Base 10

[lvlastermind]

I know that our current decimal system has a base ten.

My question is, how can i figure out fractions to decimals with different bases?

basically, i would like to know how to use different bases other than 10.

[chroot]

You'd use long-division, just like you've always done in decimal.

Consider the following division problem in binary, which has only digits: 0 and 1.

100000 / 1000

Set up your long division as usual:

______1__
1000 | 100000
- 1000
---------
000000


Of course, this makes sense: 1000 in binary is 8 in decimal. 100000 in binary is 32 in decimal. 32 / 8 = 4, or 100 in binary.

All of the normal division, multiplication, addition, and subtraction algorithms you learned in grade school work exactly the same way in any base.

If you have a specific question you're trying to solve, please let me know, and I'll help you.

- Warren

[lvlastermind]

if i have a # like 1/9 (base 10) and i want to fine what it equals with base 5...How would i go about doing that?

[Integral]

Using base 10 arithmetic, multiply your decimal number by the new base. The integer part is a digit in the new base. Repeat the process with the fractional part. Each repetition generates the next digit.

For example .1 in base 10 to base 2

2*.1 = 0.2

integer part =0 so your first digit is 0
.110~ .02

Now take the fractional part and repeat.
2 *.2 = 0.4
.110~ .002

repeat
2*.4 =0.8
.110~ .0002

repeat
2*.8 = 1.6
Finally! a non zero digit!
.110~ .00012

2*.6= 1.2
.110~ .000112

2*.2=0.4
.110~ .0001102

now you can observe that a pattern is emerging.

This same method can be used for conversion to any base.

[lvlastermind]

thanks alot

[lvlastermind]

so would .1 (base 10) equal .021262 and so fourth in base 5?

[jcsd]

so would .1 (base 10) equal .021262 and so fourth in base 5?


No, 0.1 in base 5 would equal 0.\dot{2}. Remember base 5 would only use the digits 0, 1, 2, 3 and 4 so that 6 can't be in there.

[lvlastermind]

hmm....

can you show me you work for that?

For some reason im having trouble..

[jcsd]

Sorry I left out a zero, it should be : 0.0\dot{2}_5 (out of interest (0.\dot{2}_5 would be equal to a half). As Chroot says use long divison (I'll state all numbers in base 10 unless otherwise indicated by a subscript 5, also I've used fractions rather than decimals as I feel it's probably easier to see what's going on)

(1/10)/(1/5) = 0 R 1/10
(1/10)/(1/25) = 2 R 1/50
(1/50)/(1/125) = 2 R 1/250
(1/250)/(1/625) = 2 R 1/1250
(1/1250)/(1/3125) = 2 R 1/6250

That gives us so far 0.022225 + 1/6250

Now we've probably already guessed that this is going to be a a recurring number, infact we should of seen this about the begining as 1/10 = (1/5)(1/2) and x in the equation \frac{x}{5^n} = \frac{1}{2} can never be an integer.

[lvlastermind]

alright, thx