Congruence (Number Theory) is an Equivalence
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Theorem
For all $z \in \R$, congruence modulo $z$ is an equivalence relation.
Proof
Checking in turn each of the critera for equivalence:
Reflexive
We have that Equal Numbers are Congruent:
- $\forall x, y, z \in \R: x = y \implies x \equiv y \bmod z$
so it follows that:
- $\forall x \in \R: x \equiv x \bmod z$
and so congruence modulo $z$ is reflexive.
Symmetric
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle x\) | \(\equiv\) | \(\displaystyle y \bmod z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle x - y\) | \(=\) | \(\displaystyle k z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle y - x\) | \(=\) | \(\displaystyle \left({-k}\right) z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle y\) | \(\equiv\) | \(\displaystyle x \bmod z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
So congruence modulo $z$ is symmetric.
Transitive
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle x_1\) | \(\equiv\) | \(\displaystyle x_2 \bmod z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \land\) | \(\displaystyle \) | \(\displaystyle x_2\) | \(\equiv\) | \(\displaystyle x_3 \bmod z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \left({x_1 - x_2}\right)\) | \(=\) | \(\displaystyle k_1 z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \land\) | \(\displaystyle \) | \(\displaystyle \left({x_2 - x_3}\right)\) | \(=\) | \(\displaystyle k_2 z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \left({x_1 - x_2}\right) + \left({x_2 - x_3}\right)\) | \(=\) | \(\displaystyle \left({k_1 + k_2}\right) z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \left({x_1 - x_3}\right)\) | \(=\) | \(\displaystyle \left({k_1 + k_2}\right) z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle x_1\) | \(\equiv\) | \(\displaystyle x_3 \bmod z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
So congruence modulo $z$ is transitive.
So we are justified in supposing that congruence, as we have defined it, is an equivalence.
$\blacksquare$
Sources
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 2.5$
- Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 11$: Example $11.2$
- George McCarty: Topology: An Introduction with Application to Topological Groups (1967): $\text{I}$
- C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 1.6$: Theorem $3$
- George E. Andrews: Number Theory (1971): $\S 4.1$: Theorem $4.1$
- T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 6$: Example $6.5$
- Gary Chartrand: Introductory Graph Theory (1977): Appendix $\text{A}.3$: Theorem $\text{A}.3$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 14.1$
- John F. Humphreys: A Course in Group Theory (1996): $\S 2$: Example $2.26$