Code?

[gold eagle]

Negative - that is quite the signature.

[ZeroOne]

22/7 is not even near pi, so using that one isn't such a good idea

That's what I sort of tried to make clear in my message...

Hindsight: Yes, it is "incalculable", eg. endless eg. irrational. It cannot be expressed in any "r/q" way (so not 22/7 ) where r and q both belong to rational numbers. But you can always count more decimals for it. Try PiFast, http://numbers.computation.free.fr/C...am/pifast.html . It's widely used to measure the speed of the CPU, especially within the overclocking people. It calculates the first 5.000.000 decimals and gives you the time the job took (AMD Athlon XP 1900+ some 30 sec).
And if you read my previous message carefully, you'll find the link to a page which contains the first 100.000 decimals.

Neg, you should have left the answer for "division by zero -> 1=2" as a hidden message, although I've read that a couple of times before. But other people might not have seen it before.

[pwaring]

There is a far easier way to prove that 1 = 2, Negative (although this works on the same principle):

Assume: x=y
2x=2y
x-y=0
2x-2y=0
Therefore:
x-y=2x-2y
Factorise:
1(x-y)=2(x-y)
Divide by (x-y)
1 = 2

Of course this doesn't work because x-y=0 and you can't divide by zero without getting infinity or nonsense.

Just for thought:
What is 0/0 equal to? Answer, anything you want.
Assume 0/0 = 5 then multiply both sides by 0 you get 0=0
Assume 0/0 = 200 then multiply both sides by 0 you get 0=0
etc.

[Flibberdy]

By Definition 22/7 isn't irrational.. an Irrational umber being defined as one which cannot be written as a fraction comprising of 2 integers.. If u want Pi in it's full magnificence then a Calculator's Pi button should suffice

[Easeupyo]

6. 2(a² - ab) = 1(a² - ab)
Single out the factors


2a(a - b) = a(a - b)

there are many better ways to simplify

and ofcourse 1 = 2 if a = b, or x = y...

it's like saying an apple's an orange, so deleware's in kansas

[SarinMage]

interesting

[nabylbt]

since we are all into maths ...
here is a cool one:
imagine an inifinite decreasing curve bounded by an horizonal asymptot having a precise are ...? possible or no ....