Unpack the question as: "Let epsilon > 0 be given. I want to prove that there exists a natural number n, and a (rational?) number x, such that |x| < epsilon but |x| >= 1/n." In other words, show that you can select an n for which 1/n is strictly less than epsilon. (Solving the inequality 1/n < epsilon for n might show how Archimedean property can help.)
YOU EXPLAIN BETTER THAT ALL MY DOCTORS😍 thank you for existing
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This seems handy for proving that 0.999... must be equal to 1.
For if 0.999... is not equal to 1, this will imply that the Set of Natural numbers will have a upper bound.
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It's not as well organized as most of my materials, but you can check out: czcams.com/play/PLL0ATV5XYF8BZx6_DvwgjgkjC2x9n4ZV-.html
Hello sir
I want to prove that (-epsilon,epsilon) is not a subset of (-1/n,1/n) by using Archimedean property
Please can you help by give me a hint ?
Unpack the question as: "Let epsilon > 0 be given. I want to prove that there exists a natural number n, and a (rational?) number x, such that |x| < epsilon but |x| >= 1/n." In other words, show that you can select an n for which 1/n is strictly less than epsilon. (Solving the inequality 1/n < epsilon for n might show how Archimedean property can help.)
Plz give the complete proof, not hints
My archemidean property every real number has a natural larger, let 1/epsilon be the real number and n be the natural, hence the proof