-0.000 000 000 681 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 681(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
-0.000 000 000 681(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:

|-0.000 000 000 681| = 0.000 000 000 681


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: 0.000 000 000 681.

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) 0.000 000 000 681 × 2 = 0 + 0.000 000 001 362;
  • 2) 0.000 000 001 362 × 2 = 0 + 0.000 000 002 724;
  • 3) 0.000 000 002 724 × 2 = 0 + 0.000 000 005 448;
  • 4) 0.000 000 005 448 × 2 = 0 + 0.000 000 010 896;
  • 5) 0.000 000 010 896 × 2 = 0 + 0.000 000 021 792;
  • 6) 0.000 000 021 792 × 2 = 0 + 0.000 000 043 584;
  • 7) 0.000 000 043 584 × 2 = 0 + 0.000 000 087 168;
  • 8) 0.000 000 087 168 × 2 = 0 + 0.000 000 174 336;
  • 9) 0.000 000 174 336 × 2 = 0 + 0.000 000 348 672;
  • 10) 0.000 000 348 672 × 2 = 0 + 0.000 000 697 344;
  • 11) 0.000 000 697 344 × 2 = 0 + 0.000 001 394 688;
  • 12) 0.000 001 394 688 × 2 = 0 + 0.000 002 789 376;
  • 13) 0.000 002 789 376 × 2 = 0 + 0.000 005 578 752;
  • 14) 0.000 005 578 752 × 2 = 0 + 0.000 011 157 504;
  • 15) 0.000 011 157 504 × 2 = 0 + 0.000 022 315 008;
  • 16) 0.000 022 315 008 × 2 = 0 + 0.000 044 630 016;
  • 17) 0.000 044 630 016 × 2 = 0 + 0.000 089 260 032;
  • 18) 0.000 089 260 032 × 2 = 0 + 0.000 178 520 064;
  • 19) 0.000 178 520 064 × 2 = 0 + 0.000 357 040 128;
  • 20) 0.000 357 040 128 × 2 = 0 + 0.000 714 080 256;
  • 21) 0.000 714 080 256 × 2 = 0 + 0.001 428 160 512;
  • 22) 0.001 428 160 512 × 2 = 0 + 0.002 856 321 024;
  • 23) 0.002 856 321 024 × 2 = 0 + 0.005 712 642 048;
  • 24) 0.005 712 642 048 × 2 = 0 + 0.011 425 284 096;
  • 25) 0.011 425 284 096 × 2 = 0 + 0.022 850 568 192;
  • 26) 0.022 850 568 192 × 2 = 0 + 0.045 701 136 384;
  • 27) 0.045 701 136 384 × 2 = 0 + 0.091 402 272 768;
  • 28) 0.091 402 272 768 × 2 = 0 + 0.182 804 545 536;
  • 29) 0.182 804 545 536 × 2 = 0 + 0.365 609 091 072;
  • 30) 0.365 609 091 072 × 2 = 0 + 0.731 218 182 144;
  • 31) 0.731 218 182 144 × 2 = 1 + 0.462 436 364 288;
  • 32) 0.462 436 364 288 × 2 = 0 + 0.924 872 728 576;
  • 33) 0.924 872 728 576 × 2 = 1 + 0.849 745 457 152;
  • 34) 0.849 745 457 152 × 2 = 1 + 0.699 490 914 304;
  • 35) 0.699 490 914 304 × 2 = 1 + 0.398 981 828 608;
  • 36) 0.398 981 828 608 × 2 = 0 + 0.797 963 657 216;
  • 37) 0.797 963 657 216 × 2 = 1 + 0.595 927 314 432;
  • 38) 0.595 927 314 432 × 2 = 1 + 0.191 854 628 864;
  • 39) 0.191 854 628 864 × 2 = 0 + 0.383 709 257 728;
  • 40) 0.383 709 257 728 × 2 = 0 + 0.767 418 515 456;
  • 41) 0.767 418 515 456 × 2 = 1 + 0.534 837 030 912;
  • 42) 0.534 837 030 912 × 2 = 1 + 0.069 674 061 824;
  • 43) 0.069 674 061 824 × 2 = 0 + 0.139 348 123 648;
  • 44) 0.139 348 123 648 × 2 = 0 + 0.278 696 247 296;
  • 45) 0.278 696 247 296 × 2 = 0 + 0.557 392 494 592;
  • 46) 0.557 392 494 592 × 2 = 1 + 0.114 784 989 184;
  • 47) 0.114 784 989 184 × 2 = 0 + 0.229 569 978 368;
  • 48) 0.229 569 978 368 × 2 = 0 + 0.459 139 956 736;
  • 49) 0.459 139 956 736 × 2 = 0 + 0.918 279 913 472;
  • 50) 0.918 279 913 472 × 2 = 1 + 0.836 559 826 944;
  • 51) 0.836 559 826 944 × 2 = 1 + 0.673 119 653 888;
  • 52) 0.673 119 653 888 × 2 = 1 + 0.346 239 307 776;
  • 53) 0.346 239 307 776 × 2 = 0 + 0.692 478 615 552;
  • 54) 0.692 478 615 552 × 2 = 1 + 0.384 957 231 104;

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:


0.000 000 000 681(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1110 1100 1100 0100 0111 01(2)

6. Positive number before normalization:

0.000 000 000 681(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1110 1100 1100 0100 0111 01(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:


0.000 000 000 681(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1110 1100 1100 0100 0111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1110 1100 1100 0100 0111 01(2) × 20 =

1.0111 0110 0110 0010 0011 101(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):
1.0111 0110 0110 0010 0011 101


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. 011 1011 0011 0001 0001 1101 =

011 1011 0011 0001 0001 1101


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) =
011 1011 0011 0001 0001 1101


Decimal number -0.000 000 000 681 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 011 1011 0011 0001 0001 1101

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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