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so then .9999 repeating does not come before one on the continuum.

And one more question can we say that all the reals minus all the reals equals zero.

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- #1

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so then .9999 repeating does not come before one on the continuum.

And one more question can we say that all the reals minus all the reals equals zero.

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They are equal, so they are just two representations of the same thing, so of course they are the same point.

so then .9999 repeating does not come before one on the continuum.

And one more question can we say that all the reals minus all the reals equals zero.

Your latter claim is just semantics. There is no definition for subtraction of "all something". Maybe you can define it on your own but that is quite unlikely to be accepted by others.

- #3

HallsofIvy

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Yes, that is correct.this may be a dumb question. So if .9999.....=1, then are these the same point on the number line.

so then .9999 repeating does not come before one on the continuum.

What does thisAnd one more question can we say that all the reals minus all the reals equals zero.

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yes i meant A-A=empty set . I was just wondering if we could say that.

Thanks for your answers .

Thanks for your answers .

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Hurkyl

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- #6

Char. Limit

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Probably a good idea. I'm keeping this proof handy, though. Just in case someone else comes up and tries to assert that.

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Char. Limit

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- #8

jhae2.718

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May I propose 1 = 0.999... be added to the banned topic list?

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actually, you can delete my posts from this thread IMO. I have started another on this topic in which argument will hopefully be more welcome. if you don't like it being in that thread, then don't look.May I propose 1 = 0.999... be added to the banned topic list?

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By the way, I actually did not conceive the number

0.9999999999.....

aritmetically.

I mean... what are these dots?

The only way I saw this was through a limit of a serie.

With this point of view become important the topology I put on R.

For istance if R has the discrete topology then the serie 0.9999999999... does not converge to 1.

But I just realized that we can define the number 0.999999999999..... aritmetically, without any referement to the topology.

It, in fact, can be defined by a simple property.

Let's call [a] = the largest integer lower or equal than a

(so [7,945276] = 7 or [-1,11768] = -2)

Let x be a number such that

[x] = 0

for every natural n > 0

[x *(10^n)] = 9

This property defines 0.999999999999999999........... using only order relation and aritmetical operation (and since < can be defined from aritmetic operations, we can say that the definition of 0.99999999999.... is based only on aritmetic operations of R, or only on the nature of R as a field).

From this definition, in fact, follows that x is the neutral multiplicative element of R.

So it can be PROVED that x = 1, so 0.999999999......... = 1.

- #11

gb7nash

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Most people probably won't understand this proof. The most simple proof that I've seen is:

By the way, I actually did not conceive the number

0.9999999999.....

aritmetically.

I mean... what are these dots?

The only way I saw this was through a limit of a serie.

With this point of view become important the topology I put on R.

For istance if R has the discrete topology then the serie 0.9999999999... does not converge to 1.

But I just realized that we can define the number 0.999999999999..... aritmetically, without any referement to the topology.

It, in fact, can be defined by a simple property.

Let's call [a] = the largest integer lower or equal than a

(so [7,945276] = 7 or [-1,11768] = -2)

Let x be a number such that

[x] = 0

for every natural n > 0

[x *(10^n)] = 9

This property defines 0.999999999999999999........... using only order relation and aritmetical operation (and since < can be defined from aritmetic operations, we can say that the definition of 0.99999999999.... is based only on aritmetic operations of R, or only on the nature of R as a field).

From this definition, in fact, follows that x is the neutral multiplicative element of R.

So it can be PROVED that x = 1, so 0.999999999......... = 1.

[tex]\sum_{n=1}^{\infty} 9 \left(\frac{1}{10}\right)^n=\frac{\frac{9}{10}}{1-\frac{1}{10}}=1[/tex]

Last edited:

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Char. Limit

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I think you meanMost people probably won't understand this proof. The most simple proof that I've seen is:

[tex]\sum_{n=1}^{\infty}\frac{9n}{10}=\frac{\frac{9}{10}}{1-\frac{1}{10}}=1[/tex]

[tex]\sum_{n=1}^{\infty} 9 \left(\frac{1}{10}\right)^n[/tex]

- #13

gb7nash

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whoooooops thanksI think you mean

[tex]\sum_{n=1}^{\infty} 9 \left(\frac{1}{10}\right)^n[/tex]