Q:

Convert decimal to hex/binary

I have small math question. 
Is there any way to convert decimal number (for example 3.14) to hex or binary? If it's possible, can anybody place here some links to tutorials or exaplanations?  (I don't want it for some language, I need it generally in math.) Please help.
EDIT:
Input passed in code:
0.1

Output in ASM code:
415740h

Another input:
0.058

Another output by compiler:
00415748h

But how has been this done? How can be it converted?

A:

I do not recognize your output samples as encodings of floating-point numbers or other common representations of .1 and .058. I suspect these numbers are addresses where the assembler or compiler has stored the floating-point encoding.
In other words, you wrote some text that including a floating-point literal, and the assembler or compiler converted that literal to a floating-point encoding, stored it at some address, and then put the address into an instruction that loads the floating-point encoding from memory.
This hypothesis is consistent with the fact that the two numbers differ by eight. Since double-precision floating-point is commonly eight bytes, the second address (0x415748) was eight bytes beyond the first address (0x415740).
The process for encoding a number in floating-point is roughly this:
Let x be the number to be encoded.
Set s (a sign bit) to 0 if x is positive and to 1 if x is negative. Set x to the absolute value of x.
Set e (an exponent) to 0. Repeat whichever of the following is appropriate:

If x is 2 or greater, add 1 to e and divide x by 2. Repeat until x is less than 2.
If x is less than 1, add -1 to e and multiply x by 2. Repeat until x is at least 1.

When you are done with the above, x is at least 1 and is less than 2. Also, the original number equals (-1)s·2e·x. That is, we have represented the number with a sign bit (s), and exponent of two (e), and a significand (x) that is in [1, 2) (includes 1, excludes 2).
Set f = (x-1)·252. Round f to the nearest integer (if it is a tie between two integers, round to the even integer). If f is now 252, set f to 0 and add 1 to e. (This step finds the 52 bits of x that are immediately after the “decimal point“ when x is represented as a binary numeral, with rounding after the 52nd digit, and it adjusts the exponent if rounding at that position rounds x up to 2, which is out of interval where we want it.)
Add 1023 to e. This has no numerical significance with regard to x; it is simply part of the floating-point encoding. When decoding, 1023 gets subtracted.
Now, convert s, e, and f to binary numerals, using exactly one digit for s, 11 digits for e, and 52 digits for f. If necessary, including leading zeroes so that e is represented with exactly 11 binary digits and f is represented with exactly 52 binary digits. Concatenate those digits, and you have 64 bits. That is the common IEEE 754 encoding for a double-precision floating-point number.
There are some special cases: If the original number is zero, use zero for s, e, and f. (s can also be 1, to represent a special “negative zero“. If, before adding 1023, e is less than -1022, then some adjustments have to be made to get a “denormal“ result or zero, which I do not describe further at the moment. If, before adding 1023, e is more than 1023, then the magnitude of the number is too large to be represented in floating point. It can be encoded as infinity instead, by setting e (after adding 1023) to 2047 and f to zero.

