I've been interested in representing $\sqrt 2\approx 1.414213562373095$ by an equation which uses each of $1,2,\cdots,9$ once. Suppose that the following conditions must be satisfied. Then, can we get a 'good' approximate value of $\sqrt 2$ ?
Condition 1 : Use each of $1,2,\cdots,9$ once.
Condition 2 : You can use $+,-,\times, \div, (\ ), \{\ \}, [\ ], !$ such as $$\begin{align}1+2-[4\div \{(3+5)\times (6+8)\}+7!].\end{align}$$
Condition 3 : You can use a decimal point. You can represent $0.1234$ as "$.1234$".
Condition 4 : You can use neither the radical sign nor exponential notation such as $\sqrt{3}, 4^5$.
Then, here are my examples.
Example 1 : $.436215+.978=\underline{1.41421}5\rightarrow 6$ digit.
Example 2 : $12653\div 8947=\underline{1.41421}70\cdots\rightarrow 6$ digit.
Example 3 : $3\div 7+.9856421=\underline{1.4142135}28\cdots\rightarrow 8$ digit.
So far I'm facing difficulty for finding an equation which corresponds to $\sqrt 2$ at more than $8$ digit. Can you help?
Update 1 : I've just got the following :
$\{.9\times (5+6)\}\div\{7+(3!\times 2.4)\div (8!+1)\}=\underline{1.4142135623}22\cdots\rightarrow 11$ digit.
Update 2 : I've just got the following :
$(3-.1)\times \{2-4\div (8!\div 6+7)\}\div (5-.9)=\underline{1.4142135623}82\cdots\rightarrow 11$ digit.
It can be said that this example is better than the example of update 1.