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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?

Bernard
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Nabeel Rafid
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  • Because $19-1=18$? – Théophile Oct 20 '20 at 20:44
  • Aha. So with the edit, is your question really about why $0$ disappeared? – Théophile Oct 20 '20 at 20:49
  • When a number is divided by 19, what about possible remainders? – Lion Heart Oct 20 '20 at 20:49
  • Hint: $5\cdot 4\equiv 1$, $5\cdot 4\cdot ?\equiv 11$? – Lynnx Oct 20 '20 at 20:52
  • @Théophile my bad thanks for pointing out the zero's, I know the possible remainders of an integer divided is within the set, but how does that imply the x in 5x also has to be within the set – Nabeel Rafid Oct 20 '20 at 20:52
  • I see. The $5$ doesn't matter. The theorem is really just telling you a property about all integers, including $x$. – Théophile Oct 20 '20 at 20:55
  • are you asking why does the theorem imply there exists some $l \in \mathbb{Z}$ such that $x = \frac{19l+11}{5} \in {0,1,...,19-1}$? – homosapien Oct 20 '20 at 20:57
  • @HossienSahebjame The theorem doesn't imply that. The theorem is simply saying that all integers are congruent to something modulo $19$. Whether there's a solution to the equation or not is another matter. – Théophile Oct 20 '20 at 20:59
  • that's why I asked what I asked cause I do not see how it implies that either. – homosapien Oct 20 '20 at 20:59
  • My apologies, I should have better formulated my question. In a more general case if we have a,b,m that are fixed numbers and we know there is a solution to $ax\equiv b(\mod{m})$, how can we conclude that there is a solution for $x$ such that $x\in{0,1,...,m-1}$ – Nabeel Rafid Oct 20 '20 at 21:17
  • @Nabeel Your question is formulated fine as it is. – Bill Dubuque Oct 20 '20 at 21:32

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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).

Bill Dubuque
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