Intuitively answer of $(1/1)/(1/(5^{-2}))=25$ But assuming this mathematical logic of evaluating $(a/b) /(c/d) = (a*d) / (b*c)$ equation evaluates to $1/25$.
Is there any specified rule to put parenthesis ..
What i have studied in operator associativity is / is having left to right associativity which evaluates the above to 25 .
On
The traditional method may seem strange: $$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc} $$
This isn't any magic trick. It's just that we are assuming that $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{(\frac{a}{b})}{(\frac{c}{d})}$
$$\large \frac { \frac { \color{red}{a} } { \color{blue}{b} } } { \frac { \color{green}{c} } { \color{orange}{d} } } = \frac { \color{red}{a} } { \color{blue}{b} } \div \frac { \color{green}{c} }{ \color{orange}{d} } = \frac{\color{red}a}{\color{blue}b} \times \frac{\color{orange}d}{\color{green}c} = \frac{\color{red}{a}\color{orange}{d}}{\color{blue}{b}\color{green}{c}}$$
So naturally, $$\frac{\frac{1}{1}}{\frac{1}{5^{-2}}} = \frac{1}{1} \cdot \frac{5^{-2}}{1} = 5^{-2} = \frac{1}{25}$$
Again, please do bear in mind that $$x^{-n} = (x^n)^{-1} = \frac1{x^n} \implies 5^{-2} = \frac{1}{5^2} = \frac{1}{5\times 5} = \frac1{25}$$
If you enter $1\div 1\div 1 \div 5^{-2}$ into some old calculator, it could give you the wrong answer because of the way it prefers to evaluate the statement. So it'll end up: $$ (((1\div 1)\div 1) \div 5^{-2}) = \frac{(\frac{(1/1)}{1})}{5^{-2}} = 5^2 = 25$$
Why? :: When you have $a\star b\star c \star d$ where $\star$ is some binary operation, one can only guess how the order of operation is going to be. A binary operation is made to only take in 2 guys and spit out 1 guy. If brackets are not provided to indicate how to evaluate it, we would simply choose the easiest way for us to evaluate.
We, as human beings, like symmetry and order so we naturally pair it as: $$(a\star b)\star (c \star d) $$
But a computer is impatient to look at the whole thing would evaluate the statement as it is read from left to right: $$((a\star b)\star c) \star d$$
Actually, the convenient way of thinking about it is that the computer is giving more priority to the binary-ness of the operation by focusing on creating pairs (of $x\star y$) first. So after it evaluates $a\star b$ and in memory turns it into a single entity, it then evaluates $(a\star b)\star c$ which is again stored as a single entity and finally outputes the result of $((a\star b)\star c)\star d$
This is where you're wrong:
$\dfrac{\frac{1}{1}}{\frac{1}{5^{-2}}} = \dfrac{\frac{1}{1}}{5^2} = \frac{1}{25} \neq 25$
Most likely, because you've forgotten that $5^{-2} = \frac{1}{5^2}$