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# ordered pair

For any sets $a$ and $b$, the ordered pair $(a,b)$ is the set $\{\{a\},\{a,b\}\}$.

The characterizing property of an ordered pair is:

$(a,b)=(c,d)\iff a=b\text{\ \ and\ \ }c=d,$ |

and the above construction of ordered pair, as weird as it seems, is actually the simplest possible formulation which achieves this property.

Type of Math Object:

Definition

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## Mathematics Subject Classification

03-00*no label found*70A05

*no label found*70G99

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## Comments

## explaination

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

## Re: explaination

that is, the pair is ordered by cardinality.

-apk

## Re: explaination

how to prove ?

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

## proof

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

## Re: explaination

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.

## Re: explaination

> > 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

## Re: explaination

> > 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.

## Re: explaination

> 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

## Re: explaination

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.