slightly different definition of an ordered pair In a paper I was reading an ordered pair had a slightly different definition $\langle a,b \rangle = \{a,\{a,b\}\}$ instead the normal Kuratowski definition which is that $\langle a,b \rangle = \{\{a\},\{a,b\}\}$. 
notice in the first one the $a$ is without braces.
now I know that it is wrong , but why is it wrong?
is this definition not as good as the real one?
related (a question that used the same different definition)
I need to disprove an alternate definition of an ordered pair. Why is $\langle a,b\rangle = \{a,\{b\}\}$ incorrect?
 A: How about sitting to write a proof and see where you get stuck? I am going to assume $\sf ZF$ as my set theory here, and the axiom of foundation is going to play a role, too.
Suppose that $\{a,\{a,b\}\}=\{c,\{c,d\}\}$. Note that $\in$ is a linear order on this set ($a\in\{a,b\}$ and $c\in\{c,d\}$), so $a=c$ since it is the minimal element. From this follows that $\{a,b\}=\{c,d\}$, and since $a=c$ it follows that $(\{a,b\}\setminus\{a\})=(\{c,d\}\setminus\{c\})$, namely $\{b\}=\{d\}$. Therefore $b=d$.
So we see that $a=c$ and $b=d$. And therefore this is a correct definition for ordered pairs, assuming the axiom of foundation (which was used to conclude that $\{a,b\}\notin\{a,b\}$ or $\{a,b\}\notin a$ and so on).

So why Kuratowski's definition prevails?


*

*Tradition. As time goes by this becomes more and more engraved into the basics of set theory. I always find it important to explain to my students that we can find other definitions as well, but this is the canonical one that we use.

*Because we have one, and frankly it doesn't matter that much. It is rarely the case that the choice of encoding ordered pairs into sets matters. You can think about most proofs as proof schemata that say "plug the definition of ordered pair here; plug the definition of function there; plug the definition of the real numbers here ... and combine these inference "templates" to prove the theorem".
I'm not saying that there aren't situations where you care about the definition of ordered pairs, there are such times, but those are rare, and when you get there you usually understand very well the situation, the fact that you can essentially replace the definition by another, and why the Kuratowski definition fails.
There are probably other reasons, but these two seem like plenty.
A: As Asaf Karagila already pointed out: $(a,b) := \{a, \{ a, b \}\}$ is a correct definition for ordered pairs. Following is an attempt for proving this.
The set $\{a, \{ a, b \}\}$ is an ordered pair if it obeys the property
$$ \underbrace{\{ a, \{ a, b \} \} = \{ a', \{ a', b' \} \}}_{(i)} \leftrightarrow (a = a') \wedge (b = b'). $$
Two sets are equal iff every element of one set is in the other set and vice versa. The proposition (i) is therefore equivalent to
$$ (\forall z) \: \underbrace{z \in \{ a, \{ a, b \} \} \leftrightarrow z \in \{ a', \{ a', b' \} \}}_{(ii)}. $$
The axiom of pair set requires that
$$ z \in \{ a, \{ a, b \} \} \leftrightarrow z = a \vee z = \{ a, b \}. $$
Now suppose $ z = a \wedge z = \{ a, b \} $. According to the (axiomatic) symmetric and transitive property of the equation relation we get $ a = \{ a, b \}. $ But this requires that there is an element in $a$ which is $a$. This contradicts the axiom of regularity:
$$ \neg ( a = \{ a, b \} \rightarrow a \in a ) \leftrightarrow ( a \notin a \rightarrow a \neq \{ a, b \} ). $$
From this it follows that either $z = a$ or $z = \{a, b\}$, but not both. A listing of all required implications is given as follows (it's a conjunction since (ii) has to hold for all z):
$$ 
(z = a \rightarrow z = a' \vee z = \{ a', b' \} ) \wedge (z = \{ a, b \} \rightarrow z = a' \vee z = \{ a', b' \}) 
\wedge  
(z = a' \rightarrow z = a \vee z = \{ a, b \} ) \wedge (z = \{ a', b' \} \rightarrow z = a \vee z = \{ a, b \}).
$$
Assuming $(z = a \rightarrow z = \{ a', b' \}) \wedge (z = \{ a', b' \} \rightarrow z = \{ a, b \})$, the transitive property of implication has it that $z = a \rightarrow z = \{ a, b \}$ which is equivalent to saying that $a = \{ a, b \}$. Which is false due to the axiom of regularity (see above). 

Going through all the possible permutations we find that the only valid implications are
$$ (z = a \rightarrow z = a') \wedge (z = \{ a, b \} \rightarrow z = \{ a', b' \}). $$
Or, equivalently, 
$$ (a = a') \wedge (b = b'). $$
