curiouschris said:
Infinity has always been a problem for me, apart from the concept I do not believe it exists. numbers go on to infinity and even negative infinity but they don't really exist so they don't count (hehe). some people say the universe is infinite in size. well apart from infinite in the same way as a spheres curved surface is infinite then no the universe cannot be infinite.
I raised this with a colleague, he said from a maths point of view (and presumably physics) that infinite just means immeasurably large but finite.
Is that true?
If so please call it something else other than infinite may I suggest \infty-1 (lol)
If its not true how can infinite be used within equations to describe 'real' things because its impossible.
Mathematically, infinity
must exist because there must exist the
infinitesimal.
For example, there are an infinite number of real numbers between 1 and 2 (technically the more appropriate term might be
uncountable, but by its definition, still infinite).
One might argue that any practical number, such as the \sqrt{2} which represents the diagonal of a square, might trail off with all zeros eventually, if one were to calculate out the decimal places far enough. But if that were true, it would be possible to represent \sqrt{2} as a ratio of integers, albeit very, very large integers.
But here is rough proof that this is not possible. Suppose \sqrt{2} can be represented by a ratio of integers,
M/
N (where
M and
N might be very large, yet still finite).
Suppose that we reduce
M/
N such that M and N are the smallest units possible. Now there are 3 possibilities regarding
M and
N.
(a)
M and
N are both odd.
(b)
M is odd and
N is even.
(c)
M is even and
N is odd.
Note that
M and
N cannot both be even, because then we could simply divide both by 2, and retain the same ratio.
Given the above, we could say,
\sqrt{2} = \frac{M}{N}.
Squaring both sides gives us
2 = \frac{M^2}{N^2}
which is,
2N^2 = M^2.
Here we can be confident that
M is even because 2
N is even: The the square of an odd integer gives an odd number. 2 times any integer is even thus 2
N is even. Thus if M
2 is even,
M must be even. That also implies that
N must be odd (because
M and
N cannot both be even).
Since M is even, let's make a substitution that 2
L =
M.
2N^2 = {2L}^2,
2N^2 = 4L^2,
N^2 = 2L^2.
But this means that N must also be even, since multiplying any integer by 2 creates an even number. making both M and N even, which contradicts the original proposition. Thus \sqrt{2} cannot be represented by a ratio of integers. \sqrt{2} belongs in some different class of numbers, that cannot be represented by a ratio of finite integers (namely irrational numbers).
One way to interpret this is \sqrt{2} has a truly infinite number of decimal places to represent it -- they do not trail off to all zeros eventually. Another way to interpret this is that \sqrt{2} cannot be represented by a ratio of integers -- at least not finite integers -- at least not unless both M and N are infinite.
So if one believes that squares exist, and if one believes that squares have diagonals (i.e. one believes that isosceles right triangles exist), arguably one must believe that infinity exists (at least mathematically speaking). This logic might be debatable, but it does present a very strong argument.
x/x=1 is a valid identity if x denotes a polynomial. However, if x denotes an indeterminate real number, then x/x=1 is* equivalent to the assertion x is nonzero.
