Mathematical proof: the following mathematical proof has been designed so you trust this game. it proves that, when you click the 'play' button, we display the true winning number and don't make it up. the winning number is included (encrypted) in the number 'a' as explained below. the basic formula: a = b × c the number 'a' is displayed below. the numbers 'b' and 'c' will be displayed when you click the 'play' button. the numbers 'b' and 'c' are prime numbers. therefore, both of them are unique for any specific 'a'. the number 'b' is the lowest between 'b' and 'c'. data associated with the winning number winning number: the 2nd to 6th digits of the number 'b'. the number 'a': a.=.1599803588331327720483807275359 36958906748980572530395382385232029 10043626040901853730290446259523084 10717208773542546701902716118563586 76602322355049519296900625577993339 33289956380023871659348817998314025 83378981894882464234821265949271211 28318068741358281664229958072473312 64336182573490348327311227902700213 75389019177891404769157799880973219 80952222152990143538290760286049979 15280065916435760355090220459492787 56908870779183814044887212820901288 21805056072418863077895295039183925 42322668262689923117591772495942588 66501056868231059147583096069279497 39841027021193977527279184642104203 27492082099513316718236999756250466 67701421034893348173181940663921593 87230932872292719794716872286842039 4387

for the free

< < < emily> > >

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these 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.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.