Is 0 divided by 0 equal to any number?

[MC363A]

Can anyone tell me why, or why not,.\bar{9} is equal to one?
Also, if anyone cares, I can prove that \frac{0}{0} = any number.

If 0 times any number is equal to zero, then zero divided by zero is any number.

 

[verty]

Think of the sequence with the n'th term given by '1 - 1/n'. When n = 10, the number is 0.9. When n = 100, the number is 0.99. As n grows large, the value of the term gets closer to 1. In the limit when n reaches infinity, the term 1/n becomes 0, so the limit of '1-1/n' as n tends toward infinity is 1. That's why 0.99999... is = 1.

 

[Zurtex]

Let a be any real number. We can say that:

a * 0 = 0

Dividing both sides by 0,

a = (0 / 0)

 

[Hurkyl]

Let a be any real number. We can say that:

a * 0 = 0

Dividing both sides by 0,

a = (0 / 0)

What grounds do you have for asserting that dividing a valid equation by zero yields a valid equation?

 

[Janitor]

Hurkyl,

I think Zurtex was just having his fun anticipating the invalid proof that MC was going to provide.

 

[ShawnD]

x = 0.99999999...
10x = 9.999999...
10x - x = 9.9999... - 0.999999...
9x = 9
x = 1

There's always the way of expressing repeating decimals.
0.777777... is 7/9
0.484848... is 48/99
0.674674674... = 674/999

Therefore, 0.9999... = 9/9 = 1

That might be a stretch for believability though so try this one

0.333333... = 3/9 (go ahead and check with your calculator)
3(0.33333...) = 3(3/9)
0.9999999... = 9/9
0.9999999... = 1

 

[chroot]

http://www.mathforum.org/dr.math/faq/faq.0.9999.html

 

[Bob3141592]

Can anyone tell me why, or why not,.\bar{9} is equal to one?

I think of it this way. What's \frac{1}{3} * 3 ? If you just simplify the fraction, it's simply 1. But if you write \frac{1}{3} as .3333... then the answer is .9999...

Right?

 

[Zurtex]

What grounds do you have for asserting that dividing a valid equation by zero yields a valid equation?

No grounds, in fact even if we ignore that it's really easy to show the proof is self contradicting, I've assumed 0/0 = 1 to prove that 0/0 = any real number

 

[Zurtex]

Owww sorry for the double post but I just thought of a good one.

Let a be any real number.

a*0=0

Dividing both sides by 0.

a \frac{0}{0} = \frac{0}{0}

Dividing both sides by \frac{0}{0}

a = \frac{\frac{0}{0}}{\frac{0}{0}}

a = 1

All real numbers are equal to one .

 

[matt grime]

Pah, childs play. Surely you can see the easy generalization of this idea to demonstrate that every real function is identically equal to 1. In fact I think I can demonstrate that everything is equal to 1, and I can also show 0=1 so we're really in trouble now.

 

[Zurtex]

Pah, childs play. Surely you can see the easy generalization of this idea to demonstrate that every real function is identically equal to 1. In fact I think I can demonstrate that everything is equal to 1, and I can also show 0=1 so we're really in trouble now.

Well I'm fairly new to maths and it amuses me .

Speaking of which would anyone know some good links to learn various aspects number theory? I am very interested in it but the maths course I am on covers nothing of it.

 

[matt grime]

There's a small cheap book by Baker on Number theory that's worth a look, or Le Veque's Fundatmentals of Nember Theory was available in a Dover reprint a few years ago too. I presume you mean algebraic (the nice elegant stuff) and not analytic (arguably more powerful but nasty with it).

As for the above, as there is nothing that allows us to divide by zero, yet we do it, there is nothing to stop us declaring 0*x=0 for any conceviable x, be it a number, a function or absolutely anything, and we're doing no more damage than we were before. After all, surely no lots of anything are nothing? So x=1 and I didn't even say what x is.

 

[Integral]

http://home.comcast.net/~rossgr1/Math/one.PDF a pdf I put to gather with a couple of proofs. The first simply uses the expression for the sum of a geometric series.

The 2nd however is a bit more fundamental. It uses a method which I feel gives a very good insight as to why they must be equal.

If HallsofIvy or one of the other real mathematicians on the board should look at this I would appreciate feedback on how to clean it up and finish it off better.

 

[Integral]

x = 0.99999999...
10x = 9.999999...
10x - x = 9.9999... - 0.999999...
9x = 9
x = 1

There's always the way of expressing repeating decimals.
0.777777... is 7/9
0.484848... is 48/99
0.674674674... = 674/999

Therefore, 0.9999... = 9/9 = 1

That might be a stretch for believability though so try this one

0.333333... = 3/9 (go ahead and check with your calculator)
3(0.33333...) = 3(3/9)
0.9999999... = 9/9
0.9999999... = 1

I see these type of manipulations as a good demonstration that the relationship holds but IMHO they do not constitute a proof. Any operation on a non finite digit can be called into question.

 

[matt grime]

Algebraic manipulation of infinitely long decimals is well defined.

The pdf seems far too long given the information it attempts to convey. Perhaps you ought to simply deal with the finite partial sums to explain why th infinite one exits, even if that requires you to explain the very basic analysis you are eliding, which is only a definition after all.

 

[Integral]

Algebraic manipulation of infinitely long decimals is well defined.

Sure it is, but it is not appropriate for a proof. As I said it is a fine demonstration.

 

[JonF]

There's always the way of expressing repeating decimals.

Therefore, 0.9999... = 9/9 = 1

Oh really? Proof please. To say that .9999=1 is to assume what you are trying to prove, before you prove it.

I’d also like to see a proof that all repeating decimals can be expressed as a fraction. A few examples hardly proves anything. And even if that is true, how do you know that .99999 corresponding fraction is 9/9.

0.333333... = 3/9 (go ahead and check with your calculator)
3(0.33333...) = 3(3/9)
0.9999999... = 9/9
0.9999999... = 1

Oddly enough my TI8-89 claims that .33333 is an approximation of 3/9...

 

[ShawnD]

I'd also like to see a proof that all repeating decimals can be expressed as a fraction. A few examples hardly proves anything.

Show me 1 example where a repeating decimal number cannot be expressed as a number divided by a series of 9's.

 

[JonF]

But that is not what your previous post suggest. You made it out to be that all repeating series can be express as: A/B where “A” and “B” are integers. That’s the foundation of your whole “proof”, now please prove it.

 

[ShawnD]

I did prove it. (3/9) x 3 = 1. What is so complicated?

 

[h2]

"I’d also like to see a proof that all repeating decimals can be expressed as a fraction"

Geometric series: If |r|<1 then the sum of r^k from k=0 to 00 is 1/(1-r): Sum(r^k) = 1/(1-r)

Example:
0.9999... = 9/10 + 9/100 + 9/1000 +... =
(9/10)[1 + 1/10 + 1/100 +..] =
(9/10)[ 1 / (1-1/10)] = (9/10)(10/9) = 1

 

[Integral]

http://home.comcast.net/~rossgr1/Math/decimal.pdf Is a quick demonstration that any repeating 2 digit pattern is a decimal representation of those 2 digits divided by 99. I think if one worked at it a bit the method could be generalized to show that ANY repeating decimal can be represented as the repeating digits divided by as many 9s as there are repeating digits.

It is also pretty straight forward to show that any rational number can be represented as either a finite length decimal expansion or a infinitely repeating decimal expansion.

Consider the process of long division. There are exactly 10 possible results to the multiplication of the divisor by a single digit of the quotient. The number of possible results to the subsequent subtraction is limited by the size of the divisor since the result is always smaller then the divisor. As the process is carried out there are only 2 possible outcomes, the process terminates or at some point a result of the subtraction will be repeated. This must happen because there are a finite number of possible results and the process can be repeated indefinitely. If the process does not end it must repeat on or before the number of digits equal to the size of the divisor.

A great number to look at this with is 1/7 =.142857...

I believe that this concept could be written up as a formal proof of the fact that rational numbers can be represented as either a terminating decimal or an infinitely repeating pattern.

 

[Muzza]

Oddly enough my TI8-89 claims that .33333 is an approximation of 3/9...

Did you type in an infinite number of threes?

 

[cookiemonster]

I think bringing calculators into this is a bad, bad, bad idea.

cookiemonster

 

[Integral]

I think bringing calculators into this is a bad, bad, bad idea.

cookiemonster

My thoughts also.

 

[matt grime]

I thought I'd posted a proof for Jon, but it ain't here so:

let z be an eventually recurrent decimal, that is z = x+y where x is terminating (and trivially rational) and y is a recurring decimal, say with period n. Then y*10^n - y is a terminating decimal, and hence rational, thus y is rational, and it follows z is rational too.

 

[Integral]

I thought I'd posted a proof for Jon, but it ain't here so:

let z be an eventually recurrent decimal, that is z = x+y where x is terminating (and trivially rational) and y is a recurring decimal, say with period n. Then y*10^n - y is a terminating decimal, and hence rational, thus y is rational, and it follows z is rational too.

It is not clear to me how this proves y is rational, could you please expand on this? It looks to me like y*10ⁿ -y is an integer. (since y is repeating period n) Example
y=.242424...
n=2

y*10ⁿ= 24.242424...

24.242424... - .242424...=24



Thanks

 

[matt grime]

If it is an integer it is rational. It doesn't need to be an integer. Try it with .000012121212... with the repeating12 pattern. you get .001200000... which is rational. It just produces from a recurrent decimal y, a terminating decimal, r, ie rational, satisfying y(10^n-1)= r hence y is a rational divided by an integer, thus a rational.

 

[Organic]

What is the result of 9.9999.../0.9999...?

 

[Zurtex]

What is the result of 9.9999.../0.9999... ?

Well as 0.9999... = 1 and 9.9999... = 10 * 0.9999... = 10 * 1 = 10

Then you are asking what the result of 10 / 1 is.

Edit:

I suppose another way of looking at the problem is saying that it is the same as:

\lim _{x \rightarrow \infty} \frac{10 - 10^{-x}}{1-10^{-x}}

Where x is a natural number.

 

[Organic]

9.9999.../0.9999... = 10 , which is the ratio between 9.9999... and 0.9999...
9.9999 .../1.0000... = 9.9999... , which is the ratio between 9.9999... and 1.0000...

Now let us check this arithmetic:

x = 0.99999999...
10x = 9.999999...
10x - x = 9.9999... - 0.999999...
9x = 9
x = 1

A question: How we can be sure that the result of 0.99999... - 0.99999...
is exactly 0 where there is no right side to begin the subtraction operation?

 

[matt grime]

Because addition and subtraction can be put into an algorithm for infinitely long decimals if you need to do so. However as you're asking what the result of x-x is it doesn't depend on x having a nice form, it is zero by definition. Why do people confuse numbers and their decimal representations?

 

[matt grime]

Or if you don't like that answer then how about doing it in terms of Cauchy sequences, which is after all the most useful construction of the Real numbers as a mathematical object.

 

[Organic]

Please prove that there is no connection between a number and its stuctural represention upon infinitely many scales.

 

[matt grime]

I didn't say there wasn't a connection (though what you mean by that is unclear) but that the addition of two real numbers is independent of the choice of decimal expansion, should it have two. The simple proof of this fact follows from the definition of the real numbes as the completion of the rationals. Go and get a basic analysis book. Just because you do not know it, Organic, does not mean it is not true or known by other better informed people. Proof: let x_n and y_n be two equivalent Cauchy sequences. This means x_n-y_n converges to zero. Let w_n and v_n ba any other pair of equivanlent cauchy sequences.

then the element of R that [x_n-w_n] coresponds to is the same as the class (real number) [y_n-v_n]

proof: we are to show x_n-w_n-y_n+v_n converges to zero, but that is trivially true since |x_n-w_n_y_n+v_n| < |x_n-y_n| +|w_n-v_n| and both those terms can, be made arbitrarily small by hypothesis,a nd we have proved subtraction of two real numbers is indpendent of the Cauchy sequences we pick to represent them. OK?

 

[Organic]

What you show is a rough jump that forces infinitely long sequence to become finitely long, and than you use subtraction after you created an artificial right side (which cannot exist in infinitely long sequence) in a non-logical way for your own purpose.

 

[matt grime]

No, this is the rigorous proof that addition is well defined on the Real numbers. Perhaps you ought to go and learn some mathematics?

 

[Zurtex]

Have you ever just thought about converting 0.3333333... to base 3 to get 0.1 which when multiplied by 3 you get 1.

 

[Janitor]

Zurtex,

I would anticipate that Organic's objection to your suggestion is that "there is no right side to begin the base-conversion operation."

 

[Organic]

Hi Zurtex,

This is exactly my point of view on this case, a number is not just a quantity but has also an internal structute that cannot be ignored, for example look at this paper:

http://www.geocities.com/complementarytheory/Complex.pdf

Perhaps you ought to go and learn some mathematics?

I cannot agree with mathematics which is based on forcing methods.

 

[Integral]

If it is an integer it is rational. It doesn't need to be an integer. Try it with .000012121212... with the repeating12 pattern. you get .001200000... which is rational. It just produces from a recurrent decimal y, a terminating decimal, r, ie rational, satisfying y(10^n-1)= r hence y is a rational divided by an integer, thus a rational.

very good! Thank you.
This gets us the next gem also, observe that:

y = \frac r {10^n -1}

The denominator will consist of n 9's

 

[matt grime]

Organic, the question was posed in the field of real numbers with their mathematical definition. Your opinions as to whether that is not the correct object are irrelevant to the question, and its answer. Or would you like to point out where the proof that 0.99999 =1 is wrong when working in the usual definition of the real numbers as equivalences classes of cauchy sequences? I do not need to use all the things that are true about some object to prove things about it. For instance, every digit in 0.99999... is a prefect square, I didn't use that fact. I didn't use the fact that considered as curves embedded in the plane the digits involved all have fundamental groups that aren't trivial. I didn't use the fact that 9 is 6+3 where 3 is the smallest odd prime and 6 is the smallest order of a non-abelian group. If you aren't prepared to learn what mathematics involves then how can you possibyl answer questions about it? I mean, there is a theory where 0.9999... is not equalt to 1. Perhaps you want to learn about Abraham Robinson's non-standard analysis? Whereof you do not know do not speak?




Integral, yes, but, there's no need for the r in there to be an integer, which is what you wanted originally I seem to recall. And ignores the x. Example .011111 = 1/90, arguably a recurrent decimal and not consisting eintrely of 9s in the denominator.

 

[matt grime]

Have you ever just thought about converting 0.3333333... to base 3 to get 0.1 which when multiplied by 3 you get 1.

In base 3 one has other problems such as .22222222... =1

Forget decimals, or any other system of representation like that. Just operate with the definitions of the real numbers. That's how mathematics works, practically.

 

[Integral]

Matt,
Perhaps now you are beginning understand why my proof does not perform operations on non finite digits. People such as Organic, who should be restricted to posting in Theory Development, simply will not accept any proof you can provide that non finite operations are allowable. Beyond that I was taught in my analysis courses that such operations should not be included in fundamental proofs.

Organic is special in that he has his own number system which he cannot separate from the Reals that the rest of us use.

 

[Integral]

Integral, yes, but, there's no need for the r in there to be an integer, which is what you wanted originally I seem to recall. And ignores the x. Example .011111 = 1/90, arguably a recurrent decimal and not consisting entirely of 9s in the denominator.

So simply factor out the non repeating part.

.0111... = .111... x 10^{-1} = \frac 1 9 x 10^{-1}

So we have a multiple of 10 and a rational with 9's in the denominator. This is validation of the methods mentioned up thread using 9's in the denominator. We have shown that every repeating decimal can be represented as a factor of 10 and the repeating portion over 9s.

 

[Integral]

In base 3 one has other problems such as .22222222... =1

Forget decimals, or any other system of representation like that. Just operate with the definitions of the real numbers. That's how mathematics works, practically.

Matt,
The Reals are not base dependent, a base 2 or 3 or 16 system is the same as a base 10 representation as far as the real system is concerned. Yes, different bases have different rationals as repeating decimals. For example

.1 (base 10) = .0001100110011...(base 2) (I think I got the right number of leading zeros). This is of significance because it means your computer must round off .1 .

EDIT: Opps, Matt I just reread your post, I got different meaning the 2nd time. I am trying to say the same thing you are. The Reals are Base independent.

 

[matt grime]

Where did I say the reals are base dependent? If everyone remembered what the real numbers acutally are then none of these recurring (pun intended) nightmares would happen. It's amazing how often this question comes up, isn't it?

 

[ShawnD]

Have you ever just thought about converting 0.3333333... to base 3 to get 0.1 which when multiplied by 3 you get 1.

That's sort of how the fraction representation works. 1/3 x 3 = 1.

 

[Organic]

Organic is special in that he has his own number system which he cannot separate from the Reals that the rest of us use.

Let us take the circle's equation: (x-h)^2 + (y-k) = r^2
http://www.xavierhs.org/departments/Mathematics/PreCal/Conics/conics.htm

solid is a "one piece" state

r=radius

h=x center

k=y center

But the interesting variables are x and y, where x is the entire x-axis and y is the entire y-axis.

x-axis or y-axis are "actual form of infinity" as we can see in this model:
http://www.geocities.com/complementarytheory/RiemannsLimits.pdf

To construct the circle we have to break the solid states of both x-axis and y-axis and define a sequence of unique pairs of R members, which are used as x,y coordinates of the circle.

The point here is that we have no R members before we break the solid state of x-axis and y-axis, and only after we break them we get R members.

The same state is an information form of, for example, 0.9999999...

It cannot be in both states of finite and infinite sequence of non-zero values upon infinitely many scales.

Therefore there is a XOR condition between 1.0 and 0.9999... exactly as there is a XOR condition between a solid state and a broken state.