# Controversial math question

1. Mar 25, 2004

### 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.

2. Mar 25, 2004

### 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.

3. Mar 25, 2004

### Zurtex

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

a * 0 = 0

Dividing both sides by 0,

a = (0 / 0)

4. Mar 25, 2004

### Hurkyl

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

5. Mar 25, 2004

### Janitor

Hurkyl,

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

6. Mar 26, 2004

### 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

3(0.33333.....) = 3(3/9)
0.9999999...... = 9/9
0.9999999...... = 1

Last edited: Mar 26, 2004
7. Mar 26, 2004

### chroot

Staff Emeritus
8. Mar 26, 2004

### Bob3141592

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?

9. Mar 27, 2004

### Zurtex

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

10. Mar 27, 2004

### 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 .

11. Mar 27, 2004

### 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.

12. Mar 27, 2004

### Zurtex

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

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.

13. Mar 27, 2004

### 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.

14. Mar 27, 2004

### Integral

Staff Emeritus
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.

Last edited by a moderator: Apr 20, 2017
15. Mar 27, 2004

### Integral

Staff Emeritus
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.

16. Mar 27, 2004

### 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.

17. Mar 27, 2004

### Integral

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

18. Mar 27, 2004

### JonF

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.

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

19. Mar 27, 2004

### ShawnD

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

Last edited: Mar 27, 2004
20. Mar 27, 2004

### 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.