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There isn’t a proof because it’s actually kind of arbitrary that 0.999... = 1. Fundamentally, this is true because we chose it to be true. Now, there are good
by addicted 7y ago
There isn’t a proof because it’s actually kind of arbitrary that 0.999... = 1. Fundamentally, this is true because we chose it to be true.
Now, there are good reasons we chose it to be true, and that’s what people usually use as proofs. If it’s not true then a bunch of mathematical expressions become more inconvenient. But there is no reason as such why 0.999... could not have been defined as something that was always < 1.
Fundamentally, 0.999... has no intrinsic meaning, and it’s value depends on the meaning we decide to give this representation.
- twiceaday 7y ago.999... and 1 exist on a continuous line. If they are different numbers, name a number between them.
- chottocharaii 7y ago? Just because there is nothing between two numbers does not mean the two numbers are equivalent. What nonsense is this
- MockObject 7y agoEither two numbers are equal, or they're not equal. (Unless we're calling into question the law of excluded middle, but must we?)
- momentbruh 7y agoThat is true for the reals
- Dylan16807 7y agoWithout using 9999..., please name any two real numbers that don't have a number in between.
- ben509 7y ago{ 0.999... | 1 } using surreal number notation.
- chithanh 7y agoThe Wikipedia article says how, you need a definition of real numbers that includes nonzero infinitesimals (IOW does not satisfy the Archimedian property). So let there be an ω with 0.999... = 1 - 1/ω. Then a number between 0.999... and 1 would be 1 - 0.5/ω.
- effie 7y agoHow do you know that 1/ω is not 0?
- chithanh 7y agoYou are downvoted, but you are actually correct. 0.999... != 1 can be true in nonstandard analysis. So if using standard analysis over nonstandard analysis is a convention, then ultimately 0.999... = 1 is a convention too. (The Wikipedia article even reproduces that argument)