Consider this addition rule in modular arithmetic: If I take n=10, a=3, b=7
Then, on the left hand side, we have (3)mod10 + (7)mod10= 3 + 7 = 10
On the right hand side, we have 10 mod10, which is zero.
LHS is therefore not equal to RHS. What am I doing wrong?
$\mod 10$ is not the remainder function. $3 \mod 10$ does not equal the number $3$ and $3$ exactly.
$3 \pmod {10}$ means $3$ is a representative of any of the infinite numbers that are equivalent to $3$. (That is any one of $.....-17, -7,3,13, 23.....$etc.) And we aren't saying anything about specific numbers. We are saying things what classes of infinite numbers all have in common and who they relate.
If you have a number that is in the class $3$ represents. (Let's say the number $-2,317$) And you have a number in the class that $7$ represents. (Let's say $5,892,407$) and you add them together then the sum is going to be in the same class that $3+7$ is in.
And indeed $-2,317 + 5,892,407= 5,890,090$ and both $5,890,090$ and $10$ are both in the class $\{....., -30, -20, -10, 0, 10, 20, 30,....\}$. That is the class represent by $0$. (ALthough it could just as easily have bee represented by $10$, or by $20$, or $-130$. or....)
That is what the statement $3\pmod {10} + 7 \pmod {10} \equiv (3+7) \pmod {10}$ means.
Also notice that that is not an equal sign with two stripes; it is an equivalence sign with three stripes.
When we do modular arithmetic we can think of what we are doing it two equivalent ways:
1) (a bit abstract). We are doing math on entire classes of integers and $3$ is the set $\{...-17,-7, 3,13...\}$ and $7$ is the set $\{....-13,-3, 7,17,....\}$ and $3 + 7$ is an operation which is "set addition" where $\{...-17,-7, 3,13...\}+\{....-13,-3, 7,17,....\}= \{....., -20,-10,0,10,20...\}$ which is represented by $0$.
2) $\equiv$ is not $=$. The statement $3 + 7 \equiv 0 \pmod {10}$ is not saying anything about what $3+7$ or what $0$ is. It is saying $3+7=10$ and $0$ both belong in the same equivalence class.