0.000 000 110 008 91 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 008 91(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 000 110 008 91(10) to 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: 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;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 110 008 91.

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) 0.000 000 110 008 91 × 2 = 0 + 0.000 000 220 017 82;
  • 2) 0.000 000 220 017 82 × 2 = 0 + 0.000 000 440 035 64;
  • 3) 0.000 000 440 035 64 × 2 = 0 + 0.000 000 880 071 28;
  • 4) 0.000 000 880 071 28 × 2 = 0 + 0.000 001 760 142 56;
  • 5) 0.000 001 760 142 56 × 2 = 0 + 0.000 003 520 285 12;
  • 6) 0.000 003 520 285 12 × 2 = 0 + 0.000 007 040 570 24;
  • 7) 0.000 007 040 570 24 × 2 = 0 + 0.000 014 081 140 48;
  • 8) 0.000 014 081 140 48 × 2 = 0 + 0.000 028 162 280 96;
  • 9) 0.000 028 162 280 96 × 2 = 0 + 0.000 056 324 561 92;
  • 10) 0.000 056 324 561 92 × 2 = 0 + 0.000 112 649 123 84;
  • 11) 0.000 112 649 123 84 × 2 = 0 + 0.000 225 298 247 68;
  • 12) 0.000 225 298 247 68 × 2 = 0 + 0.000 450 596 495 36;
  • 13) 0.000 450 596 495 36 × 2 = 0 + 0.000 901 192 990 72;
  • 14) 0.000 901 192 990 72 × 2 = 0 + 0.001 802 385 981 44;
  • 15) 0.001 802 385 981 44 × 2 = 0 + 0.003 604 771 962 88;
  • 16) 0.003 604 771 962 88 × 2 = 0 + 0.007 209 543 925 76;
  • 17) 0.007 209 543 925 76 × 2 = 0 + 0.014 419 087 851 52;
  • 18) 0.014 419 087 851 52 × 2 = 0 + 0.028 838 175 703 04;
  • 19) 0.028 838 175 703 04 × 2 = 0 + 0.057 676 351 406 08;
  • 20) 0.057 676 351 406 08 × 2 = 0 + 0.115 352 702 812 16;
  • 21) 0.115 352 702 812 16 × 2 = 0 + 0.230 705 405 624 32;
  • 22) 0.230 705 405 624 32 × 2 = 0 + 0.461 410 811 248 64;
  • 23) 0.461 410 811 248 64 × 2 = 0 + 0.922 821 622 497 28;
  • 24) 0.922 821 622 497 28 × 2 = 1 + 0.845 643 244 994 56;
  • 25) 0.845 643 244 994 56 × 2 = 1 + 0.691 286 489 989 12;
  • 26) 0.691 286 489 989 12 × 2 = 1 + 0.382 572 979 978 24;
  • 27) 0.382 572 979 978 24 × 2 = 0 + 0.765 145 959 956 48;
  • 28) 0.765 145 959 956 48 × 2 = 1 + 0.530 291 919 912 96;
  • 29) 0.530 291 919 912 96 × 2 = 1 + 0.060 583 839 825 92;
  • 30) 0.060 583 839 825 92 × 2 = 0 + 0.121 167 679 651 84;
  • 31) 0.121 167 679 651 84 × 2 = 0 + 0.242 335 359 303 68;
  • 32) 0.242 335 359 303 68 × 2 = 0 + 0.484 670 718 607 36;
  • 33) 0.484 670 718 607 36 × 2 = 0 + 0.969 341 437 214 72;
  • 34) 0.969 341 437 214 72 × 2 = 1 + 0.938 682 874 429 44;
  • 35) 0.938 682 874 429 44 × 2 = 1 + 0.877 365 748 858 88;
  • 36) 0.877 365 748 858 88 × 2 = 1 + 0.754 731 497 717 76;
  • 37) 0.754 731 497 717 76 × 2 = 1 + 0.509 462 995 435 52;
  • 38) 0.509 462 995 435 52 × 2 = 1 + 0.018 925 990 871 04;
  • 39) 0.018 925 990 871 04 × 2 = 0 + 0.037 851 981 742 08;
  • 40) 0.037 851 981 742 08 × 2 = 0 + 0.075 703 963 484 16;
  • 41) 0.075 703 963 484 16 × 2 = 0 + 0.151 407 926 968 32;
  • 42) 0.151 407 926 968 32 × 2 = 0 + 0.302 815 853 936 64;
  • 43) 0.302 815 853 936 64 × 2 = 0 + 0.605 631 707 873 28;
  • 44) 0.605 631 707 873 28 × 2 = 1 + 0.211 263 415 746 56;
  • 45) 0.211 263 415 746 56 × 2 = 0 + 0.422 526 831 493 12;
  • 46) 0.422 526 831 493 12 × 2 = 0 + 0.845 053 662 986 24;
  • 47) 0.845 053 662 986 24 × 2 = 1 + 0.690 107 325 972 48;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.000 000 110 008 91(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 001(2)

5. Positive number before normalization:

0.000 000 110 008 91(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 110 008 91(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 001(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0001 001(2) × 20 =

1.1101 1000 0111 1100 0001 001(2) × 2-24


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): -24

Mantissa (not normalized):
1.1101 1000 0111 1100 0001 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

103(10) =

0110 0111(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 110 1100 0011 1110 0000 1001 =

110 1100 0011 1110 0000 1001


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) =
0110 0111

Mantissa (23 bits) =
110 1100 0011 1110 0000 1001


Decimal number 0.000 000 110 008 91 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1110 0000 1001

Share

» Convert decimal 0.000 000 110 008 82 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 minutes

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