Best Answer Anza Power, 03 August 2013 - 07:34 PM

I have a proof:

Assume the rectangle is a square with vertices (0,0) (0,1) (1,0) (1,1), we'll draw a vertical line inside the square at x=0.5.

Now let a_{n} be the y value of the point when it crosses the line for the n'th time, if the point was to cover the entire square then for every y in range [0,1] there must exist some n such that a_{n}=y, but this says that the range [0,1] is countable which we know it's not.

_{n}such that forall y in [0,1) exists n in N such that a

_{n}=y, here's what you do:

Build a 2D table that starts at the top left and extends infinitely to the right and down, now now every real number can be written as 0.539563,,, ignore the 0 and the decimal point and take the digits after the decimal point, for each a_{n} take it and put it's digits in the n'th row of the table.

Let t(i,j) be the value at row i column j of the table, now take all the digits at the diagonal of the table t(m,m) and make a new number from these digits by adding 1 to each digits and warping (0 becomes 1, 1 becomes 2 ... 8 becomes 9, 9 becomes 0), so if the table looked like this:

11704...

68700...

05978...

56752,,,

,,,,,,,,,,,,,

The number we'll get is 0.2906...

Now it is impossible for this number to be in the table, because if it were at row m then the value of t(m,m) is it's m'th digit, but it's m'th digit it t(m,m) plus 1,