Why is $\frac{1}{\frac{1}{0}}$ undefined? Is the fraction
$$\frac{1}{\frac{1}{0}}$$
undefined? 
I know that division by zero is usually prohibited, but since dividing a number by a fraction yields the same result as multiplying the number by the fraction's reciprocal, you could argue that
$$\frac{1}{\frac{1}{0}} = (1)\left(\frac{0}{1}\right) = 0$$
Is that manipulation permissible in this case? Why or why not?
 A: The expression is undefined because the value you are trying to divide $1$ by is undefined. Therefore the operation cannot take place. If you had something like $\frac{1}{\frac{1}{\ln 0}}$ then that would also be undefined because you cannot evaluate $\ln 0$. I think your misunderstanding comes from the fact that you treat $\frac{1}{0}$ as wholly separate term and ignore it's value (not even undefined). 
A: Yeah, this is undefined. When we divide some number with some N.D. number it results in N.D. number. 
I think you are making a mistake by considering $1/0$ a fraction. Its not a fraction because denominator is zero, its a number which is not defined. So, here "fraction's reciprocal" doesn't make any sense.
But taking limit of $1/{(1/x)}$ as $x\to 0$ makes sense.
A: A fraction $\frac{a}{b}$ is defined as a solution of the equation $bx=a$. Of course, the equation $0x=1$ has no solutions in $\mathbb{R}$. If you very want to solve this equation you could do as follows. 
Consider $\mathbb{R}$ as a (multiplicative) semigroup and try to embed it in such a semigroup $S$ that $\exists x\in S: 0x=1$. Of course, you get nothing, since $0=0\cdot 1=0\cdot 0x=0x=1$. Then you can consider more general situation: take as $S$ some magma (see in Wikipedia "Magma (algebra)"). Since the multiplication will be non-associative, the contradiction disappears: you get $0(0x)=0$, i.e. $0\cdot 1=0$. I believe that such a magma exists, but I didn't try to build it.
A: The algebraic identity $\frac{a}{\frac{b}{c}} = a\frac{c}{b}$ only holds when both $\frac{b}{c}$ and $\frac{c}{b}$ are defined.  This is similar to why $\sqrt{-2}\sqrt{-3} \neq \sqrt{6}$.
Think of it in terms of "priority of operations:"  $\frac{1}{\frac{1}{0}}$ can only be defined if both $1$ and $\frac{1}{0}$ are defined, because that is the ultimate operation in the expression, and both sides of it must be defined first.  Otherwise, it is undefined itself also.
A: Any expression having an undefined term somewhere inside is undefined as a whole. The rule $\frac1{\frac1x}=x$ holds only for $x\ne0$.
A: Another way to think about this is order of operations: 
$$
\frac{1}{\frac{1}{0}}=1/(1/0)
$$
I always compute what's inside the parenthesis first, which gives me undefined, and I have to stop there.
A: For $\frac{0}{0}$ one can argue that it can take any value, depending on how the $0$s are reached: $\lim _{x→0}{\frac{x}{x}} = 1$ but $\lim _{x→0} \frac{3x}{x} = 3$.  So, defining it does not make sense in general.
For $0^{0}$ you can reach $1$ ($\lim _{x→0} x^x$) or e. g. $0$ ($\lim _{x→0} 0^x$), so a general extension of the definition doesn't make sense, too.
I don't find any such "good reason for being undefined" like varying results for the example of $\frac{1}{\frac{1}{0}}$ (all approaches I could think of result in $0$, nothing else).  So I just can agree with the other answers here that it being undefined is just because rules like $\frac{1}{\frac{1}{x}} = x$ only hold valid if any sub-term also is defined.
