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ordered pair

Type of Math Object: 
Definition
Major Section: 
Reference
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Mathematics Subject Classification

03-00 no label found70A05 no label found70G99 no label found

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For all those out there who don't understand why an ordered pair must be defined like this, try exchanging a and b in the definition above, and you'll see that (a,b) != (b,a).

-apk

that is, the pair is ordered by cardinality.
-apk

how to prove ?

(a,b)=(c,d) <==> a=c, b=d

(a,b)=(c,d) <==> a=c, b=d

nctu writes:

> how to prove ?
>
> (a,b)=(c,d) <==> a=c, b=d

It's obvious that if a=c and b=d then (a,b)=(c,d). For the other direction, you can start by proving that if {x,y}={x,z} then y=z. Now suppose (a,b)=(c,d), that is, {{a},{a,b}}={{c},{c,d}}. Either {a}={c} or {a}={c,d}, both of which imply that c=a. So we have {{c},{c,b}}={{c},{c,d}}, and therefore {c,b}={c,d}, and therefore b=d.

> > how to prove ?
> >
> > (a,b)=(c,d) <==> a=c, b=d
>
> It's obvious that if a=c and b=d then (a,b)=(c,d). For the
> other direction, you can start by proving that if
> {x,y}={x,z} then y=z. Now suppose (a,b)=(c,d), that is,
> {{a},{a,b}}={{c},{c,d}}. Either {a}={c} or {a}={c,d}, both
> of which imply that c=a. So we have {{c},{c,b}}={{c},{c,d}},
> and therefore {c,b}={c,d}, and therefore b=d.

I assume you mean that, out of {c} or {c,d}, {a} = {c} is the only valid choice, as {a} = {c,d} leads to the contradiction {c,d} = {c}.

apk

> > Now suppose (a,b)=(c,d), that is,
> > {{a},{a,b}}={{c},{c,d}}. Either {a}={c} or
> > {a}={c,d}, both of which imply that c=a.

> I assume you mean that, out of {c} or {c,d},
> {a} = {c} is the only valid choice, as {a} = {c,d}
> leads to the contradiction {c,d} = {c}.

No, that's not what I meant. In fact, {c,d} = {c} is not a contradiction, it merely implies that d = c.

What I meant was that c is an element of the right-hand side (whether that is {c} or {c,d}), and therefore it must be an element of the left-hand side, and therefore it must be a.

> No, that's not what I meant. In fact, {c,d} = {c} is not a
> contradiction, it merely implies that d = c.

Ah yes, good point!

apk

There's a neat way of recovering a and b from {{a},{a,b}} using Boolean operations which can be used to prove this assertion. To make things clearer, let P denote the set {{a},{a,b}}. Then note that the intersection of P is {a} and the union of P is {a,b}. Thus, we have already recovered a and it only remains to recover b. In the case where the intersection equals the union, a=b, so find b as a. Otherwise, we take the set difference of the union minus the intersection to obtain b.

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