Words
Induction
100

It's a homomorphism that's a bijection

What is an isomorphism?

200

When we assume that the statement we are proving is false and then everything gets all wacky and goofy

What is proof by contradiction?

300

A function f: A->B is this if for every element b in B there exists element a in A with f(a)=b

What is surjective?

400

One of them relations on a set that's reflexive, symmetric, and transitive.

What is an equivalence relation?

400

Proof that if G is an abelian group then for all a, b in G and for any positive integer n we have (ab)ⁿ=aⁿbⁿ

What is:

We proceed by induction on n.

Base case: n=1

(ab)¹=ab=a¹b¹

Inductive Hypothesis: We assume (ab)ⁿ=aⁿbⁿ

Inductive Step: We want to show that (ab)ⁿ+¹=aⁿ+¹bⁿ+¹

Because G is abelian and because we assumed (ab)ⁿ=aⁿbⁿ we have:

(ab)ⁿ+¹=ab(ab)ⁿ=a(ab)ⁿb=aaⁿbⁿb=aⁿ+¹bⁿ+¹

We conclude that (ab)ⁿ=aⁿbⁿ for any positive integer n.

500
An equation for which we are only interested in integer solutions.

What is a Diophantine equation?

500

Proof that n³-n+3 is divisible by 3 for all positive integers n using mathematical induction

What is:

Base Case: n=1

(1)³-1+3=3. 3 is divisible by 3.

Inductive Hypothesis: We assume that n³-n+3=3k, where k is some integer.

Inductive Step: We want to show that (n+1)³-(n+1)+3=3j where j is some integer. 

(n+1)³-(n+1)+3=n³+3n²+2n+3=(n³-n+3)+3n²+3n

We assumed in our inductive hypothesis that n³-n+3=3k so we have

3k+3n²+3n=3(k+n²+n)

Therefore, (n+1)³-(n+1)+3 is divisible by 3 so we have proven that the n+1 case works. We conclude that n³-n+3 is divisible by 3 for all positive integers n.

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