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# 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

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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!

math complex-numbers
Aug 22, 2014

#5
+117762
+18

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

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Aug 23, 2014

#1
+8

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

Once again, good luck!

Aug 22, 2014
#2
+8

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!

Aug 22, 2014
#3
0

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

Good Luck!

Aug 23, 2014
#4
+4473
+10

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

Aug 23, 2014
#5
+117762
+18

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

Melody Aug 23, 2014
#6
+117762
+6

Mmm I just noticed that I did not use any logs

Aug 23, 2014
#7
0

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

Aug 23, 2014
#8
+5

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

3.$${{\mathtt{e}}}^{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right)} = \underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\small\textbf+\,}}{i}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{\pi}}\right)}$$            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!

Aug 24, 2014
#9
+117762
+5

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.

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

Aug 24, 2014
#10
+124524
0

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

Aug 24, 2014
#11
0

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.

Aug 25, 2014
#12
+9458
0

Good job anonymous 12-year-old. Me, a 13-year-old, couldn't have thought of that complex proof :O

MaxWong  Aug 17, 2016