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I'm not a math pro but I don't think those are good evidence to refute that claim. I have no idea whether the claim is valid or not, though I vaguely recall th

I'm not a math pro but I don't think those are good evidence to refute that claim. I have no idea whether the claim is valid or not, though I vaguely recall this discussion from school. From your examples -- one's a ratio and the other is irrational. Almost as if you suggest "The set of real numbers exists, ergo 0.999999... exists"?

umanwizard 9y ago
"0.999... = 1" isn't really a "claim", it's a definition. Well, it's technically a claim, but one that is obviously true once we agree on what 0.999... means. Anyone who doesn't agree that 0.999... = 1 has a different definition of 0.999... from the one used in mainstream math notation.
yorwba 9y ago
The battle over the question whether or not something exists in the mathematical sense has essentially been won by the formalists, who will allow any definition to claim the existence of something so long as it is consistent with the rest of mathematics. Then, given that 0.99999... exists, what useful properties could it possibly have, if it is to be treated as a number. 0.99999... = 0.9999... * 1 = 0.99999... * 9/9 = 0.99999... * (10 - 1)/9 = (0.99999... * 10 - 0.99999... * 1)/9 = (9.99999... - 0.99999...)/9 = 9.00000.../9 = 9/9 = 1 So if 0.99999... exists and has the same arithmetic properties as decimal expansions of rational numbers, its value must be 1. Now whether you think it should have these properties is an entirely different question, but most mathematicians seem to like it this way.
cgmg 9y ago
No, I'm simply saying those are other examples of numbers with infinite decimal expansions. In fact, most numbers don't have a finite decimal expansion. And the claim is correct. It is the statement that the limit of the partial sums of the inverse powers of 9 is 1, which can be rigorously proven with the epsilon-delta definition of a limit.

umanwizard 9y ago

"0.999... = 1" isn't really a "claim", it's a definition. Well, it's technically a claim, but one that is obviously true once we agree on what 0.999... means. Anyone who doesn't agree that 0.999... = 1 has a different definition of 0.999... from the one used in mainstream math notation.

yorwba 9y ago

The battle over the question whether or not something exists in the mathematical sense has essentially been won by the formalists, who will allow any definition to claim the existence of something so long as it is consistent with the rest of mathematics. Then, given that 0.99999... exists, what useful properties could it possibly have, if it is to be treated as a number. 0.99999... = 0.9999... * 1 = 0.99999... * 9/9 = 0.99999... * (10 - 1)/9 = (0.99999... * 10 - 0.99999... * 1)/9 = (9.99999... - 0.99999...)/9 = 9.00000.../9 = 9/9 = 1 So if 0.99999... exists and has the same arithmetic properties as decimal expansions of rational numbers, its value must be 1. Now whether you think it should have these properties is an entirely different question, but most mathematicians seem to like it this way.

cgmg 9y ago

No, I'm simply saying those are other examples of numbers with infinite decimal expansions. In fact, most numbers don't have a finite decimal expansion. And the claim is correct. It is the statement that the limit of the partial sums of the inverse powers of 9 is 1, which can be rigorously proven with the epsilon-delta definition of a limit.