# need help with equivalence class

• Dec 18th 2008, 12:44 AM
tukilala
need help with equivalence class
For x belonging to Z, let us denote by [x] equivalence class which contains x. Prove that if [x]=[x'] and [y]=[y'] then [x+y]=[x'+y'] and [xy]=[x'y']
Remark: When xRy people say that "x and y are congruent mod 3", and sometimes write x equivalent y (mod3)

someone can help me please here???
• Dec 19th 2008, 12:35 AM
CaptainBlack
Quote:

Originally Posted by tukilala
For x belonging to Z, let us denote by [x] equivalence class which contains x. Prove that if [x]=[x'] and [y]=[y'] then [x+y]=[x'+y'] and [xy]=[x'y']
Remark: When xRy people say that "x and y are congruent mod 3", and sometimes write x equivalent y (mod3)

someone can help me please here???

If $[x]$ denotes the equivalence class in $\mathbb{Z}$ modulo $3$ then all elements of $[x]$ have the same remainder as $x$ when divided by $3$, and any integer which has this remainder is in $[x]$.

So:

$[x]=[x']$ and $[y]=[y']$ means there are integers $a,\ b,\ d,\ e,\ r_1,\ r_2$ where $r_1,\ r_2 \in \{0,1,2\}$ such that:

$
x=3a+r_1,\ x'=3b+r_1,\ y=3d+r_2,\ y'=3e+r_2
$

Then $x+y=3(a+d)+(r_1+r_2)$, and so $[x+y]=[r_1+r_2]$

Similarly:

$x'+y'=3(b+e)+(r_1+r_2)$, and so $[x'+y']=[r_1+r_2]$

Hence: $[x+y]=[x'+y']$

CB