Conversion from binary fraction to octal Can someone verify my results?
    Binary to Octal
1) 10101.11 is 25.75

2) 0.01101 is 0.6875

3) 10110110.001 is 266.125   

 A: To convert quickly, group into chunks of three, starting at the decimal point.
Say I have the binary number:  $$1001010110100000101.10010101$$  Lets write this as 
$$001 \ 001 \ 010 \ 110 \ 100 \ 000 \ 101 \ .100 \ 101 \ 010$$  Notice I added some zeros, which don't change the number, to make sure I have chunks of three.  Now just write each of these chunks of three as a unique number in $0,1,2,3,4,5,6,7$ by converting from binary as you normally would.
So for my number above, $001=1$, $010=2$, $110=6$, $100=4$, $000=0$, $101=5$, $100=4$, $101=5$, and $010=2.$  Hence in octal we have that $$1001010110100000101.10010101=126405.452$$
Now try using this method to convert the numbers you have above.  (I won't do it for you) The decimal parts are not correct at the moment.
Why does this work?  The reason this works is because $8$ is a power of $2$.  In particular $8=2^3$, which is why we split into blocks of $3$.  Each block of $3$ in binary corresponds to a unique number in octal, and vice-versa.  Say I have a binary number $$x=a_n a_{n-1} \cdots a_2 a_1 a_0.a_{-1}a_{-2}\cdots a_{-m}.$$  Each $a_i$ here represents a digit that is either $0$ or $1$.  Then $$x=a_n 2^n +a_{n-1}2^{n-1}+\cdots +2a_1 +a_0+a_{-1} 2^{-1}+\cdots +a_{-m}2^{-m}.$$  We now want to write this in base 8.  Lets group terms:  
$$x=\cdots +8^1\left(4a_5+2a_4+a_3\right)+ 8^0\left(4a_2+2a_1 +a_0\right)+8^{-1}\left(4a_{-1}+2a_{-2}+a_{-3}\right)+8^{-2}\left(\cdots \right)\cdots .$$
Then we see that in octal, we just want to group triplets of the binary numbers.
In a very similar way, we can convert from Hexadecimal to binary, and binary to hexadecimal, by grouping into chunks of $4$.  (Hexadecimal is base $16=2^4$)
A: Wolfram alpha can check this for you.
A: $$(0.011011101)_2$$
Start selecting three Bits of numbers from left side. For example, first bit will be $011$, second bit will be $011$ and the third bit will be $101$. We started selected three Bits from left side. And after selecting all the bits convert that all the bits into octal number. So the answer of the question given above will be:
$$(0.335)_8.$$ 
