$e + τ$ is irrational proof check 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.

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.
