# Simple proof of irrationality of e

Is this reasoning correct?

Assume $e=\frac{p}{q}$ where $p$ and $q$ are natural numbers taking natural log on both sides and using the fact that natural $\ln e = 1$. We come up with $1 = \ln\left(\frac{p}{q}\right)$. Taking derivative on both sides $0 = \large\frac{1}{\frac{p}{q}}$ taking $\frac{p}{q}$ to other side results in $0 = 1$ absurdity therefore $e=\frac{p}{q}$ is false and hence $e$ is irrational.

What do you think guys?

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You cannot differentiate, remember $p$ and $q$ are fixed numbers and not variables (it would be like saying $0 = \ln 1$, and by taking derivative you get $0 = \frac1 1 = 1$). – Joel Cohen May 6 '14 at 19:43
Differentiation makes non sense at all in this case. Try to use the series expansion of $e$ to prove $e$'s irrationality. – Samrat Mukhopadhyay May 6 '14 at 19:54
Now what about 3? Following your example, 3 = p/q implies 0 = ln (3) which is absurd, so 3 = p/q is false and hence 3 is irrational. – gnasher729 May 6 '14 at 20:00
thanks guys I got excited by just seeing the result after 5 minutes of labor. I think one can do the differentiation however it would result in 0 = 0 since both sides of the equations are just constants and hence tells nothing about e – user148329 May 6 '14 at 20:04
Remember, differentiation is something you do to functions, not to numbers. It is easy to get sloppy and write 5 when what is meant is the function f(x) that equals 5 for all x. The latter you can differentiate to get "the function f(x) that is equal to zero for all x". The former -- the number 5 -- you cannot differentiate to get the number 0. – Eric Lippert May 6 '14 at 23:03

Is this reasoning correct?

No. The differentiation part makes no sense. Consider doing the same thing with $e = C$ for some constant $C$. We would prove that $e$ in fact is not a number at all. This would mean that a huge number of mathematical papers would have to be revoked, and mathematical research would really take a huge blow.

And we don't want that.

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Well, we DO want that if it were true, or yielded a useful branch of mathematics for us to research. But if it's wrong, or an extraneous solution, then I completely agree, we don't want to invalidate a few thousand years of research for no good reason. =) – corsiKa May 7 '14 at 4:19

Before differentiating, you actually come up with $1 = \ln e$. So what are you taking the derivative with respect to? $e$? I'm sure you know that differentiating with respect to a constant doesn't make sense. ;-)

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Or more precisely, "what function are you deriving?". – Jack M May 7 '14 at 0:21