-0.000 000 000 73 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 73(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 73(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 73| = 0.000 000 000 73


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 73.

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 73 × 2 = 0 + 0.000 000 001 46;
  • 2) 0.000 000 001 46 × 2 = 0 + 0.000 000 002 92;
  • 3) 0.000 000 002 92 × 2 = 0 + 0.000 000 005 84;
  • 4) 0.000 000 005 84 × 2 = 0 + 0.000 000 011 68;
  • 5) 0.000 000 011 68 × 2 = 0 + 0.000 000 023 36;
  • 6) 0.000 000 023 36 × 2 = 0 + 0.000 000 046 72;
  • 7) 0.000 000 046 72 × 2 = 0 + 0.000 000 093 44;
  • 8) 0.000 000 093 44 × 2 = 0 + 0.000 000 186 88;
  • 9) 0.000 000 186 88 × 2 = 0 + 0.000 000 373 76;
  • 10) 0.000 000 373 76 × 2 = 0 + 0.000 000 747 52;
  • 11) 0.000 000 747 52 × 2 = 0 + 0.000 001 495 04;
  • 12) 0.000 001 495 04 × 2 = 0 + 0.000 002 990 08;
  • 13) 0.000 002 990 08 × 2 = 0 + 0.000 005 980 16;
  • 14) 0.000 005 980 16 × 2 = 0 + 0.000 011 960 32;
  • 15) 0.000 011 960 32 × 2 = 0 + 0.000 023 920 64;
  • 16) 0.000 023 920 64 × 2 = 0 + 0.000 047 841 28;
  • 17) 0.000 047 841 28 × 2 = 0 + 0.000 095 682 56;
  • 18) 0.000 095 682 56 × 2 = 0 + 0.000 191 365 12;
  • 19) 0.000 191 365 12 × 2 = 0 + 0.000 382 730 24;
  • 20) 0.000 382 730 24 × 2 = 0 + 0.000 765 460 48;
  • 21) 0.000 765 460 48 × 2 = 0 + 0.001 530 920 96;
  • 22) 0.001 530 920 96 × 2 = 0 + 0.003 061 841 92;
  • 23) 0.003 061 841 92 × 2 = 0 + 0.006 123 683 84;
  • 24) 0.006 123 683 84 × 2 = 0 + 0.012 247 367 68;
  • 25) 0.012 247 367 68 × 2 = 0 + 0.024 494 735 36;
  • 26) 0.024 494 735 36 × 2 = 0 + 0.048 989 470 72;
  • 27) 0.048 989 470 72 × 2 = 0 + 0.097 978 941 44;
  • 28) 0.097 978 941 44 × 2 = 0 + 0.195 957 882 88;
  • 29) 0.195 957 882 88 × 2 = 0 + 0.391 915 765 76;
  • 30) 0.391 915 765 76 × 2 = 0 + 0.783 831 531 52;
  • 31) 0.783 831 531 52 × 2 = 1 + 0.567 663 063 04;
  • 32) 0.567 663 063 04 × 2 = 1 + 0.135 326 126 08;
  • 33) 0.135 326 126 08 × 2 = 0 + 0.270 652 252 16;
  • 34) 0.270 652 252 16 × 2 = 0 + 0.541 304 504 32;
  • 35) 0.541 304 504 32 × 2 = 1 + 0.082 609 008 64;
  • 36) 0.082 609 008 64 × 2 = 0 + 0.165 218 017 28;
  • 37) 0.165 218 017 28 × 2 = 0 + 0.330 436 034 56;
  • 38) 0.330 436 034 56 × 2 = 0 + 0.660 872 069 12;
  • 39) 0.660 872 069 12 × 2 = 1 + 0.321 744 138 24;
  • 40) 0.321 744 138 24 × 2 = 0 + 0.643 488 276 48;
  • 41) 0.643 488 276 48 × 2 = 1 + 0.286 976 552 96;
  • 42) 0.286 976 552 96 × 2 = 0 + 0.573 953 105 92;
  • 43) 0.573 953 105 92 × 2 = 1 + 0.147 906 211 84;
  • 44) 0.147 906 211 84 × 2 = 0 + 0.295 812 423 68;
  • 45) 0.295 812 423 68 × 2 = 0 + 0.591 624 847 36;
  • 46) 0.591 624 847 36 × 2 = 1 + 0.183 249 694 72;
  • 47) 0.183 249 694 72 × 2 = 0 + 0.366 499 389 44;
  • 48) 0.366 499 389 44 × 2 = 0 + 0.732 998 778 88;
  • 49) 0.732 998 778 88 × 2 = 1 + 0.465 997 557 76;
  • 50) 0.465 997 557 76 × 2 = 0 + 0.931 995 115 52;
  • 51) 0.931 995 115 52 × 2 = 1 + 0.863 990 231 04;
  • 52) 0.863 990 231 04 × 2 = 1 + 0.727 980 462 08;
  • 53) 0.727 980 462 08 × 2 = 1 + 0.455 960 924 16;
  • 54) 0.455 960 924 16 × 2 = 0 + 0.911 921 848 32;

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 73(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 0010 1010 0100 1011 10(2)

6. Positive number before normalization:

0.000 000 000 73(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 0010 1010 0100 1011 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 73(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 0010 1010 0100 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 0010 1010 0100 1011 10(2) × 20 =

1.1001 0001 0101 0010 0101 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 0001 0101 0010 0101 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 1000 1010 1001 0010 1110 =

100 1000 1010 1001 0010 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 1000 1010 1001 0010 1110


Decimal number -0.000 000 000 73 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1000 1010 1001 0010 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