>1 is arbitrary.  1=1 is redundant arbitration.

Let x = y
It follows that:
                  x - y = 0

And that:
                  2x - 2y = 0

Therefore:
                  x - y = 2x - 2y

But:
                  x - y = 1(x - y)  and...
                  2x - 2y = 2(x - y)

So:
                  1(x - y) = 2(x - y)

Dividing both sides by (x - y):
                  1 = 2

(I know it's a falcity but it's fun to look at ;)

--
Lance.


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