≡ 3 (mod 12) } is disjoint from
{ y ∈ Z | y ≡ 2 (mod 8) } (progress check 5.15)
To prove A disjoint from B
Prove A ∩ B = ∅
Prove emptiness by contradiction
Assume { x ∈ Z | x ≡ 3 (mod 12) } ∩ { y ∈ Z | y ≡ 2 (mod 8) }
≠ ∅
So there exists a such that a-3 = 12k and a-2 = 8c for integers c,k