A = {n ∈ N | n ≥ 1 and n = 4j − 3 for some j ∈ N}
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.


2 Answers 2


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 $n=2k+1 \in B$


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