How did you get this? It seems obvious that because we admitted that
the longer line *was* longer, after you have used all of the
(infinity of) points on the shorter line, there will be points on the
longer line in between the lines you drew across.
> But not all infinities are equal. Let's try to put the integers in a one to
> one correspondence with all the points in the line from 0 to 1 expressed as a
> decimal.
>
> 1 - 0.a1,a2,a3,a4,a5 ...
> 2 - 0.b1,b2,b3,b4,b5 ...
> 3 - 0.c1,c2,c3,c4,c5 ...
> 4 - 0.d1,d2,d3,d4.d5 ...
> .
> .
>
> The trouble is it doesn't work, there are decimals not included, for example,
> the point 0.A1,B2,C3,D4,E5 ... where A1 is any digit except a1, B2 is any
> digit except b2, C3 is any digit except c3 etc. This point differs in at
> least one decimal place with any point in our one to one scheme, we've used
> all the integers but there are still points remaining, so there must be more
> points on a line than integers.
But doesn't this apply to the odd number example above? If we use
what appears to be this same logic, since we have used all of the
integers, by pairing them up with only half of the *same* integers,
it follows that the infinity of the integers is equal to half of
itself. Therefore the whole argument must be wrong.
Wolfkin.