# Binary Fractions / Binary Fractions

Binary fractions use the same weighting principle as de dexerations, except that each binary digit uses a 2-based enumeration system.

**Table of contents**göster

We know that decimal numbers use the base ten numbering system so that each digit in the number ten can receive one of ten possible values in the range 0 – 9.

During a decimal number, each household will have a value ten times greater than the digit to one right.

However, as you move from right to left, each figure will be ten times larger than the previous figure, and each figure will be ten times smaller than the number of neighbors as it moves from left to right in the opposite direction.

Example : 1234**.** 567

In this example of decimal numbers, the number (number 5) just to the right of the decimal point is worth a tenth (1/10 or 0.1) of the digit just to the left of the decimal point, which means that the force after the point progresses by 1/10.1/100, as we now switch to decimal numbers according to the 1s digits.

Thus, as you move from left to right throughout the number, each subsequent household value will be one-tenth of the value according to its left position.

The decimal numbering system then produces a positional representation or uses the concept of relative weight values, where each digit represents a different weighted value, depending on the position on either side of the decimal place.

Therefore, in the mathematically standard decimal enumeration system, these values are usually written as 4 0 ,^{}^{3 1}, 2 2 ,^{} ^{1 3}for each position to the left of the decimal step in our example above.

Likewise, the weight of the number becomes more negative in order to bring fractions to the right of the dextager:^{5 -1},^{6 -2}, 7^{-3,} etc.

Thus, we can see that each step in the standard dexication system indicates the size or weight of the digit inside the figure.

The value of any de de-deger will now be equal to the sum of the number of digits multiplied by their weight, so in our example above: N = 1234.567_{10} de degerile will equal the following:

1000 + 200 + 30 + 4 + 0.5 + 0.06 + 0.007 = 1234.567_{10}

or the weight of each digit digit:

(1 x 1000) + (2 x 100) + (3 x 10) + (4 x 1) + (5 x 0.1) + (6 x 0.01) + (7 x 0.001) = 1234.567_{10}

even if we want to show it in the form of polynomials;

(1 x 10^{3}) + (2 x 10^{2}) + (3 x 10^{1}) + (4 x 10^{0}) + (5 x 10^{-1}) + (6 x 10^{-2}) + (7 x 10^{-3}) = 1234.567_{10}

We can also use this position representation idea, where each number represents a different weighted value depending on the position in the binary number system.

The difference here is that the binary number system (or binary numbers only) is a positional system in which the different weighted positions of the digits have base-2 power instead of base-10.

**Binary Fractions**

A binary numbering system is a base-2 numbering system that contains only two digits with "0" or "1".

Thus, each digit of a binary number can take the value "0" or "1", the position of 0 or 1, by specifying its weight.

We can take binary weight for values lower than 1, called unmarked fractional binary numbers.

Similar to deter fractions, binary numbers can also be shown as unmarked fractional numbers by placing deacter numbers in the right deter number, or in this case binary.

Therefore, all fractional digits to the right of the binary point have corresponding weights that create a binary fraction, which has a negative fraction of binary.

**In other words, the power of 2 is negative.**

Thus, for fractions of binary numbers to the right of the binary point, the weight of each digit becomes more negative: 2^{-1},^{2 -2},^{2 -3}, 2^{-4}, etc.

2^{0}= 1

2^{-1}= 1/2^{1 }=1/2 = 0.5

2^{-3}= 1/2^{3 }=1/8 = 0.125.

Thus, if we take the double fraction of_{0.1011 2,}the position weights of each step are taken into account by giving the decimal equivalent:

0.1011 => 1 x 2^{-4} + 1 x 2^{-3} + 1 x 2^{-2} + 1 x 2^{-1} = 0.0625 + 0.125 + 0 .5 = 0.6875_{10}

### Binary Fractions Question Example 1

Now let's say we have the following binary number: 1101.0111_{2}, what will be the de-deger equivalent?

1101.0111 = (1 × 2^{3}) + (1 × 2^{2}) + (0 × 2^{1}) + (1 × 2^{0}) + (0 × 2^{-1}) + (1 × 2^{-2}) + (1 × 2^{-3}) + (1 x 2^{-4})

= 8 + 4 + 0 + 1 + 0 + 1/4 + 1/8 + 1/16

= 8 + 4 + 0 + 1 + 0 + 0.25 + 0.125 + 0.0625 = 13.4375_{10}

Therefore, the decimal equivalent of 1101.0111_{2}is given as follows: 13.4375_{10}

Thus, we can see that fractional binary numbers with binary numbers weighing less than^{1(2 0)}can be converted to degers by dividing the binary weight factor sequentially into two values for each drop, respectively.

**Other Binary Fractions Examples**

0.11 = (1 x 2^{-1}) + (1 x 2^{-2}) = 0.5 + 0.25 = 0.75_{10}

11.001 = (1 x 2^{1}) + (1 x 2^{0}) + (1 x 2^{-3}) = 2 + 1 + 0.125 = 3.125_{10}

1011.111 = (1 × 2^{3}) + (1 × 2^{1}) + (1 × 2 0 ) (1 × 2 -1 ) + (1 ×^{2}^{0}) × 2^{-2}) + (1 × 2^{-3}) = 8 + 2 + 1 + 0.5 + 0.25 + 0.125 = 11.875_{10}

**Converting Decimals to Binary Fractions**

The conversion of a de deterrence fraction to a fractionary binary number is achieved using a method similar to the one we use for integers. However, this time multiplication is used instead of dividing with integers instead of the rest used with the hand digit, which is the binary equivalent of the fractional portion of the de deger.

When converting from decath to binary, the integer (right-to-left positive order) portion and fraction (left-to-right negative part) portion of the degerile number are calculated separately.

For an integer of a number, the binary equivalent is found by repeatedly noting in reverse order from the least meaningful bit (LSB), dividing the integer portion of the decathlete number (known as consecutive division) by 2 (÷ 2) in a row. Until the most important bit (MSB), the binary equivalent is generated until the value is "0".

Find binary equivalent of de deter integer : 118_{10}

118 (divide by 2) = 59 plus remaining 0 (LSB)

59 (divide by 2) = 29 plus remaining 1 (←)

29 (divide by 2) = 14 plus remainers 1 (←)

14 (divide by 2) = 7 plus remainers 0 (←)

7 (divide by 2) = 3 plus remaining 1 (←)

3 (divide by 2) = 1 plus remaining 1 (←)

1 (divide by 2) = 0 plus remaining 1 (MSB)

So the didial equivalent of_{118 10}is as follows: 1110110_{2} ← (LSB)

By multiplying the given fractional portion of the decimal number by 2 (x2) in a row (known as a succession of times), the fractional part of the number is indicated in the order in which it moves until the value that produces two consecutive digits is "0".

Therefore, if the multiplication process produces a result greater than 1, then the result is "1" in the hand, and if the multiplication process produces a result less than "1", then the result is "0".

Also note that if the result of consecutive multiplication operations does not appear to be progressing towards zero, the fractional number will have an infinite length, or until the equivalent number of bits is obtained, for example, 8-bit or 16 bits, depending on the degree of accuracy required, etc.

Find the binary fraction equivalent of the de deceirable fraction: 0.8125_{10}

0.8125 (multiply by 2) = 1.625 = 0.625 (1) MSB (0.6)

0.625 (multiply by 2) = 1.25 = 0.25 in hand 1

0.25 (multiply by 2) = 0.50 = 0.5 achieved 0 (↓)

0.5 (multiply by 2) = 1.00 = 0.0 hand 1 (LSB)

Therefore, the didial equivalent of_{0.8125 10}is as follows: 0.1101_{2} ← (LSB)

Using the above procedure to convert a binary fraction to the de deger equivalent, we can double-check this answer: 0.1101 = 0.5 + 0.25 + 0.0625 = 0.8125_{10}

**Binary Fraction Question Example 2**

Find the following de de-desected binary fraction equivalent: 54.6875

First, using consecutive division from above, we convert integer 54 to a binary number normally.

54 (divide by 2) = 27 remaining 0 (LSB)

27 (divide by 2) = 13 remaining 1 (←)

13 (divide by 2) = 6 remaining 1 (←)

6 (divide by 2) = 3 remaining 0 (←)

3 (divide by 2) = 1 remaining 1 (←)

1 (divide by 2) = 0 remaining 1 (MSB)

Therefore, the binary equivalent of_{54 10}is as follows: 110110_{2}

Next, we convert the decathlete fraction of 0.6875 to a binary fraction using consecutive multiplication.

0.6875 (multiply by 2) = 1.375 = 0.375 achieved 1 (MSB)

0.375 (multiply by 2) = 0.75 = 0.75 achieved 0 (↓)

0.75 (multiply by 2) = 1.50 = 0.5 achieved 1 (↓)

0.5 (multiply by 2) = 1.00 = 0.0 hand 1 (LSB)

Therefore, the binary equivalent of_{0.6875 10}is as follows: 0.1011_{2} ← (LSB)

Therefore, the binary equivalent of the de deger number: 54.6875_{10}, 110110.1011_{2.}

## Summarize

In this article, we have seen some information about the Binary Fractions, in order to convert deacticular fractions to the equivalent binary fraction, we must multiply the de deactimal fractional part by 2 and record the digit that appears to the left of the Binary figure.

This didial number with the hand digit will always be "0" or "1".

Then we must multiply the remaining de deacticular fraction by 2, repeating the above sequence using successive multiplication until the fraction is reduced to zero or until the binary bits required for a repeating binary fraction are completed.

Fractional numbers are represented by the negative forces of 2.

We have to perform two separate operations for mixed deity numbers. For an integer, there must be a succession of splits to the left of the dextroble digit and a succession of multiplications for the fractioned part to the right of the dextroque.

Note that the integer portion of a mixed deger will always be an exact binary number equivalent, but the decathlete fraction may not be the part, since if we want to show the exact number of de deactimal fractions, we can achieve a repetitive fraction that results in an infinite number of binary digits.