Let $A = \{4X + 1| X \in \mathbb{Z}\}$ and $B = \{4Y − 3| Y \in \mathbb{Z}\}$. Prove that $A = B$.
So we know need to prove that two sets are equal by showing that each set is a subset of the other.
Claim $1: A \subseteq B$ Proof of claim $1$: Let $n \in A$. Then there exists some $k \in \mathbb{Z}$ such that $n = 4X + 1$. Show that $n \in B$ by showing that there exists an integer $Y$ such that $n = 4Y − 3$. $n = 4X + 1, n = 4Y-3, \implies 4X+1 = 4Y-3$. then $Y= X+1$, therefore $n=4X+1=4(Y+1)-3$.
My questions:
- Why solving this equation can prove that $A \subseteq B$? My understand is that since we can use $Y+1$ to represents every $X$ where $X \in \mathbb{Z}$. then ______.
Claim $2: B \subseteq A$ . ........
Your ultimate goal is to show that $A \subseteq B$. We do so by taking $n \in A$ and we want to show that $n \in B$.
The way to show $n \in B$ is to show the existence of one $Y$ such that $n=4Y-3$.
Solving the equation exhibit the existence of such $Y \in \mathbb{Z}$.
Since $n \in A$, we know there exists such as $X \in \mathbb{Z}$ such that $n=4X+1$, from there we want to construct this particular $Y$ and one simple way is to let $4X+1 = 4Y - 3$ and show that you can solve for $Y \in \mathbb{Z}$.
For the reverse direction, you would start by taking an element in $B$ and show that it is in $A$ and you will end up solving the same equation again, but this time round, you started with $Y \in \mathbb{Z}$ and you want to check that $X \in \mathbb{Z}$.