Binary modulo operation

Question

Empirically, I can know that (a+b+c) mod 2 = (a-b-c) mod 2.

e.g.,)

1+2+3 = 6, 6 mod 2 = 0
1-2-3 = -4, -4 mod 2 = 0

1+2+4 = 7, 7 mod 2 = 1
1-2-4 = -5, -5 mod 2 = 1

It seems that it is only possible when we use binary modulo (mod 2).

Is there any formal proof for this?

Follows immediately from the linked congruence arithmetic rules (sum rule) and $,n\equiv -n\pmod{2},,$ and $,a\equiv b\pmod{n}\Rightarrow a\bmod n = b\bmod n$ — Bill Dubuque, Aug 10 '23 at 04:22

score 0 · Answer 1 · answered Nov 24 '17 at 13:37

0

Because $$a-b-c=a+b+c-2(b+c)\equiv(a+b+c)\operatorname{mod}2.$$

answered Nov 24 '17 at 13:37

Michael Rozenberg

194,933

score 0 · Answer 2 · answered Nov 24 '17 at 13:44

Sure! Simply add $b+c$ to both sides and note that $2x=0$ mod $2$ for all numbers $x$. So you get $a=a$ mod $2$ implies that $a+2(b+c)= a$ mod $2$. Now subtract $b+c$ on both sides to obtain $a+b+c=a-b-c$ mod $2$. Note that it really didn't matter that we had three numbers instead of $2$ in other words $a+b= a-b$ mod $2$. Another way to see it is the simple rule that addition and subtraction happen to coincide module $2$ ($+=-$ mod $2$). This last rule is often applied in for instance algebraic topology courses when people do not want to care about signs.

For higher numbers $n$ there are of course the very similar identities $a+(n-1)b=a-b$ mod $n$. Altough you won't in general have $a+b=a-b$ mod $n$ if $n>2$.

Binary modulo operation

2 Answers2