4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 4
    $\begingroup$ 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
  • 5
    $\begingroup$ @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
9
$\begingroup$

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.

$\endgroup$
14
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.