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I can't remember where I saw this but it really confuses me.

If

x = .999999999.....

10x = 9.9999999......

10x - x = 9x

x = 1?

So does that mean if that .9 recurring is 1 exactly?

Thanks for any help.

- Thread starter Ed Aboud
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I can't remember where I saw this but it really confuses me.

If

x = .999999999.....

10x = 9.9999999......

10x - x = 9x

x = 1?

So does that mean if that .9 recurring is 1 exactly?

Thanks for any help.

Mentor

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Yes, by definition and derivation. You got it.

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Another way of looking at it: Suppose a and b are real numbers with b>a. There there exists another number, c, such that a < c < b. Suppose we have the hypothesis that .9999.... is not equal to 1. Can you find a number between .999999.... and 1?

Sarcasm alert!!!!!!!i Am So Glad Someone Finally Posted About This, I Have Never Seen It Here In The Forum Before.

The question of whether .999999... is even a real number at all - the question if the sequence .9 + .09 + .009 + ... converges - is a bit more complicated and requires some basic real analysis.

Really?

The question of whether .999999... is even a real number at all - the question if the sequence .9 + .09 + .009 + ... converges - is a bit more complicated and requires some basic real analysis.

From what I know, a usual axiom appended to the real number field is that every bounded, increasing sequence converges..

Not too complicated to show that the condition holds in this case..

Sarcasm alert!!!!!!!

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And I am sorry if it was posted before, I'm relatively new to PF.

Yeah, I mean it's not really too complicated, its just probably a little beyond a standard highschool level.Really?

From what I know, a usual axiom appended to the real number field is that every bounded, increasing sequence converges..

Not too complicated to show that the condition holds in this case..

The possible "problem" would be if the real numbers weren't complete (they are complete, so it's not a problem). The analogy would be, the rational numbers are not complete, so the rational sequence 3, 3.1, 3.14, 3.1415, ..., a bounded increasing sequence, doesn't actually converge in the rational numbers. If the real numbers were not complete, .9, .99, .999, ... might converge to some non-real "infintessimal" sort of number, in the same way that 3.14, 3.141, ... converges to a non-rational number.

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The form of that axiom that I've always seen is that the real numbers are complete (i.e. every Cauchy sequence of real numbers has a limit in the real numbers). A bounded, increasing sequence is a Cauchy sequence though, so it has a limit.Really?

From what I know, a usual axiom appended to the real number field is that every bounded, increasing sequence converges..

Not too complicated to show that the condition holds in this case..

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This is true, but be fair to the young'uns. This is far beyond high school mathematics. It is even beyond the typical Calc 1/2/3/... sequence required of all science and engineering undergraduates in most colleges. Science majors do study advanced math, but it is stuff like Green's theorem, complex analysis, numerical methods, ... That [tex]0.\bar{9}=1[/tex] is even beyond a lot of high school mathThe form of that axiom that I've always seen is that the real numbers are complete (i.e. every Cauchy sequence of real numbers has a limit in the real numbers).

What we need is a standard answer to this question, and now we have the vehicle for this answer: the PF library. Unfortunately, I will be very busy for the next few weeks. Any takers on writing this up? Then when the question comes up again we can simply point to the library entry.

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That's a good idea that perhaps someone could undertake.What we need is a standard answer to this question, and now we have the vehicle for this answer: the PF library. Unfortunately, I will be very busy for the next few weeks. Any takers on writing this up? Then when the question comes up again we can simply point to the library entry.

There are also hundreds of threads on this topic that one can find by searching. I'm thus locking this thread.

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