Other post below included to show how the opportunity came about
:
slartibartfast:
:)
If I recall correctly, the downfall of greek society was due to
Hippasus' heresy of proving that applying pythogorus' theorum to a 1x1 right-angle triangle results in an impossible number... sorry, "
irrational number" [...] [/slartibartfast]
In the mood to "see" why for some large numbers x^2 = x? [...]
This isn't a proof, just a "show that", but one I think is slick. The main obstacle is showing x+1 = x. We can count the number of atoms in the universe:
1 mole -> 10^27[*]
1 star -> 10^33 kg of H atoms -> 10^(33+3+27) = 10^63 atoms
1 obserable universe [several years old data] -> 10^12 galaxys -> 10^(12+12) stars -> 10^87 atoms
(let) 1 islandthing = 10^13 obserable universes -> 10^100 = 1 googol atoms
Does it make any difference if we add 1 more atom to an island-thing? If we assert that 1+googol = googol then the rest follows as a proof.
I have little choice now but to [do a cheap trick]. 1 followed by as many zeros as the number of atoms in an islandthing = 10^ (10^100) = googolplex. I will, however, stay a bit more physical by considering the number of subsets of an islandthing (taking every atom to be unique).
Observe the correspondence between the subsets of {a,b,c} and three bit binary numbers, where each bit represents the inclusion/exclusion of a particular member:
000 {}
001 {c}
010 {b}
011 {b,c}
100 {a}
101 {a,c}
110 {a,b}
111 {a,b,c}
When we include 0, an n bit number enumerates 2^n numbers (a 3 bit number enumerated 8 numbers above), so the number of subsets of an n element set is 2^n. The number of subsets of an islandthang is 2^googol.
See what happens when we double 2^googol:
2^googol * 2 = 2^(googol+1)
We get more abstract when we consider 2^ (2^googol). Let's multiply this number by itself:
2^ (2^googol) * 2^ (2^googol) = 2^ (2^googol + 2^googol) = 2^ (2^(googol+1))
* edit: whoops, one mole is 6.022 x 10^23 ~= 10^24, don't know why I included another factor of 1,000