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 be multiple proofs. Good Luck!
+118687 $$\\e^{ix}=cosx+isinx\\\\
e^{i\frac{\pi}{2}}=cos\frac{\pi}{2}+isin\frac{\pi}{2}\\\\
e^{i\frac{\pi}{2}}=0+i\times 1\\\\
e^{i\frac{\pi}{2}}=i\\\\
\mbox{raise each side to the power of i}\\\\
\left(e^{i\frac{\pi}{2}}\right)^i=i^i\\\\
e^{i^2\frac{\pi}{2}}=i^i\\\\
e^{(-1)\frac{\pi}{2}}=i^i\\\\
e^{\frac{-\pi}{2}}=i^i\\\\
therefore\\
i^i=e^{\frac{-\pi}{2}} \\\\therefore\\
i^i \mbox{ is a real number }$$
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!
+4473 The following video lists the claim and proves it beautifully (literally!): https://www.youtube.com/watch?v=PxViWDgAZ7Q
+118687 $$\\e^{ix}=cosx+isinx\\\\
e^{i\frac{\pi}{2}}=cos\frac{\pi}{2}+isin\frac{\pi}{2}\\\\
e^{i\frac{\pi}{2}}=0+i\times 1\\\\
e^{i\frac{\pi}{2}}=i\\\\
\mbox{raise each side to the power of i}\\\\
\left(e^{i\frac{\pi}{2}}\right)^i=i^i\\\\
e^{i^2\frac{\pi}{2}}=i^i\\\\
e^{(-1)\frac{\pi}{2}}=i^i\\\\
e^{\frac{-\pi}{2}}=i^i\\\\
therefore\\
i^i=e^{\frac{-\pi}{2}} \\\\therefore\\
i^i \mbox{ is a real number }$$
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!
+118687
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.
$$\begin{array}{rll}
e^{\pi i}&=&cos\pi+isin\pi\\
e^{\pi i}&=&-1+0\\
e^{\pi i}&=&-1\qquad \\
ln(e^{\pi i})&=&ln(-1)\qquad \\
ln(-1)&=&\pi i\qquad \\
now&&\\
x&=&i^i\\
lnx&=&lni^i\\
lnx&=&ilni\\
lnx&=&iln(-1)^{1/2}\\
lnx&=&\frac{i}{2}ln(-1)\\
lnx&=&\frac{i}{2}\times \pi i\\
lnx&=&\frac{\pi\times i^2}{2}\\
lnx&=&\frac{\pi\times -1}{2}\\
lnx&=&\frac{-\pi}{2}\\
e^{lnx}&=&e^{\frac{-\pi}{2}}\\
x&=&e^{\frac{-\pi}{2}}\\
x&\in&Z\\
therefore&&\\
i^i&\in&Z\\
\end{array}$$
