# binary and floating point representation

suppose that we have following binary digits

$00011001.110$,we can do following thing

$00011001.110=1\cdot2^4+1\cdot2^3+1\cdot2^0+1\cdot2^{-1}+1\cdot2^{-2}=25.75$

then what does means?

We then “float” the binary point:

$00011001.110 \Rightarrow 1.1001110 \cdot 2^4$

mantissa = 1.1001110, exponent = 4

i did not understand this part,generally for sign we have $1$ bit,for exponent we have $8$ bit and for mantissa we need $23$ bit,so i can't understand last part,from floating to binary point,please help me

• Are you referring to IEEE format for floating point numbers? Jul 3, 2014 at 18:20
• IEEE-754 format i meant Jul 3, 2014 at 18:21
• Does this answer help? math.stackexchange.com/a/305901/23353 (I don't know a lot about IEEE-754, but that answer looked pretty nice.) Jul 3, 2014 at 18:21
• a bit explanation please,in my case how it would be Jul 3, 2014 at 18:40
• Please: Using an asterisk for ordinary multiplication in MathJax is vulgar. I edited it. Jul 3, 2014 at 19:16

I'm not sure I really get your question, but are you confused about what "mantissa" really means - it should be a 2-complement integer, but in normalized representation it has a leading "1."? In the standard, the leading "1." is implied: the mantissa bits represent, in your case, only 100111. If this isn't your point of confusion, kindly elaborate more.

• yes it is related to mantissa and exponent,in my example how it is calculated last part? Jul 3, 2014 at 18:47
• You have a typo in your translation which I thought was just that: do you see that 11001.11 = 1.100111 times $2^{4}$? (You have 24) then you just forget the 1., and 100111 is your m, and $2^{4}$ is your exponent. Jul 3, 2014 at 18:53
• You can think of $2^n$ as rotating n times to the left. Jul 3, 2014 at 18:56
• sorry i am lost now Jul 3, 2014 at 19:03
• @datodatuashvili: Me too probably. :) If you have 11.01 = 5.25, then it is the same as 1.101 * 2 = $2.625_d$ * $2_d$. Or think of it as the position left to any position is twice the value of that position; so $6_d$ = 110 = 11 * 2 = $3_d$ * $2_d$; $12_d$ = 1100 = 11 * $2^2$ = $3_d$ times $4_d$. I'm not quite sure where you're stuck. Jul 3, 2014 at 19:10

You start with the binary number $11001.110_2$ We want to express it as something between $1$ and $2_{10}=10_2$ times the appropriate power of two. So $11001.110_2=1.1001110 \cdot 2^4$ Note that this is the same as your expression except that you used $x$ for times and lost the exponent sign. The mantissa is the value between $1$ and $10_2$, so is $1.100111$. The exponent is the exponent of $2$ in the expression, so $4$. We have not yet packed it into the $32$ bit representation. Now if we want to go to IEEE single precision floating point, the sign bit is $0$ (positive), the exponent is $4+127=131=10000011_2$ and the significand is $11001110000000000000000$, so the whole number is $1\ 10000011\ 11001110000000000000000$ where I put the spaces in to show where the pieces go.