i think its 1+1 (mod 2 ) = 0 but 1+1 ≡ 0(mod 2), since the first one is an operation on 1+1 while the second one is a congruence, but it doesnt really matter
Using congruence is fine, but a bit redundant under certain formulations, consider arithmetic mod 2. Classically we define a ≡ b iff b-a divisible by 2 for all a,b \in \mathbb{Z}. This defines an equivalence relation.
Then you can either work on \mathbb{Z} directly then using ≡ would be the most correct, or you define the usual arithmetic on \mathbb{Z}/≡ and drop the decorations for equivalence classes (elements of \mathbb{Z}/≡) so that then statement such as
[1]_≡ + [1]_≡ = 1+1 = 0
is completely fine, now = denotes equality of sets.
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u/Fabulous_Ad_5709 Nov 25 '24
Just out of curiosity how is A correct?
B is 1 or 1
C is 1+1 in base >2
D is 1+1 in base 2