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Two problems with this question.
1. Circle with infinite circumference is impossible.
2. You can't do normal arithmetic with "infinity".
So, if (inf) = pi D, we cannot conclude that
That therefore pi = 1 since D = (inf) as well.
It would be JUST as valid (or invalid) to conclude
pi = 1/2, since if the diameter is infinite, so is the radius.
and C = 2 pi r. (inf) = 2 pi (inf), "pi = 1/2". (NOT)
Here is a case in point:
The definition of equality in counting is:
if two sets can be put in 1-1 correspondence
then they have an equal number of elements.
Consider the integers: 1, 2, 3, 4, 5, ......
There are an infinite number of them, because,
like the Energizer Bunny, they just keep going.
Now consider the even integers: 2, 4, 6, 8, 10, .....
One "obvious" conclusion is that there are only half as
many because the even integers are a subset of
the integers, and well ... what about all the odd integers ?
There are just as many of them, because we can make
a 1-1 correspondence:
1 3 5 7 9 ....
2 4 6 8 10 .....
Even(i) = Odd(i) + 1
So (#Evens) + (#Odds) = (#Integers)
(#Evens) * 2 = (#Integers)
(#Evens) = (#Integers) / 2
But .... not so fast ...
We can also make this 1-1 correspondence:
1 2 3 4 5 .....
2 4 6 8 10
Even(i) = 2 * Integer(i)
So we conclude that there are JUST AS MANY Evens as Integers (never mind the Odds !)
So for this (Inf) we have 2 * (Inf) = (Inf) !!
(And that does not prove 2=1, either)
(Now it turns out the Real numbers, or just the irrational
numbers CANNOT be put in 1-1 correspondence with
the integers, but that's another story.)
Even with the "lowest" infinity, called Aleph-0, the normal rules of arithmetic do not apply.
I must confess that I have no clue what the fuck they are talking about because I found this on google
