Pi 3.14....

[Shining Knights]

It matters not whether you win or lose; what matters is whether I win or lose.

[WetWired]

No. Irrational numbers have infinite length, therefore what you propose is almost impossible.

[Shining Knights]

It matters not whether you win or lose; what matters is whether I win or lose.

[Demosthenes]

If its infinite length than...therfore it must have the other irattional numbers????

No. An irrational number if of infinite length. A number of infinite length is not going to be embedded in another number of infinite length by definition.

[Shining Knights]

It matters not whether you win or lose; what matters is whether I win or lose.

[!King_Amazon!]

This thread does not deliver.

[Demosthenes]

But if it goes on forever than how come it cant forever go on with another irational number embedded into it? thats saying 1+1=2 but...3-1 (does not equal 2) If it goes on forever than somewhere in the forever theres going to be another irrational number in the forever (like sqr.2). LOL if that makes sense:p

Well, assuming that pi is completely random there may be large strings of numbers in there that much up exactly with the string of number in sqrt(2). There may be a 100 straight numbers, a thousand straight numbers, a million straight numbers somewhere...but to have an infinite amount of numbers match one would have to design another irrational number specifically for that purpose. For instance, (pi-1) is an irrational number that will match pi for an infinite number of digits. For the string of digits in one irrational number to be entirely embedded in another their digits have to match forever after a certain point. Unless specifically designed, it is highly unlikely that this will happen.