1 = 0.99... (RossGr)

[Parsec]

I saw a link in RossGr's siggy about his proof that 1 = .99 repeating. It was awfully long and I didn't quite follow it all the way through, but it is wrong to just say this:
1/3 = 0.33...
2/3 = 0.66...
Since 1/3 + 2/3 = 1
and 0.33... + 0.66... = 0.99...
then 1 = 0.99...

 

[JSSheridan]

I would say that 0.33... + 0.66... = 1;

But as an engineer, 0.99... is close enough to be 1 at appropiate times, and sometimes it is not. Engineering is the art of approximation. Peace.

 

[jjlsevil]

x=.9(repeating)
10x=9.9(repeating)

10x
- x
-------
9x

9.9(repeating)
- .9(repeating)
--------------------
9.0

9x=9.0

x=1


For the non believers out there.

.9(repeating) is not an irrational number.

That like saying that .3, .6, or anything for that matter(repeating) is an irrational number.

I will use the above proof to prove that any repeating decimal is NOT an irrational number.

Using the above 1/3+2/3 proof I will clarify using the proof I used.

1/3 = .3(repeating) correct, use long division to prove this
2/3 = .6(repeating) correct, use long division to prove this

These are, now, both proven to be rational numbers.

Basic arithmetic rules apply to rational numbers.

Given x=.3 repeating, and 10x=3.3 repeating; this is only moving the decimal place by multiplying by 10. Whoever said that 9x = 8.9 repeating is wrong, go back to school; flame on.
10x
- x
-------
9x

3.3 repeating
- .3 repeating
-----------------
3.0 Arithmetic rules of subtraction apply to rational number; by defenition, no repeating decimal is irrational.

9x = 3
x = 3/9 = 1/3

Given y=.6 repeating, and 10y=6.6 repeating

10y
-y
-----
9y

6.6repeating
-.6 repeating
-----------------
6.0
9y=6
y=6/9=2/3

x=1/3=.3(repeating)
y=2/3=.6(repeating)

x+y=1
1/3+2/3=1 if you try to say this is false, you're an idiot.

Stop trying to disprove this.

 

[MrDudeMan]

Originally posted by: jjlsevil
x=.9(repeating)
10x=9.9(repeating)

10x
- x
-------
9x

9.9(repeating)
- .9(repeating)
--------------------
9.0

9x=9.0

x=1

this is not a good example. you can manipulate a lot of things to be how you want it with given situations/resources

this is not an accurate representation of how it does/doesnt equal 1, and this very problem has been discussed here already

 

[RossGr]

Originally posted by: Parsec
I saw a link in RossGr's siggy about his proof that 1 = .99 repeating. It was awfully long and I didn't quite follow it all the way through, but it is wrong to just say this:
1/3 = 0.33...
2/3 = 0.66...
Since 1/3 + 2/3 = 1
and 0.33... + 0.66... = 0.99...
then 1 = 0.99...

Mathematically, the trouble with this computation is that it cannot be completed. You are adding an infinite number of digits, this is impossible. This computation falls into the catogory of demonstration, it is not a proof. The fact that the result is valid does not justify the questionable methods used to obtain it.

My proof is a valid mathematical approach to the problem. I essentially show that .999... exists in every neighborhood on the real line that contains 1. That is you cannot find an interval around 1 which does not contain .999.... No matter how small an interval you choose. That is the defintion of equality on the real line. So mathematically 1= .999...

I feel that my proof contain exellect concepts for see why they are equal. It shows clearly that adding any small incremet to .999... results in 1 + . {a bunch of zeros}999.....

you will always have 9s left over, on any number you add to .999... results in 1 + some more and that is the definition of 1.

 

[jeremy806]

That is the defintion of equality on the real line.

Careful there boss. Because for every number that you pick there is still a neighborhood that does not contain the chosen number........

The flaw is when it is stated that x = .999...

.999... is not defined. In you define it as the limit... then you are already defining x as 1.


jeremy806...

 

[Mday]

Originally posted by: Parsec
I saw a link in RossGr's siggy about his proof that 1 = .99 repeating. It was awfully long and I didn't quite follow it all the way through, but it is wrong to just say this:
1/3 = 0.33...
2/3 = 0.66...
Since 1/3 + 2/3 = 1
and 0.33... + 0.66... = 0.99...
then 1 = 0.99...

not quite.

if you want to say .9999... = 1, and use .666... = 2/3 and 1/3 = .333... to back it up, then you're pretty much using circular reasoning. cuz you still have to prove that the repeating part is a rational number.

 

[Dinominant]

It is because we are using a base 10 math system. It is the same as saying in binary, 1.1... equals 10. If we used a base 9 math system, then yes 0.9... in base 10 equals 1 in base 9. Using algebra in a fancy way doesn't explain anything when you use irrational numbers that can be explained rationally in another system.

BTW

x=0.9?
9x=8.9... not 9.0

0.9? does not equal 1

 

[Shalmanese]

Originally posted by: RossGr
My proof is a valid mathematical approach to the problem. I essentially show that .999... exists in every neighborhood on the real line that contains 1. That is you cannot find an interval around 1 which does not contain .999.... No matter how small an interval you choose. That is the defintion of equality on the real line. So mathematically 1= .999...

I feel that my proof contain exellect concepts for see why they are equal. It shows clearly that adding any small incremet to .999... results in 1 + . {a bunch of zeros}999.....

you will always have 9s left over, on any number you add to .999... results in 1 + some more and that is the definition of 1.

What about the interval (0.9999+1)/2 to 1?

If you can play funny tricks with numbers, so can I.

 

[RossGr]

Originally posted by: jeremy806
That is the defintion of equality on the real line.

Careful there boss. Because for every number that you pick there is still a neighborhood that does not contain the chosen number........

The flaw is when it is stated that x = .999...

.999... is not defined. In you define it as the limit... then you are already defining x as 1.


jeremy806...

Since when is .999... undefined? The ellipsis is accepted notation for an infinite repetition of the preceding pattern when it appears at the end of a number. It represents the existance of a unspecified but finite number when it appears inside a number. So there are no limits involved. .999.... is by definition short hand for an infinite number of 9s there is no need to take a limit. This notaion indicates that the limit has been taken.

 

[RossGr]

Originally posted by: Shalmanese

Originally posted by: RossGr
My proof is a valid mathematical approach to the problem. I essentially show that .999... exists in every neighborhood on the real line that contains 1. That is you cannot find an interval around 1 which does not contain .999.... No matter how small an interval you choose. That is the defintion of equality on the real line. So mathematically 1= .999...

I feel that my proof contain exellect concepts for see why they are equal. It shows clearly that adding any small incremet to .999... results in 1 + . {a bunch of zeros}999.....

you will always have 9s left over, on any number you add to .999... results in 1 + some more and that is the definition of 1.

What about the interval (0.9999+1)/2 to 1?

If you can play funny tricks with numbers, so can I.

I have shown, in my proof that

1- 10^n < .999... < 1+10^n

choose n=5, I believe that will be smaller then your interval. With simple algebra you can swap place with the .999... and the 1.

Did you even take the time to look at the proof?

 

[jjlsevil]

What is wrong with my proof?
Was it not explained perfectly for you.

Given, for example. 1.9(repeating) - .9(repeating) = 1.0??????

Do I really have to repeat this?

 

[RossGr]

Originally posted by: jjlsevil
What is wrong with my proof?
Was it not explained perfectly for you.

Given, for example. 1.9(repeating) - .9(repeating) = 1.0??????

Do I really have to repeat this?

Any manipulation which involves IMPLIED operations on an unspecified infinitly large digit is not a proof it is a DEMONSTRATION. This is the same trouble with the method of adding the fractions, operations on unspecified digits is not allowed.

 

[jjlsevil]

RossGr

So you're saying that fractions cannot be proven as infinitely repeating decimals and vice versa.
Given that statement is true, I am wrong.










But I don't believe that statement to be true. It's been a while since I went to math class, so I could be wrong. I doubt it, though.

 

[RossGr]

Originally posted by: jjlsevil
RossGr

So you're saying that fractions cannot be proven as infinitely repeating decimals and vice versa.
Given that statement is true, I am wrong.

But I don't believe that statement to be true. It's been a while since I went to math class, so I could be wrong. I doubt it, though.

When did I say that? I said that for a valid mathematical proof you cannot use implied operations on a digit "at infinity". There are other methods, such as what I used in my proof, to prove this type of result.

 

[bwanaaa]

Although the proof, 1/3 +2/3 =1, is tantalizing---it bothers me that .9repeating=1.

Is it possible that simple addition of irrational numbers is not clearly defined? I would be more comfortable with the addition of 2 infinite series-can each of the irrational numbers above, be expressed as the sum of an infinite series?

 

[PowerEngineer]

Not another thread on this....

.999999.... is definitely equal to 1.0

Moderator, do us all a favor and lock this thread!!!

:beer:

 

[illusion88]

Originally posted by: PowerEngineer
Not another thread on this....

.999999.... is definitely equal to 1.0

Moderator, do us all a favor and lock this thread!!!

:beer:

come on!
It will be fun!
/me sticks some popcorn in the airpopper. You wanna get down on this?

 

[RossGr]

Originally posted by: bwanaaa
Although the proof, 1/3 +2/3 =1, is tantalizing---it bothers me that .9repeating=1.

Is it possible that simple addition of irrational numbers is not clearly defined? I would be more comfortable with the addition of 2 infinite series-can each of the irrational numbers above, be expressed as the sum of an infinite series?

Oh dear... how can I begin to share knowledge when someone makes this sort of comment, Must I teach even the most basic elements. Why do you call .999... irrational? is .333.... irrational? I think it is about the 5th grade where you learn that infinitly repeating and terminating decimals represent rational numbers. If you do not even know that much about the number system I do not think you should contribute to this thread.

BTW an irrational number is one which there is NO repeating patterns, clearly .999.... is not of this type.

 

[Wiktor]

And what about the number 1 - 0.(9), by (9) I mean reapeting 9.
It is 0.0.....................1, but what kind of a pattern is that?? No pattern. Is it a rational number then? Can you go outside rational numbers just by using the operator "-" and rational numbers?
Rational numbers are such that can be represented as fractions: p/q
1/3 = 0.(3) etc. where p, q are integers.
but 0.(9)?
that would be:
999.../1000... - what is that? Infinity / Infinity?

OK, let's assume 0.(9) is a rational number, 0.000........................1 must be a rational number as well (look above).

Now you said that because there is no number between 0.(9) and 1 - they are equal. (Something about neighbourhood).
The same holds for 1.000................1 and 1.
By transitivity we quickly end up with the following:
0.(9) = 1 = 1.000..................1 that implies that:
0.(9) = 1.000................1

But is it equal?

To show that it is not equal I can try to find a number that is between 0.(9) and 1.000.......1, why wouldn't I pick the number 1 then?

Or can't you see that this can be easly extended to all rational numbers? (All of them are equal?). Well no.


I know this discussion is pointless so I won't go on. I belive that 0.(9) has a limit of 1, but is not equal to 1 as a number. I see no trouble I could run into this way.

EDIT:
Right, I gave it more thought and I think it all depends on what set on numbers we are working on:
If it is RATIONAL numbers, then:
If we consider 0.(9) to be a rational number (because there is a pattern - 9 is always followed by 9), then:
Consider this operation:
1 - 0.(9) = ?
it is equal to 0, that implies that 1 = 0.(9), there I convinced myself , why not 0.000...............1 ? Because that is not a rational number - no repeating pattern!
Now we end up with this lack of symmetricy (sp?), 0.(9) is a rational number, but 0.000..........1 isn't.
If we assume that 0.000................1 is a rational number, 0.(9) != 1.

If it is REAL numbers, then:
0.000...............1 is a real number and it is the solution to
1 - 0.(9) = ?
and therefore 1 != 0.(9).


Now I have to think if 0.000...........1 is a number at all (if it ain't, then 0.(9)=1 both under rational and real numbers!).

 

[Jeff7]

Originally posted by: RossGr

Originally posted by: Parsec
I saw a link in RossGr's siggy about his proof that 1 = .99 repeating. It was awfully long and I didn't quite follow it all the way through, but it is wrong to just say this:
1/3 = 0.33...
2/3 = 0.66...
Since 1/3 + 2/3 = 1
and 0.33... + 0.66... = 0.99...
then 1 = 0.99...

Mathematically, the trouble with this computation is that it cannot be completed. You are adding an infinite number of digits, this is impossible. This computation falls into the catogory of demonstration, it is not a proof. The fact that the result is valid does not justify the questionable methods used to obtain it.

My proof is a valid mathematical approach to the problem. I essentially show that .999... exists in every neighborhood on the real line that contains 1. That is you cannot find an interval around 1 which does not contain .999.... No matter how small an interval you choose. That is the defintion of equality on the real line. So mathematically 1= .999...

I feel that my proof contain exellect concepts for see why they are equal. It shows clearly that adding any small incremet to .999... results in 1 + . {a bunch of zeros}999.....

you will always have 9s left over, on any number you add to .999... results in 1 + some more and that is the definition of 1.

How can you accurately plot 0.99repeating on a number line? It would need to be an infinitessimally small distance in front of the 1.0 point. How would this be accomplished? I just wonder that current math has any good way (other than fractions) of dealing with an infinite string of numbers. Conventional math seems to rely on having a set amount of digits to work with.

 

[rjain]

has no one here taken high school algebra?

 

[rjain]

Originally posted by: Jeff7

I just wonder that current math has any good way (other than fractions) of dealing with an infinite string of numbers. Conventional math seems to rely on having a set amount of digits to work with.

Infinite sums. Simple algebra. Even shows up on SATs.

 

[ZeroNine8]

It seems like what was proven is that the limit of 0.999... = 1 as the number of repeated digits goes to infinity. I have no problem believing that 1 is the limit of the convergent series 0.999..., however just because it is the limit does not mean it is equal. Because infinity isn't a number, merely a concept, you can't claim that 0.999... with infinite digits actually equals 1, only that it approaches it as the limit with more and more 9's.

In plenty of cases, the limit of a convergent series is never actually reached by that series.

 

[RossGr]

Originally posted by: Wiktor
And what about the number 1 - 0.(9), by (9) I mean reapeting 9.
It is 0.0.....................1, but what kind of a pattern is that?? No pattern. Is it a rational number then? Can you go outside rational numbers just by using the operator "-" and rational numbers?
Rational numbers are such that can be represented as fractions: p/q
1/3 = 0.(3) etc. where p, q are integers.
but 0.(9)?
that would be:
999.../1000... - what is that? Infinity / Infinity?

OK, let's assume 0.(9) is a rational number, 0.000........................1 must be a rational number as well (look above).

Now you said that because there is no number between 0.(9) and 1 - they are equal. (Something about neighbourhood).
The same holds for 1.000................1 and 1.
By transitivity we quickly end up with the following:
0.(9) = 1 = 1.000..................1 that implies that:
0.(9) = 1.000................1

But is it equal?

To show that it is not equal I can try to find a number that is between 0.(9) and 1.000.......1, why wouldn't I pick the number 1 then?

Or can't you see that this can be easly extended to all rational numbers? (All of them are equal?). Well no.


I know this discussion is pointless so I won't go on. I belive that 0.(9) has a limit of 1, but is not equal to 1 as a number. I see no trouble I could run into this way.

EDIT:
Right, I gave it more thought and I think it all depends on what set on numbers we are working on:
If it is RATIONAL numbers, then:
If we consider 0.(9) to be a rational number (because there is a pattern - 9 is always followed by 9), then:
Consider this operation:
1 - 0.(9) = ?
it is equal to 0, that implies that 1 = 0.(9), there I convinced myself , why not 0.000...............1 ? Because that is not a rational number - no repeating pattern!
Now we end up with this lack of symmetricy (sp?), 0.(9) is a rational number, but 0.000..........1 isn't.
If we assume that 0.000................1 is a rational number, 0.(9) != 1.

If it is REAL numbers, then:
0.000...............1 is a real number and it is the solution to
1 - 0.(9) = ?
and therefore 1 != 0.(9).


Now I have to think if 0.000...........1 is a number at all (if it ain't, then 0.(9)=1 both under rational and real numbers!).

Did you bother to read my rules for the use of the ellipsis? .000...1 implies that a finite but unspeficed number of zeros has been left out.


There is NO NUMBER with an infinite number of zeros followed by a 1. This statement is self contradictory. If the string of zeros is followed by a 1 (or anything other then a zero!) it is not infinite, it terminates at the 1 and can be counted.

in general your .000........1 = 10^-n where you have not specified n, this is a rational number.