$e + τ$ 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.

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. 
