binary #'s and base 8#'s

[chrissyjen]

Question:
binary number 1001 is =to what base 8 number?

sorry not a lot of room to show what I have tried so I will try to explain some.
8(raise0)=1, 8(raise1)=8, 8(raise2)=16 and so on
then set up into collums of UNITS, TWOS, FOURS......
tried some form of division
then on to converting binary using place-value system (so so wrong)
can you help me find the right formula?
thank you
cj

 

[stapel]

Question:
binary number 1001 is =to what base 8 number?

To learn how number bases work and how to convert between them, try here. :wink:

Once you've learned the basic background material, the following should be helpful:

. . . . .1001[sub:3uyaw6nr]2[/sub:3uyaw6nr] = 1(2[sup:3uyaw6nr]3[/sup:3uyaw6nr]) + 0(2[sup:3uyaw6nr]2[/sup:3uyaw6nr]) + 0(2[sup:3uyaw6nr]1[/sup:3uyaw6nr]) + 1(2[sup:3uyaw6nr]0[/sup:3uyaw6nr]) = 8 + 1 = 9

. . . . .9 = 8 + 1 = 1(8[sup:3uyaw6nr]1[/sup:3uyaw6nr]) + 1(8[sup:3uyaw6nr]0[/sup:3uyaw6nr])

 

[soroban]

[size120]Hello, chrissyjen![/size]

Binary number 1001 is equal to what base-8 number?

\(\displaystyle \text{Since we are going from base-2 to base-8 (where }8 = 2^3\text{), there is a trick.}\)


\(\displaystyle \text{Example: convert }11,\!110,\!010_2\text{ to base-8.}\)

\(\displaystyle \text{Starting at the right, group the digits in sets-of-}three:\;\;(11)(110)(010)\)

\(\displaystyle \text{Convert each group to base-ten: }\;\underbrace{11}_3 \underbrace{110}_6 \underbrace{010}_2\)

. . \(\displaystyle \text{And }there\text{ is our answer! }\;\;11,\!110,\!010_2 \;=\;362_8\)



\(\displaystyle \text{Example: Convert }101,\!100,\!100,\!110,\!111_2\text{ to base-16.}\)

\(\displaystyle \text{Starting at the right, group the digits in sets-of-}f\!our:\;\;(101)(1001)(0011)(0111)\)

\(\displaystyle \text{Convert each group to base-ten: }\;\underbrace{101}_5 \underbrace{1001}_9 \underbrace{0011}_3 \underbrace{0111}_7\)

. . \(\displaystyle \text{Therefore: }\;101,\!100,\!100,\1110,\1111_2 \;=\;5937_{16}\)