Why there is a unique combination of binary number for every decimal number?

Answer 1

every number has an equivalent in any other base.
the number is an absolute
how you state it (base = language) is variable

Answer 2

Take any base 10 number, x, to start with. This number will be greater then or equal to 2^n and less then 2^(n+1) for some number n. In the binary expansion of x, the (n+1)th spot from the left will be 1. Now replace x with (x-2^n) and repeat the process. Eventually all that will be left is zero and you be done.

Answer 3

For n bits the place n ? N, unsigned integer values selection from 0 to (2^n)-a million inclusive and signed integer values selection from 0-(2^(n-a million)) to (2^(n-a million))-a million inclusive, the place ^ is the flexibility operator. changing fractions from decimal into binary would be somewhat no longer trouble-free because of the fact some terminating decimals repeat infinitely in binary (0.a million is an occasion). of direction, you are able to start up by changing the integer area and putting that on the left-hand edge of the radix element, considering the fact which you comprehend neither facet can overflow to the different. as a fashion to transform the fractional area, nonetheless, you will desire to make certain whether it is an integer quotient of a organic selection power of two, whether it is then you definately in simple terms convert and shift the bits over in this occasion, whether it is not then it will be a repeater and you will desire to locate the backside organic selection m of the form (2^n)-a million with n being a organic selection such that the fractional area is an integer quotient of m, then convert, shift the bits and upload a similar set of bits repeating on the tip. it is somewhat complicated, yet once you artwork by some trouble-free examples (inclusive of 0.a million) you will see how the belief works.