Solution to congruence I am studying abstract mathematics and I came across this in my textbook.
Example: Find a solution to the congruence $$5x\equiv11\pmod{19}$$
It starts off the solution with: If there is a solution, then, by Theorem 3.1.4, there is a solution within the set $\{0,1,2,\ldots,18\}$.
Theorem 3.1.4 states: For a given modulus $m$, each integer is congruent to exactly one of the numbers in this set $\{0,1,2,\ldots,m-1\}$.
My question is: how does the theorem imply that $x$ must be in this above set?
 A: Suppose there is a solution for some integer $x$. The theorem states:

For a given modulus m, each integer is congruent to exactly one of the numbers in this set {0,1,2,…,m−1}.

Replace "each integer" with $x$.
A: Good critique. The theorem does not imply that by itself. Rather, we need to use further properties of congruences to make that inference. Let's derive that inference in further detail.
Working $\bmod 19,\,$ if an integer $\,n\,$ is a root of $\,5x\equiv 11$ then the theorem implies that $\,\color{#c00}{n\equiv \bar n}\,$ for $\,0\le n < 19\ $ (viz. $\,\bar n = n\bmod 19).\,$ Then by the Congruence Product Rule $\, 11\equiv 5\color{#c00}n\equiv 5 \color{#c00}{\bar n},\,$ so $\,5\bar n\equiv 11\,$ by transitivity of congruence, i.e if $\,n\,$ is a root then so too is every integer $\,\bar n\equiv n.$
More generally, if $\,f(x)\,$ is a polynomial with integer coef's and $n$ is a root of $\,f(x)\,$ then so too is every $\bar n\equiv n,\,$ e.g. $\,\bar n = n\bmod 19$, which follows by applying the Polynomial Congruence Rule $\,\bar n\equiv n\Rightarrow f(\bar n)\equiv f(n)\ $ (which holds because polynomials are compositions of sums and products, so we can inductively apply the Congruence and Power Rules - see the linked proof).
