Equivalence Relations and Partial Order - Symmetry and Anti-Symmetry

Question

• I am confused on how to check symmetry and anti symmetry
• I came to a conclusion that '==' is symmetric, but can it be anti-symmetric?
• '>=' function is reflexive and transitive. But I cannot determine whether is symmetric or anti-symmetric
• Is mod function symmetric or anti-symmetric
• I am a bit confused with symmetry and antisymmetry and on how to determine if a function is symmetric or anti-symmetric.
• Please advice me on how to go about this.
• Thanks in advance

Answer 1

A relation $R$ is symmetric if whenever you have $xRy$ you also have $yRx$

For example lines being parallel is symmetric. People having the same birthday is symmetric.

Anti-symmetric is whenever we have $xRy$ and $yRx$ then we have $x=y$

For example if two numbers $x$ and $y$ satisfy $x\le y$ and $y\le x$ then we have $y=x$

Or if two positive integers $m$ and $n$ are such that $m|n$ and $n|m$ then we have $m=n$

Thus relations $\le$ or $|$ are anti-symmetric.

Answer 2

I am a bit confused with symmetry and antisymmetry and on how to determine if a function is symmetric or anti-symmetric.

A (bivariant) function, say $f$, is symmetric if $\forall x\forall y~[f(x,y)=f(y,x)]$.

A (bivariant) function, say $f$, is antisymmetric if $\forall x\forall y~[f(x,y)=f(y,x)\to x=y]$

Equivalently:

A functional relation, say $\def\R{\mathop{\rm R}}\R$, is symmetric if $\forall x~\forall y~[x\R y\to y\R x]$

A functional relation, say $\def\R{\mathop{\rm R}}\R$, is antisymmetric if $\forall x~\forall y~[x\R y\land y\R x\to x=y]$

In words: A function is symmetric if it always has the same values whatever the order of the arguments. A function is antisymmetric if this is only the case when the arguments actually have the same value.

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I came to a conclusion that '==' is symmetric, but can it be anti-symmetric?

(To clarify, == is the equality relation, or simply $=$. This is a functional relation.)

So it is true that both symmetry and antisymmetry hold. $$\begin{align}&\forall x~\forall y~[x=y\to y=x]&\qquad&\text{true}\\[2ex]&\forall x~\forall y~[(x=y\land y=x)\to x=y]&\qquad&\text{true}\end{align}$$

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Try the same with the relation $\geqslant$ and function ${\bmod{}}$.

$$\begin{align}&\forall x~\forall y~[x\geqslant y\to y\geqslant x]&\qquad&\text{true/false}\\[2ex]&\forall x~\forall y~[(x\geqslant y\land y\geqslant x)\to x=y]&\qquad&\text{true/false}\end{align}$$

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$$\begin{align}&\forall x~\forall y~[x\bmod y = y\bmod x]&\qquad&\text{true/false}\\[2ex]&\forall x~\forall y~[(x\bmod y ~=~ y\bmod x)\to x=y]&\qquad&\text{true/false}\end{align}$$