-0.000 000 000 72 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 72(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 72(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 72| = 0.000 000 000 72


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 72.

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 72 × 2 = 0 + 0.000 000 001 44;
  • 2) 0.000 000 001 44 × 2 = 0 + 0.000 000 002 88;
  • 3) 0.000 000 002 88 × 2 = 0 + 0.000 000 005 76;
  • 4) 0.000 000 005 76 × 2 = 0 + 0.000 000 011 52;
  • 5) 0.000 000 011 52 × 2 = 0 + 0.000 000 023 04;
  • 6) 0.000 000 023 04 × 2 = 0 + 0.000 000 046 08;
  • 7) 0.000 000 046 08 × 2 = 0 + 0.000 000 092 16;
  • 8) 0.000 000 092 16 × 2 = 0 + 0.000 000 184 32;
  • 9) 0.000 000 184 32 × 2 = 0 + 0.000 000 368 64;
  • 10) 0.000 000 368 64 × 2 = 0 + 0.000 000 737 28;
  • 11) 0.000 000 737 28 × 2 = 0 + 0.000 001 474 56;
  • 12) 0.000 001 474 56 × 2 = 0 + 0.000 002 949 12;
  • 13) 0.000 002 949 12 × 2 = 0 + 0.000 005 898 24;
  • 14) 0.000 005 898 24 × 2 = 0 + 0.000 011 796 48;
  • 15) 0.000 011 796 48 × 2 = 0 + 0.000 023 592 96;
  • 16) 0.000 023 592 96 × 2 = 0 + 0.000 047 185 92;
  • 17) 0.000 047 185 92 × 2 = 0 + 0.000 094 371 84;
  • 18) 0.000 094 371 84 × 2 = 0 + 0.000 188 743 68;
  • 19) 0.000 188 743 68 × 2 = 0 + 0.000 377 487 36;
  • 20) 0.000 377 487 36 × 2 = 0 + 0.000 754 974 72;
  • 21) 0.000 754 974 72 × 2 = 0 + 0.001 509 949 44;
  • 22) 0.001 509 949 44 × 2 = 0 + 0.003 019 898 88;
  • 23) 0.003 019 898 88 × 2 = 0 + 0.006 039 797 76;
  • 24) 0.006 039 797 76 × 2 = 0 + 0.012 079 595 52;
  • 25) 0.012 079 595 52 × 2 = 0 + 0.024 159 191 04;
  • 26) 0.024 159 191 04 × 2 = 0 + 0.048 318 382 08;
  • 27) 0.048 318 382 08 × 2 = 0 + 0.096 636 764 16;
  • 28) 0.096 636 764 16 × 2 = 0 + 0.193 273 528 32;
  • 29) 0.193 273 528 32 × 2 = 0 + 0.386 547 056 64;
  • 30) 0.386 547 056 64 × 2 = 0 + 0.773 094 113 28;
  • 31) 0.773 094 113 28 × 2 = 1 + 0.546 188 226 56;
  • 32) 0.546 188 226 56 × 2 = 1 + 0.092 376 453 12;
  • 33) 0.092 376 453 12 × 2 = 0 + 0.184 752 906 24;
  • 34) 0.184 752 906 24 × 2 = 0 + 0.369 505 812 48;
  • 35) 0.369 505 812 48 × 2 = 0 + 0.739 011 624 96;
  • 36) 0.739 011 624 96 × 2 = 1 + 0.478 023 249 92;
  • 37) 0.478 023 249 92 × 2 = 0 + 0.956 046 499 84;
  • 38) 0.956 046 499 84 × 2 = 1 + 0.912 092 999 68;
  • 39) 0.912 092 999 68 × 2 = 1 + 0.824 185 999 36;
  • 40) 0.824 185 999 36 × 2 = 1 + 0.648 371 998 72;
  • 41) 0.648 371 998 72 × 2 = 1 + 0.296 743 997 44;
  • 42) 0.296 743 997 44 × 2 = 0 + 0.593 487 994 88;
  • 43) 0.593 487 994 88 × 2 = 1 + 0.186 975 989 76;
  • 44) 0.186 975 989 76 × 2 = 0 + 0.373 951 979 52;
  • 45) 0.373 951 979 52 × 2 = 0 + 0.747 903 959 04;
  • 46) 0.747 903 959 04 × 2 = 1 + 0.495 807 918 08;
  • 47) 0.495 807 918 08 × 2 = 0 + 0.991 615 836 16;
  • 48) 0.991 615 836 16 × 2 = 1 + 0.983 231 672 32;
  • 49) 0.983 231 672 32 × 2 = 1 + 0.966 463 344 64;
  • 50) 0.966 463 344 64 × 2 = 1 + 0.932 926 689 28;
  • 51) 0.932 926 689 28 × 2 = 1 + 0.865 853 378 56;
  • 52) 0.865 853 378 56 × 2 = 1 + 0.731 706 757 12;
  • 53) 0.731 706 757 12 × 2 = 1 + 0.463 413 514 24;
  • 54) 0.463 413 514 24 × 2 = 0 + 0.926 827 028 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).


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 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 0111 1010 0101 1111 10(2)

6. Positive number before normalization:

0.000 000 000 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 0111 1010 0101 1111 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 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 0111 1010 0101 1111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 0111 1010 0101 1111 10(2) × 20 =

1.1000 1011 1101 0010 1111 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):
1.1000 1011 1101 0010 1111 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 0101 1110 1001 0111 1110 =

100 0101 1110 1001 0111 1110


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 0101 1110 1001 0111 1110


Decimal number -0.000 000 000 72 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 0101 1110 1001 0111 1110

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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