#1**+10 **

change 220.59 base 10 to base 8

Mmm, the whole number part is easy enough

220/8 = 27 remainder 4

27/8=3 remainder 3

3/8=0 remainder 3

so $$220_{10}=334_8$$

Now the decimal part - I have never done this before.

There is no power of 8 that is also a power of 10 so it will have to be an approximation.

I will work it out to 3 base 8 places and then round it back to 2 I think.

$$\\\frac{59}{10^2}=\frac{x}{8^3}\\\\

59*8^3/100 = 302.08\\\\

so\;\;\;\; X\approx 302\\\\

\frac{59}{10^2}\approx \frac{302}{8^3}\\\\

$now change $302_{10}$ to base 8$\\\\$$

302/8=37 remainder 6

37/8=4 remainder 5

4/8= 0 remainder 4

so

$$\\302_{10}=456_{8}\\\\

so\\\\

0.59 \;(base10)\approx 0.456\; (Base8) \approx 0.46 \;\;$to 2 base 8 places$$$

SO

$$220.59_{10}\approx 334.46_{8}$$

I just checked it with Wolfram alpha and it is correct

http://www.wolframalpha.com/input/?i=change+220.59+to+base+8

Melody
Aug 2, 2015

#1**+10 **

Best Answer

change 220.59 base 10 to base 8

Mmm, the whole number part is easy enough

220/8 = 27 remainder 4

27/8=3 remainder 3

3/8=0 remainder 3

so $$220_{10}=334_8$$

Now the decimal part - I have never done this before.

There is no power of 8 that is also a power of 10 so it will have to be an approximation.

I will work it out to 3 base 8 places and then round it back to 2 I think.

$$\\\frac{59}{10^2}=\frac{x}{8^3}\\\\

59*8^3/100 = 302.08\\\\

so\;\;\;\; X\approx 302\\\\

\frac{59}{10^2}\approx \frac{302}{8^3}\\\\

$now change $302_{10}$ to base 8$\\\\$$

302/8=37 remainder 6

37/8=4 remainder 5

4/8= 0 remainder 4

so

$$\\302_{10}=456_{8}\\\\

so\\\\

0.59 \;(base10)\approx 0.456\; (Base8) \approx 0.46 \;\;$to 2 base 8 places$$$

SO

$$220.59_{10}\approx 334.46_{8}$$

I just checked it with Wolfram alpha and it is correct

http://www.wolframalpha.com/input/?i=change+220.59+to+base+8

Melody
Aug 2, 2015

#2**+5 **

Very nice , Melody.....

I'm having trouble wrapping my head around this part.....

59 / 10^{2} = x / 8^{3}

How did you know to do this???

CPhill
Aug 2, 2015

#3**+5 **

Thanks Chris :))

When working in base ten, the first place after the decimal point is tenths, the second is tenths squared, the third is tenths ^3 etc

When working in base eight, the first place after the decimal point is eigths, the second is eigths squared, the third is eigths ^3 etc

So I needed to change 59/10^2 to something over eight to the power of some positive integer.

I just chose to do it to three decimal places so I used 8^3.

Does that make sense?

Melody
Aug 3, 2015