# Prove that A ⊆ B. Here, N = {0, 1, 2, 3, . . . }

Let
A = {n ∈ N | n ≥ 1 and n = 4j − 3 for some j ∈ N}
and
B = {n ∈ N | n ≥ 0 and n = 2k + 1 for some k ∈ N}.

Prove that A ⊆ B. Here, N = {0, 1, 2, 3, . . . }.

So I'm preparing for an exam in Discrete math, I came up across this question and can't seem to get the answer. I've tried using some of the set identities but I don't know how to get the answer.

Clearly $$A$$ is non-empty, so take some arbitrary $$n\in A$$. Then $$n=4j-3$$ for some $$j>0$$ (convince yourself that if $$j=0$$ then $$n$$ could not be in $$A$$). It follows that: $$n=4j-3=4(j-1)+1=2(2j-2)+1$$ Since $$j>0$$, $$(2j-2)\in N$$, so $$n\in B$$. Since $$n$$ was arbitrary, $$A\subseteq B$$.
In order to show $$A\subseteq B$$ we have to show that every element of $$A$$ is an element of $$B$$.
Let $$n=4j-3$$ be an arbitrary element of $$A$$. Note that $$4j-3 =4(j-1)+2= 2(2j-2)+1$$
Thus if you let $$k=2j-2$$ then we have $$n=2k+1 \in B$$