Understanding the Relationship Between Repeating Decimals and the Number Line

[cragar]

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

 

[netheril96]

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

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

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.

 

[HallsofIvy]

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.

Yes, that is correct.

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

What does this mean? When you say "all the reals" do you mean the sum of all real numbers? If so your question makes no sense because that sum does not exist. You could mean the set of all real numbers and be talking about the "difference" of sets- in that case, if by "0" you mean "the empty set" then your statement is correct. A- A= empty set for any set A. Of course, that has nothing to do with arithmetic operations.

 

[cragar]

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

 

[Hurkyl]

(moderator's note: I've removed someone asserting that 0.999... is not equal to one, along with the followups)

 

[Char. Limit]

(moderator's note: I've removed someone asserting that 0.999... is not equal to one, along with the followups)

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

 

[Char. Limit]

The owner of the forum is Greg Bernhardt. Feel free to speak to him. Considering we get people saying what you're saying every three days, though, I don't think he'll be very supportive of your argument.

 

[jhae2.718]

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

 

[Curd]

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

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.

 

[Take_it_Easy]

I think that my previous reply has been deleted, or I never was able to post it and thought I did.
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.

 

[gb7nash]

I think that my previous reply has been deleted, or I never was able to post it and thought I did.
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.

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

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

 

[Char. Limit]

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

\sum_{n=1}^{\infty}\frac{9n}{10}=\frac{\frac{9}{10}}{1-\frac{1}{10}}=1

I think you mean

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

 

[gb7nash]

I think you mean

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

whoooooops thanks