# Beating an old horse .(9) question

1. Mar 11, 2010

### CjStaal

I know, and conform to the ideal that .9 infinitely repeating does in fact equal 1 but I have a thought that keeps bothering me.
How can.9999999999999999999999999999999999999999999999999999999(repeating forever) multiplied by 2, equal 2?

2. Mar 11, 2010

### Mensanator

What do you think it should be?

3. Mar 11, 2010

### dodo

Is that the right question? Because, if .9999... equals 1, then .9999... multiplied by 2 equals 1 multiplied by 2.

Maybe you're trying to ask how is multiplication of periodic decimals defined; or if it is defined at all, other than by the procedure in the first paragraph. (Not that I know the answer, but I'm interested in the question.)

Edit: sorry, I was answering the original post, not the second; the second poster was (again) faster than me.

4. Mar 11, 2010

### CjStaal

Well I guess look at it this way. .(9)^2 =1? while 9^2 =81... That just confuses me.

5. Mar 11, 2010

### dodo

It's an optical illusion. :) You may want to try a few more nines, like 99^2, 999^2 or 99999999^2.

6. Mar 11, 2010

### CjStaal

Yea, they always execute with the length of the original number but with 8 at the end then the length of the number, with a 1 at the end terminating it.
ie. 9*9=81
99*99=9801
999*999=998001
9999*9999=99980001
99999*99999=9999800001

7. Mar 11, 2010

### dodo

And this decoration looks like an artifact caused by the number of nines being finite.

8. Mar 12, 2010

### Hurkyl

Staff Emeritus
0.99 * 0.99 is less than 0.99(9) * 0.99(9), right?

But 0.99(9) * 0.99(9) cannot be greater than 0.99(9), right?

So, can you tell me what the first digit after the decimal point of 0.99(9) * 0.99(9) is?

0.999 * 0.999 is less than 0.999(9) * 0.999(9), right?

But 0.999(9) * 0.999(9) cannot be greater than 0.999(9), right?

So, can you tell me what the second digit after the decimal point of of 0.999(9) * 0.999(9) is?

9. Mar 12, 2010

### JSuarez

This is a (very) small ambiguity in positional systems, but first, to answer your question:

$$0.9\left(9\right) = \sum^{+\infty}_{n=1}\frac{9}{10^n}=9\frac{0.1}{1-0.1}=1$$

Therefore:

$$2\times0.9\left(9\right) = 2\sum^{+\infty}_{n=1}\frac{9}{10^n}=18\frac{0.1}{1-0.1}=2$$

But this is a general feature of any positional numerical system: for example, $0.1\left(1\right) = 1$ in base 2. In general, given a base b > 1, its highest digit is d = b - 1 and you have:

$$0.d\left(d\right) = 1$$

Note that 1 has the same representation relative to every basis > 1.

10. Mar 13, 2010

### Monimonika

Allow me to demonstrate on how the multiplication can be done.

Now, traditionally when multiplying the following (the *s are used for space-filling, please adjust fonts or copy-paste into a simple text editor to line up characters) (I'm a newbie and can't figure out the LaTeX codes that would make this look even semi-coherent):

*0.999
*x 2
*------

we were taught to start from the right and work our way left:

*0.999
*x 2
*------
*0.018
*0.18
+1.8
--------
*1.998

However, since "2 x 0.999" is really just "2 x (0.009 + 0.09 + 0.9)", we can use the commutative property of addition/multiplication to change the order to "2 x (0.9 + 0.09 + 0.009)" and still get the same answer.

Therefore we can calculate from left to right, instead:

*0.999
*x 2
*------
*1.8
*0.18
+0.018
--------
*1.998

Using this method, we can multiply 0.999... by 2 and get the following:

*0.999999999999...
*x 2
*-----------------
*1.8
*0.18
*0.018
*0.0018
*0.00018
*0.000018
*0.0000018
*0.00000018
..
...
.....
+
------------------
*1.999999999999999....

2 x 0.999... = 1.999... = 1 + 0.999... = 1 + 1 = 2

Now, onto 0.999... x 0.999... (in my next comment)!

11. Mar 13, 2010

### Monimonika

Here is a (long and convoluted) way to multiply 0.999... by 0.999...:

**0.999...
x 0.999...
-------------

I will break this into separate components that will all be added back together at the end. Since 0.999... can be expressed as the geometric series "0.9 + 0.09 + 0.009 + 0.0009 +...", let's multiply each of those components by 0.999...:

**0.999...
x 0.900...
-------------
*0.81
*0.081
*0.0081
*0.00081
*0.000081
.
..
...
+
------------------
*0.8999999999....

and

**0.9999...
x 0.0900...
-------------
*0.081
*0.0081
*0.00081
*0.000081
*0.0000081
.
..
...
+
------------------
*0.08999999999....

and

**0.99999...
x 0.00900...
-------------
*0.0081
*0.00081
*0.000081
*0.0000081
*0.00000081
.
..
...
+
------------------
*0.008999999999....

and so on and so forth...

Yeah, I hope you recognize how the pattern goes. Anyway, let's try adding the multiplied components up:

*0.89999999...
*0.08999999...
*0.00899999...
*0.00089999...
*0.00008999...
.
..
...
+
--------------------

Eeek! Guess I'll have to break even this one down a bit. Remember, the above is the result of "0.999... x (0.9 + 0.09 + 0.009 +...)".

Let's add one row at a time, shall we? Here are the first two rows added together:

*0.89999999...
+0.08999999...
------------------

Let's see...since we know that there are only 9s from the 1000th place onward to the right for both rows, we can reaarange this from "(0.89 + 0.00999...) + (0.08 + 0.00999...)" to "(0.89 + 0.08) + (0.00999... + 0.00999...)".

So, broken down:

*0.89
+0.08
--------
*0.97

plus

*0.00999...
+0.00999...
------------
*0.01999... = 0.00999... x 2

(Remember the "0.999... x 2" thing? Same stuff.)

And then we add the two results together:

*0.97000000...
+0.01999999...
-------------
*0.98999999...

Phew! First two rows done. Now to add the third row (look back if you've lost track of what we were originally doing):

*0.98999999...
+0.00899999...
---------------

Okay, repeat of what we did in the previous steps, except this time it's from the 10000th place:

*0.989
+0.008
---------
*0.997

and

*0.000999...
+0.000999...
------------
*0.001999... = 0.000999... x 2

*0.99700000...
+0.00199999...
-------------
*0.99899999...

*0.99899999...
+0.00089999...
---------------

You should know the procedure by now:

*0.9989
+0.0008
---------
*0.9997

and

*0.00009999...
+0.00009999...
------------
*0.00019999... = 0.0000999... x 2

*0.99970000...
+0.00919999...
-------------
*0.99989999...

Fifth row:

*0.99989999...
+0.00008999...
---------------
*0.99998999...

Okay, so I skipped typing out the work here, but this comment is getting way too long. Just to refresh your (and my) memory, this is what the 0.999... x 0.999... equation looked like when converted it into an addition equation:

*0.89999999...
*0.08999999...
*0.00899999...
*0.00089999...
*0.00008999...
.
..
...
+0.00000000...
--------------------

I then started adding up the rows one by one (albeit I started by adding the first two rows). The results of each addition are as follows:

1st row: 0.89999999...
etc.

Well, the pattern should be obvious here, with that "8" going further to the right with the addition of each row. And since this pattern will go on without end, the result would be:

*0.89999999...
*0.08999999...
*0.00899999...
*0.00089999...
*0.00008999...
.
..
...
+
--------------------
*0.99999999...

And thus I show that 0.999... x 0.999... = 0.999...! Go me!

12. Mar 13, 2010

### Staff: Mentor

The more I look, the more nines I see. I wonder if that means you are all my creation.

13. Mar 13, 2010

### Tinyboss

Dodo's right: if you believe that .(9)=1, then 2*.(9)=2*1=2, end of story. That's the only answer you need...anything else just boils down to more arguments that .(9)=1, which you claim to believe already.

14. Mar 14, 2010

### ramsey2879

n*2 -n = n
.9(9) * 2 -.9(9) = 1.9(9) -.9(9) = 1 = .9(9)

15. Mar 14, 2010

### HallsofIvy

By the definition of our decimal number system, .99999... means the limit of .9, .99, .999, .9999, ....

Multiply each of those by 2:
1.8, 1.98, 1.998, 1.9998, ...

The limit of that is 2.

16. Jun 19, 2010

### hoodle

CjStaal wrote "I know, and conform to the ideal that .9 infinitely repeating does in fact equal 1 but I have a thought that keeps bothering me. How can.999...(repeating forever) multiplied by 2, equal 2?"

It can't, and doesn't. Simple. Why do you conform to the ideal, since it is clearly false? You've noticed the obvious. Those values aren't equal. It is illogical to pretend they are equivalent. Calculator programmers know the answer is wrong even though it's close to the right one, and they therefore program calculators to return a 1, which is the right answer, instead of the .999... that is actually calculated.

Decimal fractions are good for giving us fast close estimates in most cases, but this system is incapable of accurately representing most fractional values. Specifically, any values that are not evenly divisible by ten or its factors of two and five, are to a greater or lesser extent inaccurately represented in decimal fraction format. Some inaccurately represented values terminate, but are in the wrong "place value" positions; the others are the endless repeating or non-repeating representations for values. These "infinite" representations of values are clear indications of a falsely represented value. The value is exact and finite. The attempted representation for it is not, and can never be accurately arrived at through this calculation technique.

The more use of decimal fractions, the greater the likelihood of increased error and deviation in representing any values which include fractions. Those who use them a lot, like engineers, and some scientists, etc. know this, and adjust their decimal fraction calculation to be closer to accurate. There are established techniques for doing this. This still doesn't necessarily fully correct the value representation, rendering it accurate, however. The calculation to correct the error inherent in the decimal fraction calculation system would be more complex, and more work than just doing the calculations accurately in the first place, which can be done using "vulgar" old fashioned fractions.

The decimal fraction system is a buggy computing system, but useful enough for quickly estimating values that are pretty close to the value sought, if that's all that's required. If accuracy is important, old fashioned "vulgar" fractions will do the job with perfect accuracy, but are far more difficult and time-consuming to use. Also, the accurate statement of some important values remains unknown, like pi, for instance. We only have the inaccurate estimate that includes a decimal fraction. This presents difficulties and an impediment, but hopefully some diligent mathematicians will apply themselves to the problem.

Sorry if I'm tardy for the party, but hope this helps.

17. Jun 19, 2010

### Staff: Mentor

Yes it does. 0.(9) = 1, It can be proven in zillion ways. Search the forums for proofs as they were posted on many occasions. The key is in word "forever".

18. Jun 19, 2010

### HallsofIvy

No, it doesn't help at all. Pretty much everything you have said is wrong. "Diligent mathematicians" have applied themselves to the problem (hundreds of year ago) and, in particular, defined exactly what is meant by an "infinite series" and how to calculate them.

In particular, "0.999999.."= $.9+ .9(.1)+ .9(.01)+...= .9(1+ .1+ .1^2+ .1^3+ ...)$ is a "geometric series" which has sum
$$\frac{.9}{1- .1}= \frac{.9}{.9}= 1$$
exactly.

I suggest you learn some mathematics before you make statements like "Also, the accurate statement of some important values remains unknown, like pi, for instance." We know exactly what pi is equal to, thank you. The fact that it cannot be written in decimal form with a finite number of decimal places is irrelevant. The decimal numeration system is only one way of writing numbers- and not a particulary important one.

19. Jun 19, 2010

### Count Iblis

Multiplication by two is a continuous mapping. To get someting less trivial (or ambiguous), you need to consider a mapping that is not continuous. E.g., define the function

f(x) = x for |x|<1,

f(1) = 0

Then what is f(0.999999999999999999999999999....) ?

This of course depends on whether you should take the limit first and then evaluate the function or the other way around. My opinion is that it is not natural to take the limit inside the argument of the function first.

20. Jun 19, 2010

### Hurkyl

Staff Emeritus
Why not? That's what you wrote. If you meant something else, you should have written something else!

I've seen this notation:
f(1-)​
when someone wants the value of f(x) as x approaches 1 from the left, rather than the value of f at 1. Such ideas can even be built into a number system (although it's somewhat subtle and I don't remember how exactly it works). Of course, you could replace use hyperreals or operator algebras to get the effect too.