Prove that$ A \subseteq B$ where $A = \{ x \in Z \vert \ x\equiv 4\pmod 9 \}\ $ and $\ B = \{ x \in Z \vert \ x\equiv 1\pmod 3 \}$

Question

I am really confused that how can I demonstrate please help me to explain this

Prove that$ A \subseteq B$ where $A = \{ x \in Z \vert \ x\equiv 4\pmod 9 \}\ $ and $\ B = \{ x \in Z \vert \ x\equiv 1\pmod 3 \}$.

How it is possible to demonstrate that $A$ is a subset of $B$ ?

Answer 1

Suppose $x \in A$. Then $x = 9n + 4$ for some $n \in \mathbb{Z}$.

But then we can also write $x$ as $$x = 9n + 4 = 9n + 3 + 1 = 3(3n+1) + 1.$$ Now as $n$ is an integer, so too is $3n+1$, which we put $m$.

Thus, $$x = 3m+1,$$ where $m = 3n+1 \in \mathbb{Z}$. So $$x \cong 1 ( \mod 3),$$ showing that $x \in B$. Hence $A \subset B$.

Answer 2

Here we will show $A \subseteq B$ by inclusion. Let $a \in A$ be an arbitrary element. By the definition of modular arithmetic, we know since $x \equiv 4 \text{ (mod} \ 9)$ there must exist some integer $z_1 \in \mathbb{Z}$ such that $$x = 4 + 9z_1.$$ Notice that this implies $$x = 1 + 3 + 9z_1 = 1 + 3(1+3z_1).$$

Since the quantity $(1 + 3z_1)$ is an integer (being a linear combination of integers), and all integer multiples of 3 are equivalent to 0 modulo 3, it must be that $$x \equiv 1 \text{ (mod} \ 3)$$