Prove that $a + 2b \equiv 0 \bmod 3$ is an equivalence relation

Question

Would this proof work to show that $R$ ={(x, y) ∈ ℤ²: x + 2y ≡₃ 0} is an equivalence relation?

Reflexive:

Let x ∈ ℤ. x + 2x= 3x, and 3x ≡ 0 mod 3 since 3 divides 3x with remainder 0.

Symmetric:

Let x, y ∈ ℤ. Suppose (x, y) ∈ $R$. Then x + 2y ≡ 0 mod 3 -> x ≡ -2y ≡ y ≡ 2x -> y + 2x ≡ 0 mod 3.

Transitive:

Edit: Oops definitely did not do the transitive part of the proof properly -- does this work instead?

Let x, y, z ∈ ℤ and suppose (x, y), (y, z) ∈ $R$. Then x + 2y ≡ y+ 2z ≡ 0 mod 3. x + 2y = 3m and y + 2z = 3n for some m, n in ℤ. Then x + 2(3n-2z) = 3m, so simplifying that shows x + 6n - 4c = 3n...? Not sure how to get the sign for c to be positive.

Transitive means that "$a\sim b$ and $b\sim c$ implies $a\sim c$" — lulu, Apr 23 '21 at 15:28
Transitive isn't supposed to be that if a~b and b~c then a~c? — Evangelos Yfantis, Apr 23 '21 at 15:32
$\bmod 3!:\ a+2b\equiv 0 \iff a\equiv -2b\equiv b,$ so you are done if you already know congruence is an equivalence relation (proof here) — Bill Dubuque, Apr 23 '21 at 15:34
In fact, $a+2b \equiv 0 \pmod 3$ is equivalent to $a \equiv b \pmod 3$. — Geoffrey Trang, Apr 23 '21 at 15:34
Or, w/o congruences note $,a+2b = (a-b)+3b,$ so $,3\mid a+2b\iff 3\mid a-b, $ then use the linked dupe to show the equivalent latter standard form is an equivalence relation. — Bill Dubuque, Apr 23 '21 at 15:53

score 1 · Answer 1 · answered Apr 23 '21 at 15:30

1

For proving that the relation is transitive, $$ a+2b = 3m ~~~;a,b,m \in \mathbb{Z}$$ $$ b+2c = 3n ~~~;c, n \in \mathbb{Z}$$ Add both the equations, $$a+3b+2c=3(m+n) $$ $$ \implies a+2c=3(m+n-b)$$ $$ \implies a+2c \equiv 0 \pmod{3}$$

answered Apr 23 '21 at 15:30

Ankit Saha

1,742

Thank you, I definitely did not have the transitive part right! – Apr 23 '21 at 15:35

Prove that $a + 2b \equiv 0 \bmod 3$ is an equivalence relation

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