Help me with this Math Equation!

[iGuest 3]

I studies at Bakayaro University in Tokyo, a university with hundred years of history. Many celebrities came out of here, such as Tokugawa Ieyasu, Oda Nobunaga, Ayumi Hamasaki and et cetera. I'm in the Rocket Science program. Today, my professor, Bob le Blanc (He is a Italian), have given out this mathematical equation to us. He states this equation as the basis foundation of all the complex mathematics. He said, in order to work this out, you must understand the concept of Hokkaido Mathelogy, which is originally developed by Utada Hikaru.At last, he told us less than three people in history have figure out this equation. Two of those three happened to be dead extremly early due to lack of brain oxygens.I do not understand this equation. Anyone here wise enough to explain me this mysterious mathmetical equation?>>> " 1 + 1 = ? ">I'd thankful, no, grateful if any of you knows the answer. There will be rewards for people who can explain to me.・りがとうございます！Sincerely,バナナ子( Originally posted on Heavengames Forum "Outside Discussions" :[urll]http://www.heavengames.com/cgi-bin/forums/display.cgi?action=ct&f=1,323999,,all[/url]:

[iGuest 3]

3 people out of billions of people solved it, heh, what are the chances of you finding a fourth here lolz.

[Bikerman 2]

He is pulling your leg.Utada Hikaru is a Japanese singer who has a single called A Kaleidoscope Of Mathematics on her album.Hokkaido Mathelogy is a nonsense phrase. Hokkaido has a journal of mathematics but matheology is meaningless.As for the equation 1+1 = ?OK....if you want a mathematically rigorous answer then you have to start by defining the symbols and operators. This is done as follows:We start with the Peano Postulates:P1. 1 is in N.P2. If x is in N, then its "successor" x' is in N.P3. There is no x such that x' = 1.P4. If x isn't 1, then there is a y in N such that y' = x.P5. If S is a subset of N, 1 is in S, and the implication (x in S => x' in S) holds, then S = N.That defines our number series.Now to define addition:Definitions : D1. Let a and b be in N. If b = 1, then define a + b = a' (using P1 and P2 above). D2. If b isn't 1, then let c' = b, with c in N (using P4), and define a + b = (a + c)'.Now define '2'D3. Definition : 2 = 1'Now propose the theoremTheorem: 1 + 1 = 2Finally prove with reference to definitions:Proof:Since a+b=a' (using D1)LET a=b=1 It follows that1+1=1' (D1)and1'=2 (D3)Therefore1+1=2QED.Note that this uses S1 version of the Peano Postulates. The proof for S0 is similar but has a couple of extra steps. That, however, should do you.