symmetry vs antisymmetry

Question

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.

Answer 1

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, $[=]_X\subseteq R.$

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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.

Answer 2

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?

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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?

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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?

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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?

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Is this help enough for you to solve your problem?