- #1

- 20

- 0

This is a little reductio ad absurdum proof that .999'=1 that I came up with, and I'm just wondering what everyone thinks with regards to the soundness of it.

Assume .999' < 1

Now, for any 2 real numbers, a and b, such that a < b, a number (a+b)/2 exists such that a < (a+b)/2 < b

So, letting a=.999' and b=1:

(a+b)/2 = (.999' + 1)/2

= .999'/2 + 1/2

= .4999' + .5

= .999'

And, since a < (a+b)/2:

.999' < .999'

So, .999' is less than itself!

Adding an arbitary constant to both sides:

.999' + c < .999' + c

Hence, no number is equal to itself.