1. ## congruence

Recall that 3/k if and only if there exist nEz at k=3n. Let a and b be integers.

Define (equal sign with three bars and subscript 3) as a(equal sign with three bars and subscript 3) b if and only if 3/(a-b)

Prove the fallowing

Let [0]={xEz: 0 (equal sign with three bars and subscript 3) x}
[1]={xEz: 1(equal sign with three bars and subscript 3)x}
[2]={xEz: 2(equal sign with three bars and subscript 3)x}

Part B is

Show that the sets [0] [1] [2] partition Z

2. Originally Posted by xomichelleybelly
Recall that 3/k if and only if there exist nEz at k=3n. Let a and b be integers.

Define (equal sign with three bars and subscript 3) as a(equal sign with three bars and subscript 3) b if and only if 3/(a-b)

Prove the fallowing

Let [0]={xEz: 0 (equal sign with three bars and subscript 3) x}
[1]={xEz: 1(equal sign with three bars and subscript 3)x}
[2]={xEz: 2(equal sign with three bars and subscript 3)x}

Part B is

Show that the sets [0] [1] [2] partition Z
to start you off. as the definition tells us. $a \equiv _n b$ means $n \mid (b - a)$, which means $b - a = nk$ for some $k \in \mathbb{Z}$.

Thus, $[0] = \{ x : x = 3k \mbox{ for }k \in \mathbb{Z} \}$ $= \{ \mbox{ all multiples of 3} \} = \{ \cdots , -6, -3, 0, 3, 6, \cdots \}$

do the others. why do these partition $\mathbb{Z}$? how can we express any integer in terms of 3 times something, plus a remainder perhaps?