# Discrete mathematics subsets

Suppose I have two sets A and B:

$$A = \lbrace 2k-1 : k \in \mathbb{Z}\rbrace$$ $$B = \lbrace 2l+1 : l \in \mathbb{Z}\rbrace$$

I need to prove that A = B.

I know that to prove equality between two sets I need to prove both:

$$A \subseteq B$$

and

$$A \supseteq B$$

Suppose x is an element of A, then $$x = 2k - 1$$

EDIT : Which we can rewrite as

$$x = 2k + 1 - 2$$ $$x = 2(k-1) + 1$$

Because $$(k-1) \in \mathbb{Z}$$

We know that x is also in B.

Is this the correct way of approaching this problem?

• You know that $x = 2k-1$ for some integer $k$. Can you find a (possibly different) integer $l$ for which $x = 2l + 1$? If so, this will prove that $x \in B$.
– Dave
Sep 2, 2014 at 4:52
• Edited my original post, does it look like I'm headed in the right direction? Sep 2, 2014 at 5:01
• Yes, that's good. An improvement would be to relate your equation to the definition of the set $B$ by using the letter $l$, but your answer is correct as it is to show $A \subseteq B$. The main improvement would be to add "for some integer $k$" after you first introduce $k$. You never say what $k$ is.
– Dave
Sep 2, 2014 at 5:14

An alternate way of doing the same thing is to use the set definitions :

$A=\{2k-1:k\in \mathbb{Z}\}$

    $=\{2(l+1)-1:l\in \mathbb{Z}\}$

    $=\{2l+2-1:l\in \mathbb{Z}\}$

    $=\{2l+1:l\in \mathbb{Z}\}$

    $=B$

This actually takes care of both $A\subseteq B$ and $B\subseteq A.$

Hint:

• $2k-1=2l+1$ where $l=(k-1)$;
• $2l+1=2k-1$ where $k=l+1$.

Your approach is correct, but just to help give you an example in writing out your mathematical argument to give you some confidence, I would write it like this:

Suppose $x \in A$. Then there exists a $k \in \mathbb{Z}$ such that $x = 2k - 1 = 2l + 1$ where $l = k - 1$ (following the hint above/what you wrote). Since $k \in \mathbb{Z}$, $l \in \mathbb{Z}$, so $x \in B$.

See if you can write out the $B \subseteq A$ direction. Hope that helps!

Let $x\in A.$ Then $x=2k-1$ for some $k\in \mathbb{Z}.$ Also since $k\in \mathbb{Z}, k=l+1$ for some $l\in \mathbb{Z}.$

So $x=2(l+1)-1=2l+2-1=2l+1$ so that $A\subseteq B.$ Similarly,

let $x\in B,$ then $x=2l+1$ for some $l\in \mathbb{Z}.$ Again by same arguement as above we can show that $B\subseteq A$ and hence $A=B.$