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The Archimedean property: any point x before the finish line lies between two of the points P n (inclusive).. It is possible to prove the equation 0.999... = 1 using just the mathematical tools of comparison and addition of (finite) decimal numbers, without any reference to more advanced topics such as series and limits.
The Wiki page introduces the proof thus: "There is an elementary proof of the equation 0.999... = 1, which uses just the mathematical tools of comparison and addition of (finite) decimal numbers, without any reference to more advanced topics such as series, limits, formal construction of real numbers, etc. ".
Bertrand's postulate and a proof; Estimation of covariance matrices; Fermat's little theorem and some proofs; Gödel's completeness theorem and its original proof; Mathematical induction and a proof; Proof that 0.999... equals 1; Proof that 22/7 exceeds π; Proof that e is irrational; Proof that π is irrational
I don't feel strongly about "exactly", so I'm going to take this issue and run with it. Among non-mathematicians, real numbers often are decimals, in which case the equality of nu
As the methods I employed are almost literally those used in your 0.9999...<1 proof (only changing "is less than 1" into "is an element of the set of rationals") , you should have severe difficulties to discredit my "proof" without renouncing your own. Thirdly, you seem to claim there are real numbers without any decimal representation.
A: 1 can be written many ways: 1/1, 2/2, cos 0, ln e, i 4, 2 - 1, 1e0, 1 2, and so forth. Another way of writing it is 0.999...; contrary to the intuition of many people, decimal notation is not a bijection from decimal representations to real numbers.
How can he be wrong? He is referring to sequences with positive sums. What are you talking about? He is correct in stating it will never exceed 1. In fact it will never reach 1 ei
If (0.999...) is a real number, then it MUST be equal to 1, since there is no such thing as "the number prior to 1" by definition of the reals. Then, let's all agree ...
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