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I'm learning modular arithmetic right now and having a small hang-up in the application of the definition of a congruence.

The book states that a congruence $c \equiv d$ (mod $n$) can be written if

  • $(c - d) = kn$, i.e., $(c - d)$ is divisible by $n$

The problem I am having arises during the proof of the Chinese Remainder Theorem. Specifically, the line that states Let $x_{1} = 1 - sa = tb$ then we see that:

  • $x_{1} \equiv 1$ (mod $a$)
  • $x_{1} \equiv 0$ (mod $b$)

Take the first congruence: Wouldn't it be $x_{1} \equiv -1$ (mod $a$) i.e., $a-1$ (mod $a$)?

  • Since the expression $x_{1} = 1 -sa$ can be rewritten as $1 - x_{1} = sa$ and thus we could equate $1 = c$ and $x_{1} = d$ with the definition.

Since I've been writing this I reviewed the rules of congruence and noticed the rule:

  • $\forall c,d \in \mathbb{Z_{n}}$, $a \equiv b$ (mod $n$) iff $b \equiv a$ (mod $n$)
  • Would this rule be the reason why we can say $x_{1} \equiv 1$ (mod $n$)?

Or would it simply just be the case that if something is divisible by $kn$ then it is also divisible by $-kn$? Hence we could just write the congruence in question as $x_{1} - 1 = -kn$ and have that still apply without having to fiddle around with the $c$ and $d$ from the original definition?

Numerical Disintegration
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  • Yes, anything divisible by $r$ is also divisible by $-r$. $x\equiv1\pmod a$ because $x-1=-sa$ is a multiple of $a$. – Gerry Myerson Dec 25 '21 at 17:22
  • Thanks, Gerry, for the verification. – Numerical Disintegration Dec 25 '21 at 17:24
  • $!\bmod, \color{#c00}a!:,\ 1-sa\equiv 1\ $ (proof: $,\color{#c00}{a\equiv 0},\Rightarrow 1-s,\color{#c00}a\equiv 1-s,\color{#c00}0\equiv 1,$ by Congruence Sum & Product Rules) thus $x,\equiv 1-sa\equiv 1\Rightarrow x\equiv 1,$ by transitivity of congruence (being an equivalence relation). It is better to quickly master such equational congruence arithmetic (vs. divisibility relational arithmetic) since it allows us to reuse our well-practiced intuition on arithmetical equations - going back to grade school. – Bill Dubuque Dec 25 '21 at 17:56
  • Otoh, if you keep unwinding congruences into their definition in terms of the divisibility relation then that will obfuscate that innate arithmetical key ideas at the heart of many modular arithmetic applications. – Bill Dubuque Dec 25 '21 at 18:01

1 Answers1

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Let $x_1=1−sa=tb$.

Then $x_1-1 = (-s)a$ and so $x_1\equiv 1\mod a$.

Wuestenfux
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