# What if our numbering system was not based on 10?

1. Jan 17, 2005

### Gamish

What if our numbering system was not based on 10? What if it was based in 12, or 99, who knows. Would math theoreticly still work just as our mathematical system of 10 seems to work flawlessly?

BTW, what is .9i/2? is it .49i5? i=infinite!

2. Jan 17, 2005

### chroot

Staff Emeritus
Arithmetic can be done in any base you'd like, with no differences in functionality; the numbers would just look a little different.

People commonly use binary (base 2), octal (base 8) and hexadecimal (base 16) when working with computers. Computers themselves do everything in binary.

0.9 i / 2 = 0.45 i.

i does not mean "infinite." It means "the square root of negative one."

- Warren

3. Jan 17, 2005

### Gamish

Ya, I just use "i" because I don't know what else to use. Can you please tell me what to use to represnt infinite? And I though about it, there is NO square root if -1, hehe. Because a negative times a negative equals a positive, and 1 is the standard unit, so I guess i means "imaginary". I think I remember something about negative roots in school, I forget.

So if we have 0 1 2 3 4 5 6 7 8 9 a b, whats b/2? Or something like that...

4. Jan 17, 2005

### chroot

Staff Emeritus
How about the infinity symbol, $\infty$.
Of course there is -- it's i. It happens that i is not one of the reals, but that doesn't make it any less valid. It's unfortunate that the words "real" and "imaginary" have led so many people to think that complex numbers are somehow less valid than purely real numbers.
Well, I think you were originally asking what "0.9 times infinity, divided by two" is. The answer is infinity. 0.9 times infinity is still infinity. One-half of infinity is still infinity.

Now, your new example, with {0,1,2,3,4,5,6,7,8,9,a,b} as digits in a non-decimal base. If I am to assume that there are twelve digits, then b/2 is halfway between 6 and 7, which would be represented as 6.6 in this number system.

- Warren

5. Jan 17, 2005

### Manchot

Basically, every number can be expressed in any base numbering system you want. We usually choose base 10 to be our numbering system (as you well know). This means that every digit represents a certain number of power of 10s. For example, 172 = 1*100 + 7*10 + 2*1. Thus, in the base 12 system (a.k.a the duodecimal system), every number is expressed in terms of powers of 12. However, all arithmetic stays the same! b/2 in duodecimal means 11/2 in decimal = 6 and a half. Converting back to duodecimal, we know that .6 would equal one half, and 6 stays the same. Our result, therefore, is 6 + .6 = 6.6.

6. Jan 17, 2005

### Zurtex

Just to give you guys a heads up I think Gamish means 0.9 recurring when he talks about 0.9i and his 0.49i5 is an attempt to display infinitesimals, just so you know where this thread is probably going

7. Jan 17, 2005

### Janitor

Just to add a bit to what Manchot posted, I think it can be said that using a different base would not change any of the deeper results of number theory. There are some fun problems in "recreational mathematics" that do depend on the base, and of course they would have to be re-stated if you switched to some other base. An example would be issues involving the sum of the digits in a number. For instance, in base 10, the sum of the digits of 305 is 8. The same quantity written in terms of some base other than 10 could have different digit sum.

8. Jan 18, 2005

### Gamish

i

So, what is the text-based or ASCII form of expressing $\infty$? Let me give this example.

.9/2 = .45
.99/2 = .495
.999/2 = .4995
.9999/2 = .49995
.9$\infty$ = ? perhaps .49$\infty$ with a 5 at the end. Is this wrong? Is it inccorect to assume that a number can be after infinit?

9. Jan 18, 2005

### Cosmo16

To answer the original question, absolutly nothing would change, just how it was written. look

Binary (Base-2) Base 10
0=1
10=2
11=3
100=4
101=6
110=6

10. Jan 18, 2005

### dextercioby

Yap,folks,he meant recurring decimal...

Well,the notation i use is the one i've been taught in school,the one woth round brackets...
$$0.499...=0.4(9)=\frac{49-4}{90}=\frac{1}{2}$$

There,are u satisfied??

Daniel.

PS.There's a 5 at the end,but not in the sense u meant it...

11. Jan 18, 2005

### poolwin2001

$\infty$ is not a real number and arithmetic is not defined on it.
It is meaningless to think about $\infty$*2or $\infty$/54000

12. Jan 18, 2005

### dextercioby

BTW,
$$0.999...=0.(9)\equiv 1$$

Daniel.

13. Jan 18, 2005

### Gamish

Well, I did not put $\infty$ by itsself. What I meant by .9$\infty$
is .999...~

So, let me ask this. If $\frac {.999\infty} {2}$ does not equal $\frac {1} {2}$, instead, 4(9)5, then .999... does not equal 1?

Last edited: Jan 18, 2005
14. Jan 18, 2005

### dextercioby

Yep,it's because recurring 9 is taken as unity that everything does make sense...

Daniel.

15. Jan 18, 2005

### matt grime

why do you insist on reinvineting the wheel in this manner? there is a perfectly good notation of recurring decimals without you inventing bizarre and conflicting new uses of symbols.

you're also doing the usual mistake of thinking that you can have a decimal point, a four, an infinite number of 9's and then a 5. you can't whilst talking about decimal expansions of real numbers, so don't.

16. Jan 18, 2005

### Gamish

OK, can you please tell me then what .999.../2 is? I though that dextercioby confirmed my theory, unless you know the answer to this rather simply math problem?

17. Jan 18, 2005

### matt grime

yes, it is, in the real numbers , (equivalent to) 1/2, as we established quite a while ago. When he said there's a five at the end but not inthe sense you mean, i suppose he means that it is also represented by 0.5

18. Jan 18, 2005

### Integral

Staff Emeritus
Fortunately .4999... =.5 so all of your concerns are for naught.

Think about it, how can you have an infinite number of digits followed by anything other then that same digit. As soon as you place a different digit you have ended the series of digits, so it is not infinite.

19. Jan 18, 2005

### Gamish

Sorry for the double post, it is just that I posted it, and then it did not show up when I refreshed the page, because the post was on page 2 of the thread, lol.

OK, so we have established that .999.../2 = .4999...? And the "5" at the end is rather imaginary, because you cannot have a number at the end of infinite, which means that the (9) was not really infinite. So, with all this said, it would be innacurate to assume a number can be after infinite, so .999.../2 would have to equal .1/2 :tongue2:

Last edited: Jan 18, 2005
20. Jan 18, 2005

### matt grime

what 5 at the end? who on earth apart from you thinks there is a 5 at the end in any way shape or form?

steady on there, no one has said any such thing at all. that sentence doesn't even make an logical, mathematical sense. in fact i'd even go so far as to say that it was self contradictory.

you appear to be doing maths as no one else does, who knows what is going on in your system.