Convert the Number -0.833 333 333 4 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number. Detailed Explanations

Number -0.833 333 333 4(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

The first steps we'll go through to make the conversion:

Convert to binary (to base 2) the integer part of the number.

Convert to binary the fractional part of the number.


1. Start with the positive version of the number:

|-0.833 333 333 4| = 0.833 333 333 4

2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


0(10) =


0(2)


4. Convert to binary (base 2) the fractional part: 0.833 333 333 4.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.833 333 333 4 × 2 = 1 + 0.666 666 666 8;
  • 2) 0.666 666 666 8 × 2 = 1 + 0.333 333 333 6;
  • 3) 0.333 333 333 6 × 2 = 0 + 0.666 666 667 2;
  • 4) 0.666 666 667 2 × 2 = 1 + 0.333 333 334 4;
  • 5) 0.333 333 334 4 × 2 = 0 + 0.666 666 668 8;
  • 6) 0.666 666 668 8 × 2 = 1 + 0.333 333 337 6;
  • 7) 0.333 333 337 6 × 2 = 0 + 0.666 666 675 2;
  • 8) 0.666 666 675 2 × 2 = 1 + 0.333 333 350 4;
  • 9) 0.333 333 350 4 × 2 = 0 + 0.666 666 700 8;
  • 10) 0.666 666 700 8 × 2 = 1 + 0.333 333 401 6;
  • 11) 0.333 333 401 6 × 2 = 0 + 0.666 666 803 2;
  • 12) 0.666 666 803 2 × 2 = 1 + 0.333 333 606 4;
  • 13) 0.333 333 606 4 × 2 = 0 + 0.666 667 212 8;
  • 14) 0.666 667 212 8 × 2 = 1 + 0.333 334 425 6;
  • 15) 0.333 334 425 6 × 2 = 0 + 0.666 668 851 2;
  • 16) 0.666 668 851 2 × 2 = 1 + 0.333 337 702 4;
  • 17) 0.333 337 702 4 × 2 = 0 + 0.666 675 404 8;
  • 18) 0.666 675 404 8 × 2 = 1 + 0.333 350 809 6;
  • 19) 0.333 350 809 6 × 2 = 0 + 0.666 701 619 2;
  • 20) 0.666 701 619 2 × 2 = 1 + 0.333 403 238 4;
  • 21) 0.333 403 238 4 × 2 = 0 + 0.666 806 476 8;
  • 22) 0.666 806 476 8 × 2 = 1 + 0.333 612 953 6;
  • 23) 0.333 612 953 6 × 2 = 0 + 0.667 225 907 2;
  • 24) 0.667 225 907 2 × 2 = 1 + 0.334 451 814 4;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.833 333 333 4(10) =


0.1101 0101 0101 0101 0101 0101(2)


6. Positive number before normalization:

0.833 333 333 4(10) =


0.1101 0101 0101 0101 0101 0101(2)


The last steps we'll go through to make the conversion:

Normalize the binary representation of the number.

Adjust the exponent.

Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Normalize the mantissa.


7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.833 333 333 4(10) =


0.1101 0101 0101 0101 0101 0101(2) =


0.1101 0101 0101 0101 0101 0101(2) × 20 =


1.1010 1010 1010 1010 1010 101(2) × 2-1


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)


Exponent (unadjusted): -1


Mantissa (not normalized):
1.1010 1010 1010 1010 1010 101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


-1 + 2(8-1) - 1 =


(-1 + 127)(10) =


126(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


126(10) =


0111 1110(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.


b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =


1. 101 0101 0101 0101 0101 0101 =


101 0101 0101 0101 0101 0101


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)


Exponent (8 bits) =
0111 1110


Mantissa (23 bits) =
101 0101 0101 0101 0101 0101


The base ten decimal number -0.833 333 333 4 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 0111 1110 - 101 0101 0101 0101 0101 0101

(32 bits IEEE 754)

Number -0.833 333 333 5 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

Number -0.833 333 333 3 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

Convert to 32 bit single precision IEEE 754 binary floating point representation standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

Available Base Conversions Between Decimal and Binary Systems

Conversions Between Decimal System Numbers (Written in Base Ten) and Binary System Numbers (Base Two and Computer Representation):


1. Integer -> Binary

2. Decimal -> Binary

3. Binary -> Integer

4. Binary -> Decimal