how can I prove the set $A=\{4k+1: k \in \mathbb Z\}$ is equal to set $B = \{4k-3: k \in \mathbb Z\}$

Question

I want to know how I can approach this problem. I know I have to show A and B are subsets of one another by picking an arbitrary element x from one set and show its a member of the other set, but idk where to go from there. Sorry, if this question seems very trivial to some of you, this is my first time doing proofs

Answer 1

Yes, you want to prove that set A is a subset of set B and that set B is a subset of set A.

To prove "set A is a subset of set B" you need to prove that if n is in A then it is also in B. If n is in A then n= 4k+ 1 for some integer k. Let j= k+ 1. Then k= j- 1 so n= 4(j- 1)+ 1= 4j- 4+ 1= 4j- 3. So n is in B.

To prove "set B is a subset of set A" you need to prove that if n is in B then it is also in A. If n is in B then n= 4k- 3 for some integer k. Let j= k- 1. Then k= j+ 1 so n= 4(j+ 1)- 3= 4j+ 4- 3= 4j+ 1.

Answer 2

For simplicity lets change variable in definition of $B$: $B = \{4k' - 3| k' \in \mathbb Z\}$.

Assume $x \in A$. Then for some $k$ we have $x = 4k + 1$. Then for $k' = (k + 1)$ we have $x = 4k' - 3$. So $x \in B$.

Can you finish for the other direction?

Answer 3

For every real $x$ we have $$x\in A \iff \frac {x-1}{4}\in \Bbb Z \iff $$ $$\iff \frac {x-1}{4}+1\in \Bbb Z \iff$$ $$\iff \frac {x+3}{4}\in \Bbb Z \iff$$ $$\iff x\in B.$$