Someone help me with the 0.9 repeating = 1 proof

[kevinthenerd]

Originally posted by: PowerEngineer

It still remember my high school freshman algebra test in which we were asked to rationalize 2.9999...

I ran through the process: X=2.9999... 10X=29.9999... 9X=27.0 X=3.0

My teacher saw the stunned look on my face, and after seeing my answer asked me to think about why I was surprised. It dawned on me then that this proved that 0.9999... was actually equal to 1.0! (It seems to me that these kind of learning experiences while taking a test are the most powerful.) I can't understand why others aren't completely convinced by this demonstration.

Very true! I'm guess I'm a nerd right now because, throughout growing up, I've become addicted to such sudden rushes in my brain when I learn such things on my own. I don't find it at all entertaining to learn things out of books; learning them as though I'm the world's first mathematician (or scientist or engineer) is the most fun to me. Deriving things on my own (and occasionally learning new things along the way) is fascinating to me. It's cool when you can start with basic math and physics and work your way up to the Navier Stokes and stuff. It's much easier to remember a derivation than to memorize formulas, IMHO, and I also believe that the derivation gives you a deeper understanding of the physical phenomena at hand (and knowledge of when to assume certain things, etc.)

I think my favorite derivation, to this day, is that of centripetal acceleration in a circular path: a = v^2/r. Physics forms such a beautiful clockwork. Once you unlock the mystery of how it works, it unfolds an either deeper mystery before you: WHY does it work? That's the mystery that becomes even more glaring and draws that much more of your curiosity once you begin to learn HOW it works. WHY?

Anyway, I really disagree with the way most of the math and science courses have been taught to me. The focus has been on working standardized problems, not on synthesizing original thought out of basic concepts. It's a shame. Tests that really bend your mind are good in so many ways.

 

[pcy]

Hi,


I'd be very interested to see how you multiply 2.9999* by 9



Peter

 

[CSMR]

pcy, doing arithmetic on infinite expansions is not impossible. It is not necessary here, so the arguments given above are not useful, BUT they are correct.
2.99999...=2+sum 9*10^-i
10*2.999...=20+sum(9*10^(-i+1)=29+sum(9*10^-i)
so 9*2.999...=(10-1)*2.999...=27.
Obviously you can caluclate more easily by using the identity 2.999...=3.
Or you can do it more directly:
9*(2+sum 9*10^-i)=18+sum (81)*10^-i
=18+sum 8*10^(-i+1)+sum 10^-i
=26+sum 8*10^-i+sum 10^-i
=26+sum 9*10^-i
=26.999...
That's three ways of doing it, one of which had already been given.

 

[pcy]

Hi,


I thnink we may be in agreement here.


All the methods you describe involve performing Mathemactical Operations on the infinite series that these infinite decimal expansions represent. In particular they involve the concept of a limit, or of an identity derived from or dependant on the concept of a limit.

By contrast arithmetic is actually a typographical process whose rules have been constructed in such a way that they yield true results when interpreded as statements about numbers - and arithmetic is defined only for finite typographical strings.


What is happening here is that proofs are being offered using a version of arithmetic extended to infinite strings (without proof of validity, though in an intuitive fashion) which perport to show that (for instance - the original question):
0.9* = 1

The result is true, of course; but the proof is invalid, because the extension of arithmetic to infinite expansions depends on the body of mathematics and the very results we are trying to explain/prove in this thread.



Peter

 

[BrownTown]

real number

DEFINITION - A real number is any element of the set R, which is the union of the set of rational numbers and the set of irrational numbers. In mathematical expressions, unknown or unspecified real numbers are usually represented by lowercase italic letters u through z. The set R gives rise to other sets such as the set of imaginary numbers and the set of complex numbers. The idea of a real number (and what makes it "real") is primarily of interest to theoreticians. Abstract mathematics has potentially far-reaching applications in communications and computer science, especially in data encryption and security.

If x and z are real numbers such that x < z, then there always exists a real number y such that x < y < z. The set of reals is "dense" in the same sense as the set of irrationals. Both sets are nondenumerable. There are more real numbers than is possible to list, even by implication.

The set R is sometimes called the continuum because it is intuitive to think of the elements of R as corresponding one-to-one with the points on a geometric line. This notion, first proposed by Georg Cantor who also noted the difference between the cardinalities (sizes) of the sets of rational and irrational numbers, is called the Continuum Hypothesis. This hypothesis can be either affirmed or denied without causing contradictions in theoretical mathematics.

No number exists between .9999... and 1, therefore they cannot be seperate numbers by deffinion of the real numbers. Proving this via math is nice, but the fact that it was defined as such is what makes it true.

 

[pcy]

Hi,


True...


But I fail to see the relevance.



Peter

 

[RossGr]

It has always been my contention that these little algebraic "proofs" of 1=.999... are not proofs at all. Simply because they assume that the arithmetic works on infinite decimal expansions. These are demonstrations not proofs.

Here is a proof or at least some semblance of a proof which relies on something called the Nested Interval Theorem. Essentially it says that there exists a single number in the intersection of an infinite number of nested intervals. That means a series of intervals such that each successive interval is contained in the previous.

My proof constructs a single nested set of intervals which contains both 1 and .999... by the nested interval theorem they must be the same point.

 

[pcy]

Hi,

We certainly agree on that point.


I did post a proof earlier:

Define N(n) = 0.999999999.... (i.e Nine repeated n times)

Clearly N(n) = 1 - 1/10*n

Then 0.999* (recurring indefinately) = Lt {x -> infinity} N(n)
= Lt {x -> infinity} 1 - 1/10*n
= 1 - Lt {x -> infinity} 1/10*n
But: Lt {x -> infinity} 1/10*n = 0
so:
0.999* = 1


The key step is:
Lt {x -> infinity} 1/10*n = 0
which IMo can be established by the definion of a limit i.e.
For all Epsilon >0 there esists N such that 0 < 1/10*n < Epsilon for all n >N
<=>
Lt {x -> infinity} 1/10*n = 0


But it's clear that this is very closely associated with your Nested Interval Theorem.


Peter

 

[RossGr]

Pcy,
You are absolutely correct. In fact any proof of the nested interval theorem relies on the definition of a limit.

To All:
I feel that if you make an effort to understand the significance of my construction of the inequality .999... + .1^N > 1. And if you come to grips with the meaning of infinity. you will be able to fully understand the equality of 1 and .999... .

As for coming to grips with infinity, remember that the construction which the symbolic .999... represents is not a limit. That symbolic represents a decimal number with the digit 9 in all decimal places. You are not strolling down the series represented by .999...9 dropping a 9 into the next slot with with empty slots ahead. You are standing beside a line of 9's stretching as far as you can see in every direction. There is a difference.

 

[CSMR]

RossGR: you need your intervals to tend to zero in length or else your theorem is not true. Your method works, though again, it is somewhat circuitious.

 

[RossGr]

Yes, that is correct, the length of the intervals must tend to zero. Notice that this is ture for the set of intervals I have constructed.

The diffiulty with proving this is comming up with something that a non mathematician can grasp. That is what motivated my starting point. I feel that it makes it pretty clear that adding any number to .999... resultes in something greater then 1.

 

[kevinthenerd]

Originally posted by: Eeezee
She still wouldn't believe me though! She keeps saying that since there is always something at the end of the string of numbers (despite being infinite) then it can be added together with something else to get 1 and is therefore not 1. I'm certain that you can't add anything to 0.9 repeating and get 1, but how can I prove this to her?

Ok, I'm a nerd. (See screenname.)
Nerds wrote the textbooks.
I read those textbooks.
0.9 repeating is 1 because I said so.


/thread

 

[pcy]

Hi,

Originally posted by: RossGr
The diffiulty with proving this is comming up with something that a non mathematician can grasp...

Indeed. The Lady clearly has a bit of trouble with Infinity - see below. It looks to me like she sees it as if it were some unimaginably big, but still finite, value.

kevinthenerd may have the right approch:

Originally posted by: kevinthenerd

Originally posted by: Eeezee
She still wouldn't believe me though! She keeps saying that since there is always something at the end of the string of numbers (despite being infinite) then it can be added together with something else to get 1 and is therefore not 1. I'm certain that you can't add anything to 0.9 repeating and get 1, but how can I prove this to her?

Ok, I'm a nerd. (See screenname.)
Nerds wrote the textbooks.
I read those textbooks.
0.9 repeating is 1 because I said so.

Decimal EXpansions are actually defined in Arithmetic only if they are finite.

0.999* is not a finite Atithmetic string - it is a finite Mathematical notation indicating an infinite string. That means that Mathematicians are free to define what it means. They chose to define it in terms of the limit we have been discussing, because it gave useful, consistant results. All Mathematicains agree on this point. It is a piece of Matematical notation: it is their call. So, indeed:
0.9 repeating is 1 because I said so.

There are two other arguements tha Lady might find convincing.

The first is this: any number (>0) that Lady might come up with to add to 0.999* to make 1 is clearly too large. This is esentially the lay version of the Matehmatic concept of a limit.

The second: The idea that:
"there is always something at the end of the string of numbers"
is incorrect. Lady wants to add 0.00*1 (an infinite number of 0s followed by a 1) to 0.999* to make 1. I can sympathise. I wanted to do that too when I first saw this stuff in school.

I initially saw an infinite sequence as one whose end I could never reach, but which still had an end. I now see it as a sequence which has no end. The difference is subtle, but crucial. If it has an end but I just can't reach it, there could be a solitary 1 after all those 0s. But if it has no end, there is no place for the 1 to be, so it simply cannot exist.

The "no end" view is correct, because if 0.00*1 were permitted to exist, it would be the smallest number greater than 0, leading immediately to the contradiction that 1/10th of it would be smaller still, but also greater than 0.

It's important that this chain of reasoning is part of the motivation for defining what 0.999* means in the way that matematicians chose.

So it comes back to:
0.9 repeating is 1 because I said so.
yet again.



Peter