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...?