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
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
Very nice , Melody.....
I'm having trouble wrapping my head around this part.....
59 / 102 = x / 83
How did you know to do this???
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?