32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 5.082 197 684 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 5.082 197 684(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)

1. First, convert to binary (in base 2) the integer part: 5.
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;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.


5(10) =

101(2)


3. Convert to binary (base 2) the fractional part: 0.082 197 684.

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) 0.082 197 684 × 2 = 0 + 0.164 395 368;
  • 2) 0.164 395 368 × 2 = 0 + 0.328 790 736;
  • 3) 0.328 790 736 × 2 = 0 + 0.657 581 472;
  • 4) 0.657 581 472 × 2 = 1 + 0.315 162 944;
  • 5) 0.315 162 944 × 2 = 0 + 0.630 325 888;
  • 6) 0.630 325 888 × 2 = 1 + 0.260 651 776;
  • 7) 0.260 651 776 × 2 = 0 + 0.521 303 552;
  • 8) 0.521 303 552 × 2 = 1 + 0.042 607 104;
  • 9) 0.042 607 104 × 2 = 0 + 0.085 214 208;
  • 10) 0.085 214 208 × 2 = 0 + 0.170 428 416;
  • 11) 0.170 428 416 × 2 = 0 + 0.340 856 832;
  • 12) 0.340 856 832 × 2 = 0 + 0.681 713 664;
  • 13) 0.681 713 664 × 2 = 1 + 0.363 427 328;
  • 14) 0.363 427 328 × 2 = 0 + 0.726 854 656;
  • 15) 0.726 854 656 × 2 = 1 + 0.453 709 312;
  • 16) 0.453 709 312 × 2 = 0 + 0.907 418 624;
  • 17) 0.907 418 624 × 2 = 1 + 0.814 837 248;
  • 18) 0.814 837 248 × 2 = 1 + 0.629 674 496;
  • 19) 0.629 674 496 × 2 = 1 + 0.259 348 992;
  • 20) 0.259 348 992 × 2 = 0 + 0.518 697 984;
  • 21) 0.518 697 984 × 2 = 1 + 0.037 395 968;
  • 22) 0.037 395 968 × 2 = 0 + 0.074 791 936;
  • 23) 0.074 791 936 × 2 = 0 + 0.149 583 872;
  • 24) 0.149 583 872 × 2 = 0 + 0.299 167 744;

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...)


4. 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:


0.082 197 684(10) =

0.0001 0101 0000 1010 1110 1000(2)


5. Positive number before normalization:

5.082 197 684(10) =

101.0001 0101 0000 1010 1110 1000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


5.082 197 684(10) =

101.0001 0101 0000 1010 1110 1000(2) =

101.0001 0101 0000 1010 1110 1000(2) × 20 =

1.0100 0101 0100 0010 1011 1010 00(2) × 22


7. 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 0 (a positive number)

Exponent (unadjusted): 2

Mantissa (not normalized):
1.0100 0101 0100 0010 1011 1010 00


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(8-1) - 1 =

(2 + 127)(10) =

129(10)


9. 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;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

129(10) =

1000 0001(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 010 0010 1010 0001 0101 1101 000 =

010 0010 1010 0001 0101 1101


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

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
1000 0001

Mantissa (23 bits) =
010 0010 1010 0001 0101 1101


The base ten decimal number 5.082 197 684 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0001 - 010 0010 1010 0001 0101 1101

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

» Number 5.082 197 683 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

» Number 5.082 197 685 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

The latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

Number -64 646 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 06:37 UTC (GMT)
Number -407 671 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 06:36 UTC (GMT)
Number -25.313 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 06:36 UTC (GMT)
Number 111.111 116 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 06:36 UTC (GMT)
Number -189 931 478 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 06:36 UTC (GMT)
Number -1 601.5 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 06:36 UTC (GMT)
Number 9.445 38 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 06:36 UTC (GMT)
Number -11 451 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 06:36 UTC (GMT)
Number -0.333 3 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 06:36 UTC (GMT)
Number 1 000 000 110 110 011 066 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 06:36 UTC (GMT)
All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

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

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

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

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

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

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111