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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
The trouble with 0.000...1 and 0.999...9 is, of course, context. ... at present the proof for 0.999.... equaling 1 seems to rely on an intuition as did the intuitive ...
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
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.
It is perfectly true that 0.999...999 with any natural number of nines denotes a real number that is less than 1. But "0.999..." is shorthand for saying that there are an infinite number of nines.
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
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