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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 2007-05-18 11:09 ID:ijhPKrg1

>>101
Copypasta:

2.3 Definition If there exists a 1-1 mapping of A onto B, we say that A and B can be put in 1-1 correspondence, or that A and B have the same cardinal number, or, briefly, that A and B are equivalent, and we write A ~ B. This relation clearly has the following properties:
It is reflexive: A ~ A.
It is symmetric: If A ~ B, then B ~ A.
It is transitive: If A ~ B and B ~ C, then A ~ C.
Any relation with these three properties is called an quivalence relation.

2.4 Definition For any positive integer n, let J_n be the set whose elements are the integers 1, 2,..., n; let J be the set consisting of all positive integers. For any set A, we say:
(a) A is finite if A ~ J_n for some n (the empty set is also considered to be finite).
(b) A is infinite if A is not finite.
(c) A is countable if A ~ J.
(d) A is uncountable if A is neither finite nor countable.
(e) A is at most countable if A is finite or countable.

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copied straight from the first book I found, Rudin, Principles of Mathematical Analysis.  Honestly, finite vs infinite ain't hard.  Just wait till you get to toposes without NNOs.  Then it gets hard.

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