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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!

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  • $\begingroup$ I am confused. You mean "not defined" because it is $0/0$? That would be a valid concern, but isn't generally the case. It is true that there is an exception when $a=b=0$, but there are many cases of the equations holding without that being the case, e.g., $a=1$, $b=\log_3(2)$, and $c=\dfrac{\log_3(2)}{1+\log_3(2)}$. $\endgroup$ – Jonas Meyer Jul 11 '13 at 7:40
  • $\begingroup$ @JonasMeyer Yes because it comes out to be 0/0. I think I get the idea now. I thought there could be only one possible solution for a, b and c. $\endgroup$ – tusharmath Jul 11 '13 at 7:52
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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.

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  • $\begingroup$ Thanks Brian. I did not consider not integral solutions. I guess I need to take the night grade lessons again :( $\endgroup$ – tusharmath Jul 11 '13 at 7:53
  • $\begingroup$ @Tushar: You’re welcome. It’s an easy mistake to make: I’ll admit that I at first assumed integral solutions, but then I decided that the question was just too silly if that were intended and thought about what might happen in general. $\endgroup$ – Brian M. Scott Jul 11 '13 at 8:01
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$$2^a=3^b=6^c=k\text{(say)}$$

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

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  • $\begingroup$ are you saying that it will be true for values of a,b,c equal to a value other than 0 ? $\endgroup$ – tusharmath Jul 11 '13 at 7:45
  • $\begingroup$ @TusharMathur, yes. $0/0$ is not defined,right? $\endgroup$ – lab bhattacharjee Jul 11 '13 at 8:11
  • $\begingroup$ 0/0 is not defined. $\endgroup$ – tusharmath Jul 11 '13 at 8:30
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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.

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