Congruence Relation/Examples/Equal Sine of pi x over 6 on Integers for Addition
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Example of Congruence Relation
Let $\Z$ denote the set of integers.
Let $\RR$ denote the relation on $\Z$ defined as:
- $\forall x, y \in \Z: x \mathrel \RR y \iff \sin \dfrac {\pi x} 6 = \sin \dfrac {\pi y} 6$
Then $\RR$ is not a congruence relation for addition on $\Z$.
Proof
Note that by Equivalence Relation Examples: Equal $\sin \dfrac {\pi x} 6$ on Integers, $\RR$ is an equivalence relation.
However:
\(\ds \sin \dfrac \pi 6\) | \(=\) | \(\ds \sin \dfrac {5 \pi} 6\) | \(\ds = \dfrac 1 2\) | |||||||||||
\(\, \ds \land \, \) | \(\ds \sin \dfrac {2 \pi} 6\) | \(=\) | \(\ds \sin \dfrac {4 \pi} 6\) | \(\ds = \dfrac {\sqrt 3} 2\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(\RR\) | \(\ds 5\) | |||||||||||
\(\, \ds \land \, \) | \(\ds 2\) | \(\RR\) | \(\ds 4\) |
But:
- $\sin \dfrac {\paren {1 + 2} \pi} 6 = 1$
while:
- $\sin \dfrac {\paren {4 + 5} \pi} 6 = -1$
So while we have:
- $\paren {x_1 \mathrel \RR x_2} \land \paren {y_1 \mathrel \RR y_2}$
where $x_1 = 1$, $x_2 = 5$, $y_1 = 2$, $y_2 = 4$
we have:
- $\paren {x_1 + y_1} \not \mathrel \RR \paren {x_2 + y_2}$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.1 \ \text{(b)}$