I was reading the Tau Manifesto (no offence to pi fans) and realized you could do as follows. Starting with the Euler identity for a full rotation: $$e^{iτ}=1$$ If $e+τ=\frac{p}{q}$ then: $$e^{i(p/q-e)}=1$$ $$e^{ip/q}=e^{ie}$$ $$i\frac{p}{q}=ie$$ $$\frac{p}{q}=e$$ Which we know is false, therefore $e+τ$ must be irrational. Is there any flaw in this proof? I need to know!

EDIT A possible objection is that if $e^{im}=e^{in}$ in general then since $e^{iτ}=e^{0i}$, $τ=0$. In the Euler equation we are talking about rotations, and a rotation of $τ$ is equivalent to a rotation of $0$.

EDIT1 The resolution to this is that $e^{ia/b} = e^{ie}$ decomposes to: $$\frac{a}{b}+m\tau=e+n\tau$$ $$\frac{a}{b}=e+\tau(n-m)$$ $$\frac{a}{b}=e+k\tau$$ since m and n are just integers (the information that $k=1$ having been lost). Have accepted mweiss' answer since he got close.

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    $\begingroup$ If you substitute $p/q = e + \tau$ into all of your steps you would observe that $e + \tau = e \implies \tau = 0$, which is false. $\endgroup$ – Henricus V. Dec 19 '16 at 0:17
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    $\begingroup$ It is certainly not true that if $e^{ia} = e^{ib}$ then $a=b$. If that were true, then you could prove $\tau=0$, thus: $e^0 = 1 = e^\tau.$ If $e^0=e^\tau,$ then $0=\tau. \qquad$ $\endgroup$ – Michael Hardy Dec 19 '16 at 0:23
  • $\begingroup$ @selfawareuser1: So $e+\tau=p/q$ and $e=p/q?$ Then $\tau=0,$ which we know is false. $\endgroup$ – Will R Dec 19 '16 at 0:33
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    $\begingroup$ @selfawareuser1: Yet, $0$ and $2\pi$ are not the same number; and in particular one of the is rational and the other is irrational! $\endgroup$ – hmakholm left over Monica Dec 19 '16 at 1:01
  • $\begingroup$ @selfawareuser1: And so what? You're attempting to investigate whether the number in itself is rational or not -- conflating it with a different number simply because they lead to "the same effect" in a function that you randomly chose to apply to them will not make you any wiser. As the example shows applying that function does not tell you anything about whether a number is rational or not, which makes your approach fundamentally misguided. $\endgroup$ – hmakholm left over Monica Dec 19 '16 at 14:06

As others have already noted, the complex exponential function is not one-to-one; specifically, since $e^{\tau i}=1$, for any $a, b$ with $b = a + n\tau$ for some integer $n$, we would have $e^{ai} = e^{bi}$. Therefore, if $e^{ai}=e^{bi}$ then the most we can conclude is that $ai = bi + n \tau i$ for some $n$.

In your proof, then, the argument would run like this:

If $e+\tau=\frac{p}{q}$ then: $$e^{i(p/q-e)}=1$$ $$e^{ip/q}=e^{ie}$$ $$i\frac{p}{q}=ie + n\tau i$$ $$\frac{p}{q}=e + n\tau$$

So the conclusion is that if $e + \tau$ is rational, then $e + n\tau$ is rational for some $n$. But we knew that already.


Be careful with the exponential complex. I can proof with the same argument that $2=4$.

$$e^{2\pi i } = e^{4\pi i } \Rightarrow 2\pi i = 4\pi i \Rightarrow 2=4 $$

The problem is that the real exponential function is one to one, but the complex one is periodic.

  • $\begingroup$ Interesting, will see how that affects argument... $\endgroup$ – selfawareuser1 Dec 19 '16 at 0:17

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