-0.000 000 000 723 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 723(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
-0.000 000 000 723(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:

|-0.000 000 000 723| = 0.000 000 000 723


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: 0.000 000 000 723.

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) 0.000 000 000 723 × 2 = 0 + 0.000 000 001 446;
  • 2) 0.000 000 001 446 × 2 = 0 + 0.000 000 002 892;
  • 3) 0.000 000 002 892 × 2 = 0 + 0.000 000 005 784;
  • 4) 0.000 000 005 784 × 2 = 0 + 0.000 000 011 568;
  • 5) 0.000 000 011 568 × 2 = 0 + 0.000 000 023 136;
  • 6) 0.000 000 023 136 × 2 = 0 + 0.000 000 046 272;
  • 7) 0.000 000 046 272 × 2 = 0 + 0.000 000 092 544;
  • 8) 0.000 000 092 544 × 2 = 0 + 0.000 000 185 088;
  • 9) 0.000 000 185 088 × 2 = 0 + 0.000 000 370 176;
  • 10) 0.000 000 370 176 × 2 = 0 + 0.000 000 740 352;
  • 11) 0.000 000 740 352 × 2 = 0 + 0.000 001 480 704;
  • 12) 0.000 001 480 704 × 2 = 0 + 0.000 002 961 408;
  • 13) 0.000 002 961 408 × 2 = 0 + 0.000 005 922 816;
  • 14) 0.000 005 922 816 × 2 = 0 + 0.000 011 845 632;
  • 15) 0.000 011 845 632 × 2 = 0 + 0.000 023 691 264;
  • 16) 0.000 023 691 264 × 2 = 0 + 0.000 047 382 528;
  • 17) 0.000 047 382 528 × 2 = 0 + 0.000 094 765 056;
  • 18) 0.000 094 765 056 × 2 = 0 + 0.000 189 530 112;
  • 19) 0.000 189 530 112 × 2 = 0 + 0.000 379 060 224;
  • 20) 0.000 379 060 224 × 2 = 0 + 0.000 758 120 448;
  • 21) 0.000 758 120 448 × 2 = 0 + 0.001 516 240 896;
  • 22) 0.001 516 240 896 × 2 = 0 + 0.003 032 481 792;
  • 23) 0.003 032 481 792 × 2 = 0 + 0.006 064 963 584;
  • 24) 0.006 064 963 584 × 2 = 0 + 0.012 129 927 168;
  • 25) 0.012 129 927 168 × 2 = 0 + 0.024 259 854 336;
  • 26) 0.024 259 854 336 × 2 = 0 + 0.048 519 708 672;
  • 27) 0.048 519 708 672 × 2 = 0 + 0.097 039 417 344;
  • 28) 0.097 039 417 344 × 2 = 0 + 0.194 078 834 688;
  • 29) 0.194 078 834 688 × 2 = 0 + 0.388 157 669 376;
  • 30) 0.388 157 669 376 × 2 = 0 + 0.776 315 338 752;
  • 31) 0.776 315 338 752 × 2 = 1 + 0.552 630 677 504;
  • 32) 0.552 630 677 504 × 2 = 1 + 0.105 261 355 008;
  • 33) 0.105 261 355 008 × 2 = 0 + 0.210 522 710 016;
  • 34) 0.210 522 710 016 × 2 = 0 + 0.421 045 420 032;
  • 35) 0.421 045 420 032 × 2 = 0 + 0.842 090 840 064;
  • 36) 0.842 090 840 064 × 2 = 1 + 0.684 181 680 128;
  • 37) 0.684 181 680 128 × 2 = 1 + 0.368 363 360 256;
  • 38) 0.368 363 360 256 × 2 = 0 + 0.736 726 720 512;
  • 39) 0.736 726 720 512 × 2 = 1 + 0.473 453 441 024;
  • 40) 0.473 453 441 024 × 2 = 0 + 0.946 906 882 048;
  • 41) 0.946 906 882 048 × 2 = 1 + 0.893 813 764 096;
  • 42) 0.893 813 764 096 × 2 = 1 + 0.787 627 528 192;
  • 43) 0.787 627 528 192 × 2 = 1 + 0.575 255 056 384;
  • 44) 0.575 255 056 384 × 2 = 1 + 0.150 510 112 768;
  • 45) 0.150 510 112 768 × 2 = 0 + 0.301 020 225 536;
  • 46) 0.301 020 225 536 × 2 = 0 + 0.602 040 451 072;
  • 47) 0.602 040 451 072 × 2 = 1 + 0.204 080 902 144;
  • 48) 0.204 080 902 144 × 2 = 0 + 0.408 161 804 288;
  • 49) 0.408 161 804 288 × 2 = 0 + 0.816 323 608 576;
  • 50) 0.816 323 608 576 × 2 = 1 + 0.632 647 217 152;
  • 51) 0.632 647 217 152 × 2 = 1 + 0.265 294 434 304;
  • 52) 0.265 294 434 304 × 2 = 0 + 0.530 588 868 608;
  • 53) 0.530 588 868 608 × 2 = 1 + 0.061 177 737 216;
  • 54) 0.061 177 737 216 × 2 = 0 + 0.122 355 474 432;

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:


0.000 000 000 723(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1111 0010 0110 10(2)

6. Positive number before normalization:

0.000 000 000 723(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1111 0010 0110 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:


0.000 000 000 723(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1111 0010 0110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1111 0010 0110 10(2) × 20 =

1.1000 1101 0111 1001 0011 010(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 1101 0111 1001 0011 010


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 0110 1011 1100 1001 1010 =

100 0110 1011 1100 1001 1010


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 0110 1011 1100 1001 1010


Decimal number -0.000 000 000 723 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 0110 1011 1100 1001 1010

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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