-0.000 000 000 745 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 745(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
-0.000 000 000 745(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. Start with the positive version of the number:

|-0.000 000 000 745| = 0.000 000 000 745


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: 0.000 000 000 745.

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) 0.000 000 000 745 × 2 = 0 + 0.000 000 001 49;
  • 2) 0.000 000 001 49 × 2 = 0 + 0.000 000 002 98;
  • 3) 0.000 000 002 98 × 2 = 0 + 0.000 000 005 96;
  • 4) 0.000 000 005 96 × 2 = 0 + 0.000 000 011 92;
  • 5) 0.000 000 011 92 × 2 = 0 + 0.000 000 023 84;
  • 6) 0.000 000 023 84 × 2 = 0 + 0.000 000 047 68;
  • 7) 0.000 000 047 68 × 2 = 0 + 0.000 000 095 36;
  • 8) 0.000 000 095 36 × 2 = 0 + 0.000 000 190 72;
  • 9) 0.000 000 190 72 × 2 = 0 + 0.000 000 381 44;
  • 10) 0.000 000 381 44 × 2 = 0 + 0.000 000 762 88;
  • 11) 0.000 000 762 88 × 2 = 0 + 0.000 001 525 76;
  • 12) 0.000 001 525 76 × 2 = 0 + 0.000 003 051 52;
  • 13) 0.000 003 051 52 × 2 = 0 + 0.000 006 103 04;
  • 14) 0.000 006 103 04 × 2 = 0 + 0.000 012 206 08;
  • 15) 0.000 012 206 08 × 2 = 0 + 0.000 024 412 16;
  • 16) 0.000 024 412 16 × 2 = 0 + 0.000 048 824 32;
  • 17) 0.000 048 824 32 × 2 = 0 + 0.000 097 648 64;
  • 18) 0.000 097 648 64 × 2 = 0 + 0.000 195 297 28;
  • 19) 0.000 195 297 28 × 2 = 0 + 0.000 390 594 56;
  • 20) 0.000 390 594 56 × 2 = 0 + 0.000 781 189 12;
  • 21) 0.000 781 189 12 × 2 = 0 + 0.001 562 378 24;
  • 22) 0.001 562 378 24 × 2 = 0 + 0.003 124 756 48;
  • 23) 0.003 124 756 48 × 2 = 0 + 0.006 249 512 96;
  • 24) 0.006 249 512 96 × 2 = 0 + 0.012 499 025 92;
  • 25) 0.012 499 025 92 × 2 = 0 + 0.024 998 051 84;
  • 26) 0.024 998 051 84 × 2 = 0 + 0.049 996 103 68;
  • 27) 0.049 996 103 68 × 2 = 0 + 0.099 992 207 36;
  • 28) 0.099 992 207 36 × 2 = 0 + 0.199 984 414 72;
  • 29) 0.199 984 414 72 × 2 = 0 + 0.399 968 829 44;
  • 30) 0.399 968 829 44 × 2 = 0 + 0.799 937 658 88;
  • 31) 0.799 937 658 88 × 2 = 1 + 0.599 875 317 76;
  • 32) 0.599 875 317 76 × 2 = 1 + 0.199 750 635 52;
  • 33) 0.199 750 635 52 × 2 = 0 + 0.399 501 271 04;
  • 34) 0.399 501 271 04 × 2 = 0 + 0.799 002 542 08;
  • 35) 0.799 002 542 08 × 2 = 1 + 0.598 005 084 16;
  • 36) 0.598 005 084 16 × 2 = 1 + 0.196 010 168 32;
  • 37) 0.196 010 168 32 × 2 = 0 + 0.392 020 336 64;
  • 38) 0.392 020 336 64 × 2 = 0 + 0.784 040 673 28;
  • 39) 0.784 040 673 28 × 2 = 1 + 0.568 081 346 56;
  • 40) 0.568 081 346 56 × 2 = 1 + 0.136 162 693 12;
  • 41) 0.136 162 693 12 × 2 = 0 + 0.272 325 386 24;
  • 42) 0.272 325 386 24 × 2 = 0 + 0.544 650 772 48;
  • 43) 0.544 650 772 48 × 2 = 1 + 0.089 301 544 96;
  • 44) 0.089 301 544 96 × 2 = 0 + 0.178 603 089 92;
  • 45) 0.178 603 089 92 × 2 = 0 + 0.357 206 179 84;
  • 46) 0.357 206 179 84 × 2 = 0 + 0.714 412 359 68;
  • 47) 0.714 412 359 68 × 2 = 1 + 0.428 824 719 36;
  • 48) 0.428 824 719 36 × 2 = 0 + 0.857 649 438 72;
  • 49) 0.857 649 438 72 × 2 = 1 + 0.715 298 877 44;
  • 50) 0.715 298 877 44 × 2 = 1 + 0.430 597 754 88;
  • 51) 0.430 597 754 88 × 2 = 0 + 0.861 195 509 76;
  • 52) 0.861 195 509 76 × 2 = 1 + 0.722 391 019 52;
  • 53) 0.722 391 019 52 × 2 = 1 + 0.444 782 039 04;
  • 54) 0.444 782 039 04 × 2 = 0 + 0.889 564 078 08;

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).


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:


0.000 000 000 745(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0011 0010 0010 1101 10(2)

6. Positive number before normalization:

0.000 000 000 745(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0011 0010 0010 1101 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 745(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0011 0010 0010 1101 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0011 0010 0010 1101 10(2) × 20 =

1.1001 1001 1001 0001 0110 110(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 1001 1001 0001 0110 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

96(10) =

0110 0000(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. 100 1100 1100 1000 1011 0110 =

100 1100 1100 1000 1011 0110


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) =
0110 0000

Mantissa (23 bits) =
100 1100 1100 1000 1011 0110


Decimal number -0.000 000 000 745 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 1100 1000 1011 0110

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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:

  • 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