0.000 000 008 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 008 7(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
0.000 000 008 7(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: 0.000 000 008 7.

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) 0.000 000 008 7 × 2 = 0 + 0.000 000 017 4;
  • 2) 0.000 000 017 4 × 2 = 0 + 0.000 000 034 8;
  • 3) 0.000 000 034 8 × 2 = 0 + 0.000 000 069 6;
  • 4) 0.000 000 069 6 × 2 = 0 + 0.000 000 139 2;
  • 5) 0.000 000 139 2 × 2 = 0 + 0.000 000 278 4;
  • 6) 0.000 000 278 4 × 2 = 0 + 0.000 000 556 8;
  • 7) 0.000 000 556 8 × 2 = 0 + 0.000 001 113 6;
  • 8) 0.000 001 113 6 × 2 = 0 + 0.000 002 227 2;
  • 9) 0.000 002 227 2 × 2 = 0 + 0.000 004 454 4;
  • 10) 0.000 004 454 4 × 2 = 0 + 0.000 008 908 8;
  • 11) 0.000 008 908 8 × 2 = 0 + 0.000 017 817 6;
  • 12) 0.000 017 817 6 × 2 = 0 + 0.000 035 635 2;
  • 13) 0.000 035 635 2 × 2 = 0 + 0.000 071 270 4;
  • 14) 0.000 071 270 4 × 2 = 0 + 0.000 142 540 8;
  • 15) 0.000 142 540 8 × 2 = 0 + 0.000 285 081 6;
  • 16) 0.000 285 081 6 × 2 = 0 + 0.000 570 163 2;
  • 17) 0.000 570 163 2 × 2 = 0 + 0.001 140 326 4;
  • 18) 0.001 140 326 4 × 2 = 0 + 0.002 280 652 8;
  • 19) 0.002 280 652 8 × 2 = 0 + 0.004 561 305 6;
  • 20) 0.004 561 305 6 × 2 = 0 + 0.009 122 611 2;
  • 21) 0.009 122 611 2 × 2 = 0 + 0.018 245 222 4;
  • 22) 0.018 245 222 4 × 2 = 0 + 0.036 490 444 8;
  • 23) 0.036 490 444 8 × 2 = 0 + 0.072 980 889 6;
  • 24) 0.072 980 889 6 × 2 = 0 + 0.145 961 779 2;
  • 25) 0.145 961 779 2 × 2 = 0 + 0.291 923 558 4;
  • 26) 0.291 923 558 4 × 2 = 0 + 0.583 847 116 8;
  • 27) 0.583 847 116 8 × 2 = 1 + 0.167 694 233 6;
  • 28) 0.167 694 233 6 × 2 = 0 + 0.335 388 467 2;
  • 29) 0.335 388 467 2 × 2 = 0 + 0.670 776 934 4;
  • 30) 0.670 776 934 4 × 2 = 1 + 0.341 553 868 8;
  • 31) 0.341 553 868 8 × 2 = 0 + 0.683 107 737 6;
  • 32) 0.683 107 737 6 × 2 = 1 + 0.366 215 475 2;
  • 33) 0.366 215 475 2 × 2 = 0 + 0.732 430 950 4;
  • 34) 0.732 430 950 4 × 2 = 1 + 0.464 861 900 8;
  • 35) 0.464 861 900 8 × 2 = 0 + 0.929 723 801 6;
  • 36) 0.929 723 801 6 × 2 = 1 + 0.859 447 603 2;
  • 37) 0.859 447 603 2 × 2 = 1 + 0.718 895 206 4;
  • 38) 0.718 895 206 4 × 2 = 1 + 0.437 790 412 8;
  • 39) 0.437 790 412 8 × 2 = 0 + 0.875 580 825 6;
  • 40) 0.875 580 825 6 × 2 = 1 + 0.751 161 651 2;
  • 41) 0.751 161 651 2 × 2 = 1 + 0.502 323 302 4;
  • 42) 0.502 323 302 4 × 2 = 1 + 0.004 646 604 8;
  • 43) 0.004 646 604 8 × 2 = 0 + 0.009 293 209 6;
  • 44) 0.009 293 209 6 × 2 = 0 + 0.018 586 419 2;
  • 45) 0.018 586 419 2 × 2 = 0 + 0.037 172 838 4;
  • 46) 0.037 172 838 4 × 2 = 0 + 0.074 345 676 8;
  • 47) 0.074 345 676 8 × 2 = 0 + 0.148 691 353 6;
  • 48) 0.148 691 353 6 × 2 = 0 + 0.297 382 707 2;
  • 49) 0.297 382 707 2 × 2 = 0 + 0.594 765 414 4;
  • 50) 0.594 765 414 4 × 2 = 1 + 0.189 530 828 8;

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:


0.000 000 008 7(10) =

0.0000 0000 0000 0000 0000 0000 0010 0101 0101 1101 1100 0000 01(2)

5. Positive number before normalization:

0.000 000 008 7(10) =

0.0000 0000 0000 0000 0000 0000 0010 0101 0101 1101 1100 0000 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 008 7(10) =

0.0000 0000 0000 0000 0000 0000 0010 0101 0101 1101 1100 0000 01(2) =

0.0000 0000 0000 0000 0000 0000 0010 0101 0101 1101 1100 0000 01(2) × 20 =

1.0010 1010 1110 1110 0000 001(2) × 2-27


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): -27

Mantissa (not normalized):
1.0010 1010 1110 1110 0000 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-27 + 2(8-1) - 1 =

(-27 + 127)(10) =

100(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;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 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) =

100(10) =

0110 0100(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. 001 0101 0111 0111 0000 0001 =

001 0101 0111 0111 0000 0001


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 0100

Mantissa (23 bits) =
001 0101 0111 0111 0000 0001


Decimal number 0.000 000 008 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 001 0101 0111 0111 0000 0001

Share

» Convert decimal 0.000 000 007 6 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 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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