# Math Help - Floating Point Format

1. ## Floating Point Format

Several different representations of real numbers have been proposed, but by far the most widely used is the floating-point representation. Floating-point representations have a base (which is always assumed to be even) and a precision p. If = 10 and p = 3, then the number 0.1 is represented as 1.00 × 10-1 . If = 2 and p = 24, then the decimal number 0.1 cannot be represented exactly, but is approximately 1.10011001100110011001101 × 2-4.

In general, a floating-point number will be represented as ± d.dd... d × e , where d.dd... d is called the significand and hasp digits. More precisely ±d0 . d1 d2 ... dp-1 × e represents the number .

The term floating-point number will be used to mean a real number that can be exactly represented in the format under discussion. Two other parameters associated with floating-point representations are the largest and smallest allowable exponents, emax and emin. Since there are p possible significands, and emax - emin + 1 possible exponents, a floating-point number can be encoded in
bits, where the final +1 is for the sign bit.
What I dont understand is the last part of :

How the whole expression could be denoted by that formular is bugging me and I would like to understand it if anyone can explain it.

Please any reasonable input would be well appreciated.

Thank you

2. This is much simpler than it looks.

How many bits are needed to encode n discrete pieces of information? $\lceil log_2(n) \rceil$
For example: the minimum number of bits needed to encode 3 numbers say x, y, and z would be ceil(lg(3)) = 2. E.g. let x = 00, y = 01, z = 11
Another example: the number of bits needed to encode all numbers between 100 and 1123 is ceil(lg(1123 - 100 + 1)) = lg(1024) = 10, so we need 10 bits.

To explain the formula:

The first part is the number of bits needed to encode $e_{max} - e_{min} + 1$ possible numbers for the exponents.
The second part is the number of bits needed to encode all numbers expressed by p digits in base $\beta$
The last + 1 is for the sign as mentioned.

3. First I would like to thank you for your wonderful reply, it answered my question straight away.

My next question is how do you work out ceil(lg(3)) = 2 or ceil(lg(1123 - 100 + 1)) = 10.

4. I'm using lg for log base 2. And ceil just rounds to the nearest integer.

All together ceil(lg(x)) means what is the smallest integer y s.t. 2^y >= x

Lets work out ceil(lg(1000)) as an example

2^9 = 512
2^10 = 1024

so 10 is the smallest integer s.t. 2^10 >= 1000
so ceil(lg(1000)) = 10

To do it on a calculator. You use the fact that lg(x) = log(x) / log(2) = ln(x) / ln(2)
So one way is to type ln(1000)/ln(2) and round up.

5. Junior member thank you so much, u sure solve this.

But I shall like to leave this thread open since the article am reading has a lot of theorems and proofs, and since my maths has been a bit rusty, I shall come in often for explanations.

But junior thank you.

6. Originally Posted by magrogi
Junior member thank you so much, u sure solve this.

But I shall like to leave this thread open since the article am reading has a lot of theorems and proofs, and since my maths has been a bit rusty, I shall come in often for explanations.

But junior thank you.
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