IEEE-754 General Issues I read the documents of the IEEE-754 and I could not understand some things
1) Is it possible to represent any number using this method? I know there are a number of final states, and there are infinite possibilities then no, but if there is a more detailed explanation of why?
2)Why can not represent the number zero? And how I represent it?

3) Why I dont need to represent 1 in representation of floating point number, I know its normalized, but normalized mean only 1 in the left side of the number? or everynumber, for example $8.3131e^{-1}$

Thanks!
 A: *

*You can't represent every number as the language is countable and there are uncountably many numbers.  More to the point, you can't even represent $1/10$ since
$$\frac{1}{10} = \frac{1}{2} \sum _{n=1}^{\infty } \frac{3}{16^n}.$$
This is exactly why computations like $0.1 + 0.2 - 0.3$ tend to yield results around $10^{-16}$, rather than exactly zero.
More generally, you cannot exactly represent a number with an infinite binary expansion.

*You can represent zero. The fraction and exponent fields must contain all zeros and the sign field can be zero or one.

*Of course, 1 needs (and has) a representation.
A: 1) Using $64$ bits, you can represent at most $2^{64}$ numbers exactly. Many more can be represented with sufficient accuracy for applications. Very very large numbers, such as $10^{1000}$ get represented as infinity (which may also be enough for applications). 
2) $0$ is represented by all bits being zero. 
3) I don't really understand this question, but you can use this conversion tool to find the IEEE-754 representation of $1$, or  other numbers. 
A: 1) No. Eg. if you want to represent $10^{50} + 1$ and $10^{50} + 2$ they're going to be the same in terms of floating points with not enough bits.
2) You have numbers in from $(-1)^{sign} (1 + \frac{mantisa}{2^{bitsofmantisa}}) 2^{exponent-bias}$ how would you express 0? You have to have special value of exponent for this.
3) For 1 you just don't have any positive bits in mantisa... 
