For the given set and relation below determine which define equivalence relations.

(a) $S$ is the set of all people in the world today, $a\sim b$ if $a$ and $b$ have an ancestor in common.

(b) $S$ is the set of all people in the world today, $a\sim b$ if $a$ lives within 100 miles of $b$.

(c) $S$ is the set of all people in the world today, $a\sim b$ if $a$ and $b$ have the same father.

(d) $S$ is the set of real numbers, $a\sim b$ if $a=\pm b$.

(e) $S$ is the set of integers, $a\sim b$ if both $a>b$ and $b>a$.

(f) $S$ is the set of all straight lines in the plane, $a\sim b$ if $a$ is parallel to $b$.

My solution:

a) Yes.

b) No, since transitive property fails.

c) Yes.

d) No, because reflexivity fails. $1\sim 1$ fails because $1=\pm 1$ is not true.

e) No, since reflexivity fails. It is not true that $a>a$.

f) Yes.

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    $\begingroup$ I disagree with your answer to $d$, or at least with your reading of the problem. I understand $a=\pm b$ to mean "either $a=b$ or $a=-b$". Hence I would say that $1\sim 1$ under this relation. $\endgroup$ – lulu Nov 4 '17 at 17:24
  • $\begingroup$ Yes. I misread the text. In that case this is equivalence relation. Right? $\endgroup$ – ZFR Nov 4 '17 at 17:26
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    $\begingroup$ f depends on whether you consider a line parallel with itself, which depends on what definition of parallel you use $\endgroup$ – XRBtoTheMOON Nov 4 '17 at 17:31

Your answers are right for most of them.

For (d), the notion that "$a = \pm b$" is somewhat ambiguous and can be confusing, but usually means that "$a = b$ or $a = -b$". You might more compactly notate this as $|a|=|b|$ as a result. In which case, your desired result pretty much follows from the fact that $=$ is itself an equivalence relation:

  • Reflexivity: Holds, as $|a|=|a|$ trivially
  • Symmetry: Holds, as $|a|=|b|$ implies $|b|=|a|$
  • Transitivity: Holds, as $|a|=|b|$ and $|b|=|c|$ implies $|a|=|c|$

(f) is a confusing one, because it depends on how one defines lines, parallelism, and intersections. It might seem a little pedantic, but it is noteworthy.

Suppose you define two lines to be parallel if they have no intersection point. Then a line $\ell$ is of course not parallel to itself: it intersects itself infinitely many times. Thus $\ell \not \sim \ell$ by this relation, and thus no reflexivity.

Suppose you define two lines to be parallel if and only if the shortest distance between them is constant, no matter where you measure from. Then $\ell \sim \ell$ after all, because that constant distance is $0$. You can go on to show symmetry and transitivity as well.

Since we're in a plane, we could define lines by their slope and $y$-intercept on a graph. Let $\ell = ax+b$ and $m=cx+d$. Thus, $\ell \sim m$ if $a=c$ and $b \ne d$, presumably. In which case, $\ell \not \sim \ell$. But then again, this choice of $b \ne d$ is somewhat arbitrary insofar as it just means we can only look at distinct lines; if you remove that condition, then of course $\ell \sim \ell$.

It's all about definitions in the end; you have to be careful with how you phrase things. (Or, rather, your source should have more carefully defined the notion of "parallel," but I digress.)


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