Can anyone prove that i^i is a real number using only advanced algebra and trig? I will post my solution in 24 hours if no one finds it- note that there may

$$\\e^{ix}=cosx+isinx\\\\ e^{i\frac{\pi}{2}}=cos\frac{\pi}{2}+isin\frac{\pi}{2}\\\\ e^{i\frac{\pi}{2}}=0+i\times 1\\\\ e^{i\frac{\pi}{2}}=i\\\\ \mbox{raise each side to the power of i}\\\\ \left(e^{i\frac{\pi}{2}}\right)^i=i^i\\\\ e^{i^2\frac{\pi}{2}}=i^i\\\\ e^{(-1)\frac{\pi}{2}}=i^i\\\\ e^{\frac{-\pi}{2}}=i^i\\\\ therefore\\ i^i=e^{\frac{-\pi}{2}} \\\\therefore\\ i^i \mbox{ is a real number }$$

Just an update: to clarify the above, this means NO CALCULUS!!

 Once again, good luck!

Other update:

a. $${\sqrt{-{\mathtt{1}}}} = {i}$$ , and also,  $${i}{\mathtt{\,\times\,}}{i} = -{\mathtt{1}}$$

b. by i^i I mean  $${{i}}^{{i}}$$

We're now at the one-hour mark. Good luck!

We're now at 4 hours and no one has solved it yet! Here's a hint- use logs.

Good Luck!

The following video lists the claim and proves it beautifully (literally!): https://www.youtube.com/watch?v=PxViWDgAZ7Q

Mmm I just noticed that I did not use any logs 

Great job Melody my proof is probably a bit less elegant and is also longer. You got the right answer though.

OK! Here is my proof!

Lemma:  $${{\mathtt{e}}}^{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right)} = -{\mathtt{1}}$$

Statement                                          Reason

1. $${{\mathtt{e}}}^{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right)} = {\mathtt{x}}$$                                    Given

2. $${{\mathtt{e}}}^{\left({\mathtt{k}}{\mathtt{\,\times\,}}{i}\right)} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{k}}\right)}{\mathtt{\,\small\textbf+\,}}{i}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{k}}\right)}$$               Euler's Identity

3. $${{\mathtt{e}}}^{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right)} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\small\textbf+\,}}{i}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{\pi}}\right)}$$             From 1. and 2.

4. $${{\mathtt{e}}}^{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right)} = {\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{i}{\mathtt{\,\times\,}}{\mathtt{0}}$$                  Simplify

5. $${{\mathtt{e}}}^{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right)} = -{\mathtt{1}}$$                                 Simplify

Q.E.D.

Main Proof:

Statement                                                   Reason

1. $${{i}}^{{i}} = {\mathtt{x}}$$                                                     Given

2.  $${ln}{\left({{i}}^{{i}}\right)} = {ln}{\left({\mathtt{x}}\right)}$$                                     Take log of both sides

3. $${i}{\mathtt{\,\times\,}}{ln}{\left({i}\right)} = {ln}{\left({\mathtt{x}}\right)}$$                                  Power rule of logarithms

4.  $${i} = {\sqrt{-{\mathtt{1}}}}$$                                              Given

5. $${i} = {\left(-{\mathtt{1}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}$$                                          Algebra

6. $${i}{\mathtt{\,\times\,}}{ln}{\left({\left(-{\mathtt{1}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)} = {ln}{\left({\mathtt{x}}\right)}$$                       From 5. and 3.

7.  $${i}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{ln}{\left(-{\mathtt{1}}\right)} = {ln}{\left({\mathtt{x}}\right)}$$                 Power rule of logarithms

8.  $${ln}{\left({\mathtt{x}}\right)} = {\mathtt{y}}$$  if and only if  $${{\mathtt{e}}}^{{\mathtt{y}}} = {\mathtt{x}}$$               Definition of ln

9. $${i}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right) = {ln}{\left({\mathtt{x}}\right)}$$                  Use the lemma

10.  $$\left({\frac{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\times\,}}{i}\right)}{{\mathtt{2}}}}\right) = {ln}{\left({\mathtt{x}}\right)}$$                             Algebra

11.  $$\left({\frac{\left(\left(-{\mathtt{1}}\right){\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{{\mathtt{2}}}}\right) = {ln}{\left({\mathtt{x}}\right)}$$                            Algebra

12.  $$\left({\frac{-{\mathtt{\pi}}}{{\mathtt{2}}}}\right) = {ln}{\left({\mathtt{x}}\right)}$$                                       Algebra

13.  $${{\mathtt{e}}}^{\left({\frac{-{\mathtt{\pi}}}{{\mathtt{2}}}}\right)} = {\mathtt{x}}$$                                           Take antilog of both sides and use step 8.

14. $${\frac{{\mathtt{1}}}{\left({{\mathtt{e}}}^{\left({\frac{{\mathtt{\pi}}}{{\mathtt{2}}}}\right)}\right)}} = {\mathtt{x}}$$                                           Algebra

15.  $${\frac{{\mathtt{1}}}{\left({{\sqrt{{\mathtt{e}}}}}^{\,{\mathtt{\pi}}}\right)}} = {\mathtt{x}}$$                                             Algebra

16.  $$\left({\frac{{\mathtt{1}}}{\left({{\sqrt{{\mathtt{e}}}}}^{\,{\mathtt{\pi}}}\right)}}\right){\mathtt{\,\times\,}}\left({\frac{{{\sqrt{{\mathtt{e}}}}}^{\,\left({\mathtt{4}}-{\mathtt{\pi}}\right)}}{{{\sqrt{{\mathtt{e}}}}}^{\,\left({\mathtt{4}}-{\mathtt{\pi}}\right)}}}\right) = {\mathtt{x}}$$               Rationalize the denominator

17.  $${\frac{\left({{\sqrt{{\mathtt{e}}}}}^{\,\left({\mathtt{4}}-{\mathtt{\pi}}\right)}\right)}{\left({{\sqrt{{\mathtt{e}}}}}^{\,{\mathtt{4}}}\right)}}$$                                              Algebra

18.  $${\frac{{{\sqrt{{\mathtt{e}}}}}^{\,\left({\mathtt{4}}-{\mathtt{\pi}}\right)}}{{{\mathtt{e}}}^{{\mathtt{2}}}}} = {\mathtt{0.207\: \!879\: \!576\: \!350\: \!761\: \!9}}$$    Algebra

And this is a real number.

Q.E.D.

Thanks to you all for submitting your proofs and also, please respond to tell me of any flaws in my proof. I will try to get back to WebCalc 2.0 with another puzzle proof question soon!

hi anonymous,

Why don't you join up.  It is very easy to do.  Much easier than posting anonymously every time.  

I liked you proof.  I just wrote it in LaTex as I worked through it.  

$$\begin{array}{rll} e^{\pi i}&=&cos\pi+isin\pi\\ e^{\pi i}&=&-1+0\\ e^{\pi i}&=&-1\qquad \\ ln(e^{\pi i})&=&ln(-1)\qquad \\ ln(-1)&=&\pi i\qquad \\ now&&\\ x&=&i^i\\ lnx&=&lni^i\\ lnx&=&ilni\\ lnx&=&iln(-1)^{1/2}\\ lnx&=&\frac{i}{2}ln(-1)\\ lnx&=&\frac{i}{2}\times \pi i\\ lnx&=&\frac{\pi\times i^2}{2}\\ lnx&=&\frac{\pi\times -1}{2}\\ lnx&=&\frac{-\pi}{2}\\ e^{lnx}&=&e^{\frac{-\pi}{2}}\\ x&=&e^{\frac{-\pi}{2}}\\ x&\in&Z\\ therefore&&\\ i^i&\in&Z\\ \end{array}$$

Very impressive, guys....!!!!

Melody

yes but I am just a 12 yr old and so my parents won't allow me to sign up.

the proof was written by me, a 12 year old.