-0.000 000 000 799 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 799(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
-0.000 000 000 799(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:

|-0.000 000 000 799| = 0.000 000 000 799


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: 0.000 000 000 799.

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) 0.000 000 000 799 × 2 = 0 + 0.000 000 001 598;
  • 2) 0.000 000 001 598 × 2 = 0 + 0.000 000 003 196;
  • 3) 0.000 000 003 196 × 2 = 0 + 0.000 000 006 392;
  • 4) 0.000 000 006 392 × 2 = 0 + 0.000 000 012 784;
  • 5) 0.000 000 012 784 × 2 = 0 + 0.000 000 025 568;
  • 6) 0.000 000 025 568 × 2 = 0 + 0.000 000 051 136;
  • 7) 0.000 000 051 136 × 2 = 0 + 0.000 000 102 272;
  • 8) 0.000 000 102 272 × 2 = 0 + 0.000 000 204 544;
  • 9) 0.000 000 204 544 × 2 = 0 + 0.000 000 409 088;
  • 10) 0.000 000 409 088 × 2 = 0 + 0.000 000 818 176;
  • 11) 0.000 000 818 176 × 2 = 0 + 0.000 001 636 352;
  • 12) 0.000 001 636 352 × 2 = 0 + 0.000 003 272 704;
  • 13) 0.000 003 272 704 × 2 = 0 + 0.000 006 545 408;
  • 14) 0.000 006 545 408 × 2 = 0 + 0.000 013 090 816;
  • 15) 0.000 013 090 816 × 2 = 0 + 0.000 026 181 632;
  • 16) 0.000 026 181 632 × 2 = 0 + 0.000 052 363 264;
  • 17) 0.000 052 363 264 × 2 = 0 + 0.000 104 726 528;
  • 18) 0.000 104 726 528 × 2 = 0 + 0.000 209 453 056;
  • 19) 0.000 209 453 056 × 2 = 0 + 0.000 418 906 112;
  • 20) 0.000 418 906 112 × 2 = 0 + 0.000 837 812 224;
  • 21) 0.000 837 812 224 × 2 = 0 + 0.001 675 624 448;
  • 22) 0.001 675 624 448 × 2 = 0 + 0.003 351 248 896;
  • 23) 0.003 351 248 896 × 2 = 0 + 0.006 702 497 792;
  • 24) 0.006 702 497 792 × 2 = 0 + 0.013 404 995 584;
  • 25) 0.013 404 995 584 × 2 = 0 + 0.026 809 991 168;
  • 26) 0.026 809 991 168 × 2 = 0 + 0.053 619 982 336;
  • 27) 0.053 619 982 336 × 2 = 0 + 0.107 239 964 672;
  • 28) 0.107 239 964 672 × 2 = 0 + 0.214 479 929 344;
  • 29) 0.214 479 929 344 × 2 = 0 + 0.428 959 858 688;
  • 30) 0.428 959 858 688 × 2 = 0 + 0.857 919 717 376;
  • 31) 0.857 919 717 376 × 2 = 1 + 0.715 839 434 752;
  • 32) 0.715 839 434 752 × 2 = 1 + 0.431 678 869 504;
  • 33) 0.431 678 869 504 × 2 = 0 + 0.863 357 739 008;
  • 34) 0.863 357 739 008 × 2 = 1 + 0.726 715 478 016;
  • 35) 0.726 715 478 016 × 2 = 1 + 0.453 430 956 032;
  • 36) 0.453 430 956 032 × 2 = 0 + 0.906 861 912 064;
  • 37) 0.906 861 912 064 × 2 = 1 + 0.813 723 824 128;
  • 38) 0.813 723 824 128 × 2 = 1 + 0.627 447 648 256;
  • 39) 0.627 447 648 256 × 2 = 1 + 0.254 895 296 512;
  • 40) 0.254 895 296 512 × 2 = 0 + 0.509 790 593 024;
  • 41) 0.509 790 593 024 × 2 = 1 + 0.019 581 186 048;
  • 42) 0.019 581 186 048 × 2 = 0 + 0.039 162 372 096;
  • 43) 0.039 162 372 096 × 2 = 0 + 0.078 324 744 192;
  • 44) 0.078 324 744 192 × 2 = 0 + 0.156 649 488 384;
  • 45) 0.156 649 488 384 × 2 = 0 + 0.313 298 976 768;
  • 46) 0.313 298 976 768 × 2 = 0 + 0.626 597 953 536;
  • 47) 0.626 597 953 536 × 2 = 1 + 0.253 195 907 072;
  • 48) 0.253 195 907 072 × 2 = 0 + 0.506 391 814 144;
  • 49) 0.506 391 814 144 × 2 = 1 + 0.012 783 628 288;
  • 50) 0.012 783 628 288 × 2 = 0 + 0.025 567 256 576;
  • 51) 0.025 567 256 576 × 2 = 0 + 0.051 134 513 152;
  • 52) 0.051 134 513 152 × 2 = 0 + 0.102 269 026 304;
  • 53) 0.102 269 026 304 × 2 = 0 + 0.204 538 052 608;
  • 54) 0.204 538 052 608 × 2 = 0 + 0.409 076 105 216;

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:


0.000 000 000 799(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 1110 1000 0010 1000 00(2)

6. Positive number before normalization:

0.000 000 000 799(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 1110 1000 0010 1000 00(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:


0.000 000 000 799(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 1110 1000 0010 1000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 1110 1000 0010 1000 00(2) × 20 =

1.1011 0111 0100 0001 0100 000(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):
1.1011 0111 0100 0001 0100 000


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. 101 1011 1010 0000 1010 0000 =

101 1011 1010 0000 1010 0000


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) =
101 1011 1010 0000 1010 0000


Decimal number -0.000 000 000 799 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 1011 1010 0000 1010 0000

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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