$(a,b)=(c,d)$ if and only if $a=c$ and $b=d$
AbcdIfAndOnlyIfAcAndBd
2006-09-07 03:06:54
2006-10-09 04:20:01
Proof
ordered pair
0
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Following is a proof that the ordered pairs $(a,b)$ and $(c,d)$ are equal if and only if $a=c$ and $b=d$.
\begin{proof}
If $a=c$ and $b=d$, then $(a,b)=\{\{a\},\{a,b\}\}=\{\{c\},\{c,d\}\}=(c,d)$.
Assume that $(a,b)=(c,d)$ and $a=b$. Then $\{\{c\},\{c,d\}\}=(c,d)=(a,b)=\{\{a\},\{a,b\}\}=\{\{a\},\{a,a\}\}=\{\{a\},\{a\}\}=\{\{a\}\}$.
Thus, $\{c,d\}\in\{\{a\}\}$. Therefore, $\{c,d\}=\{a\}$. Hence, $a=c$ and $a=d$. Since it was also assumed that $a=b$, it follows that $a=c$ and $b=d$.
Finally, assume that $(a,b)=(c,d)$ and $a \neq b$. Then $\{a\} \neq \{a,b\}$. Note that $\{\{a\},\{a,b\}\}=(a,b)=(c,d)=\{\{c\},\{c,d\}\}$. Thus, $\{c\} \in \{\{a\},\{a,b\}\}$. It cannot be the case that $\{c\}=\{a,b\}$ (lest $a=c=b$). Thus, $\{c\}=\{a\}$. Therefore, $a=c$. Hence, $\{\{a\},\{a,b\}\}=\{\{c\},\{c,d\}\}=\{\{a\},\{a,d\}\}$. Note that $\{a,b\} \in \{\{a\},\{a,d\}\}$. Since $\{a\} \neq \{a,b\}$, it must be the case that $\{a,b\}=\{a,d\}$. Thus, $b \in \{a,d\}$. Since $a \neq b$, it must be the case that $b=d$. It follows that $a=c$ and $b=d$.
\end{proof}