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So the problem I have is to write all the properties that a relation has (reflexive, symmetric, transitive, irreflexive, antisymmetric). The problem is the congruence relation on the set of triangles.

I know it is reflexive because any triangle is of course congruent with itself.

For symmetric I am confused. If you have triangle $ABC$ that is congruent with triangle $DEF$, then $DEF$ is congruent with $ABC$. But is this true if they are not equal? Like, to prove antisymmetry you must show that $xRy$ and $yRx$ to where $x = y$. $ABC$ isCongruentTo $DEF$ so then $DEF$ isCongruentTo $ABC$, but $ABC = DEF$ so therefore antisymmetric? Can something be both symmetric and antisymmetric?

The other properties I feel confident with.

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  • $\begingroup$ Well, that depends on what you mean by "the set of triangles" and what you mean by "triangle $ABC$ is congruent to triangle $DEF$." Can you give rigorous definitions? $\endgroup$ – Cameron Buie May 6 '15 at 22:27
  • $\begingroup$ Unfortunately I cannot give a more specific definition. I am doing some problems out of a textbook and that's all it says. It just says to write down all the properties that the given relation satisfies and the relation given says "The congruence relation on the set of triangles." I too wish the book was more specific. $\endgroup$ – generic user007 May 6 '15 at 22:31
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    $\begingroup$ Two triangles can be congruent and unequal. Consider a unit equilateral triangle in Tuscaloosa and a unit equilateral triangle in Ouagadougou. They are congruent, but they are not equal because they do not share any vertices or edges—they are not in the same place. $\endgroup$ – MJD May 6 '15 at 22:35
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Suppose $R$ is a relation, and let $X$ be the domain of $R.$ I claim that if $R$ is both symmetric and antisymmetric, then $R$ is the equality relation on $X$--that is, $$R=[=]_X:=\bigl\{\langle x,x\rangle:x\in X\bigr\}.$$

On the one hand, if $\langle x,y\rangle\in R$--that is, $x\:R\:y$--then $x\in X$ by definition. By symmetry, $y\:R\:x,$ and by antisymmetry, $x=y.$ Hence, $R\subseteq [=]_X.$

On the other hand, if $z\in[=]_X,$ then $z=\langle x,x\rangle$ for some $x\in X,$ which by definition of $X$ means $x\:R\:y$ for some $y$. Again by symmetry and antisymmetry, $x=y,$ so $x\:R\:x,$ meaning $z=\langle x,x\rangle\in R.$ Hence, $[=]_X\subseteq R.$


The upshot, here, is that unless the book is claiming that there is only one equilateral triangle (for example) of side length $1$ (pick a unit), then your relation is symmetric, but not antisymmetric.

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For symmetric I am confused. If you have triangle ABC that is congruent with triangle DEF , then DEF is congruent with ABC . But is this true if they are not equal? Like, to prove antisymmetry you must show that xRy and yRx to where x=y . ABC isCongruentTo DEF so then DEF isCongruentTo ABC , but ABC=DEF so therefore antisymmetric? Can something be both symmetric and antisymmetric?

  • If a relation is symmetric, then $aRb \leftrightarrow bRa$ including when $a=b$.
  • If a relation is antisymmetric, then $aRb \wedge bRa$ only when $a=b$.
  • If a relation is both, then $aRb$ only when $a=b$.

However, is this the case for congruence?

In the Domain of Triangles.

What does congruence mean? Well, in geometry, two shapes are congruent iff they have the same shape and size. The symbol is $\cong$ (\cong).

So...

For reflexion you need: $\forall X\; (X\cong X)$

  • This means: "Any triangle has the same shape and size as itself". Is this always so?

For symmetry you need: $\forall X\;\forall Y \;(X\cong Y \to Y\cong X)$

  • This means: "If any one triangle has the same shape and size as any other, then the other triangle has the same shape and size as the first." Is this always so?

For antisymmetry you need: $\forall X\;\forall Y\;\big((X\cong Y \wedge Y\cong X) \to X = Y\big)$

  • This means: "If any one triangle has the same shape and size as any other, and the other has the same shape and size as the first, then they are the exact same triangle." Is this always so?

For transitive you need: $\forall X\;\forall Y\;\forall Z\; \big((X\cong Y\wedge Y\cong Z) \to X\cong Z\big)$

This means: "If any one triangle has the same shape and size as any other, and that other has the same shape and size as any third, then the first has the same shape and size as the third." Is this always so?


Is this help enough for you to solve your problem?

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  • $\begingroup$ Thanks a lot for the explanation. I feel like this does help. Maybe I am just slow, but it sounds like it could be both? "if triangle $ABC$ has the same shape and size as triangle $DEF$ then triangle $DEF$ has the same size and shape as triangle $ABC$" Yes so it is symmetric. So then does that mean that they are the exact same triangle? Or would it be two triangles that are identical in size and length? $\endgroup$ – generic user007 May 7 '15 at 0:05
  • $\begingroup$ @DietDrPepsi Shapes are identical ($=$) when, and only when, they have the exact same size, shape, position, and orientation (that is: the exact same position of all the vertices). Shapes are congruent ($\cong$) when, and only when, they have identical scale (of corresponding sides and angles). Shapes are similar ($\simeq$) when, and only when, they share the same shape (size of corresponding angles). $\endgroup$ – Graham Kemp May 7 '15 at 0:27

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