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omg new proof

Name: .999...9 =/= 1 2007-05-12 6:26 ID:/qcJW6wi

Assumptions
If A = B, then
a)  A-B = 0
b) A = (A+B)/2 = B

If A > B, then
a)  A-B > 0
b) A > (A+B)/2 > B


Let A = 1, B = .999...9
a) A-B = 1-.999...9 = .000...1 =/= 0
b) (A+B)/2 = (1+.999...9)/2 = 1.999...9/2 = 1.999...95 < A

Therefore, .999...9 =/= 1

Name: Anonymous 2007-05-15 14:07 ID:Heaven

>>66
Your "definition" of the real numbers makes the real numbers and the rational numbers the same thing. Why even bother?

Furthermore, there's nothing "proven wrong" at all by a definition switch, even if I was willing to accept it. To prove anyone wrong you have to use the same definition and (gasp) PROVE them wrong. For example, if I say "every natural number is even", and you respond that 1 is not even, and I reply that the natural numbers are the set {0, 2, 4, 6, ... }, did I just prove you wrong? Of course not, because I changed a definition that is UNIVERSALLY ACCEPTED to make my argument, and even if you were to allow my new definition I haven't PROVEN you wrong because you were arguing that there exist non-even elements in the set formerly known as the natural numbers, not the set I was referring to as the natural numbers.

As for mathematicians getting upset when something doesn't work out as predicted, you're completely right. The converse, however, is not true. Just because some mathematicians disagree with you (even angrily sometimes) does not mean you are correct. Vehemently arguing that pi is exactly 3 would get responses from mathematicians, and many of them would be calling you a moron - does that mean pi is exactly 3?

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