chinese remainder theorem constructive proof
Olivia Zamora
I am trying to understand CRT constructive proof from wikipedia [
I am unable to follow it from x = $\sum\limits_{i=0}^{k} a_i * e_i$ this statements onward. Please help me with this.
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$\begingroup$That sum works out to $x = a_1e_1 + a_2e_2 + ... + a_ke_k$
To see why this is a solution of the simultaneous congruences, take the congruences in turn.
Modulo $n_1$,
$x \equiv a_1e_1 + a_2e_2 + ... + a_ke_k \pmod {n_1} \implies x \equiv a_1.1 + a_2.0 + ... + a_k.0 \pmod {n_1} \implies x \equiv a_1 \pmod {n_1}$
This is because $e_1$ has been carefully chosen to be $1$ modulo $n_1$ and $0$ modulo all other $n$ values under consideration.
Similarly, modulo $n_2$,
$x \equiv a_1e_1 + a_2e_2 + ... + a_ke_k \pmod {n_2} \implies x \equiv a_1.0 + a_2.1 + ... + a_k.0 \pmod {n_2} \implies x \equiv a_2 \pmod {n_2}$
Now, modulo arbitrary $n_r$ such that $(1 \leq r \leq k)$,
$x \equiv a_1e_1 + a_2e_2 + ... + a_re_r + ... + a_ke_k \pmod {n_r} \implies x \equiv a_1.0 + a_2.0 + ... + a_r.1 + ... + a_k.0 \pmod {n_r} \implies x \equiv a_r \pmod {n_r}$
This should allow you to see why that $x$ value solves each individual congruence and therefore all the congruences simultaneously.
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