Proving that not defined value is equal to something My younger brother (9th Grader) got the following maths problem-
Given: $$2^a = 3^b = 6^c$$
Prove:
$$c=\frac{a * b}{a+b}$$
From my elementary knowledge of mathematics it seems like a=b=c=0.Also, (ab)/(a+b) is not defined and not defined can be equal to 0. Which makes me think if the question makes any sense. They could have also asked if (ab)/(a+b) = 182 i.e. some random number.
My question is if the output of 

Not Defined == A Number

is true or false?
Does this question really makes sense?
Unfortunately the teacher is pretty arrogant and doesn't want to give an answer to this question!
 A: $$2^a=3^b=6^c=k\text{(say)}$$
So, $2=k^{\frac1a},3=k^{\frac1b},6=k^{\frac1c}$
$$\implies k^{\frac1a} \cdot  k^{\frac1b}=k^{\frac1c} $$
$$\implies k^{\frac1a+\frac1b} =k^{\frac1c}  $$
We know, $a^m=a^n\implies m=n$ if $a\ne0,\pm1$
Here if $k=1, a=b=c=0$ but $\frac{ab}{a+b}=\frac00$ i.e., undefined
A: Suppose that $2^a=3^b=6^c$, where $a\ne 0\ne b$. Take logs base $2$:
$$a=b\lg 3=c\lg 6=c(1+\lg 3)\;.$$
Then 
$$\frac{ab}{a+b}=\frac{b^2\lg 3}{b+b\lg 3}=\frac{b\lg 3}{1+\lg 3}=\frac{a}{a/c}=c\;.$$
Of course the only solution with integral $a,b$, and $c$ is $a=b=c=0$, but there are certainly non-integral solutions.
A: Think of "Not Defined" as being an element in its own right. Then "Not Defined == A Number" is consistently false, and "Not Defined == Not Defined" is consistently true. This is the viewpoint that emerges from interpreting "partial function" as "basepoint-preserving function between pointed sets." It is called Kleene equality.
Note that there are also directed equality relations that can be useful when describing partial functions. For example, let $$\lim : \mathbb{R}^\mathbb{N} \rightarrow \mathbb{R}$$ denote the partial function that returns the limit of a real-valued sequence. Then we have $\lim(a+b) \unlhd \lim(a)+\lim(b)$, where $\sigma \unlhd \tau$ means that if $\tau$ is well-defined, then so too is $\sigma$, and they're equal. 
