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I'm trying to prove that if $N_1\mod(a) = n_1$ and $N_2\mod(a) = n_2$ then $(N_1+N_2)\mod(a) = (n_1+n_2)\mod(a)$
By our assumption $N_1 = am_1 + n_1$ and $N_2 = am_2 + n_2$
So $N_1+N_2 = a(m_1+m_2) + n_1 + n_2$
So $(N_1+N_2)\mod(a) = (a(m_1+m_2) + n_1 + n_2)\mod(a)$
Now my justification was long and considered the case where $n_1+n_2 < a$ and $= a$ and $>$ then a however the solutions I am using to check simply said both $n_1$ and $n_2$ are between $0$ and $a-1$ and so the expression is true. It feels like their reasoning is missing a few steps because I'm not sure how they can go from $n_1$ and $n_2$ between $0$ and $a-1$ to the identity is true. Am I missing something?