-0.000 000 000 741 999 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 999(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
-0.000 000 000 741 999(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:

|-0.000 000 000 741 999| = 0.000 000 000 741 999


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: 0.000 000 000 741 999.

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) 0.000 000 000 741 999 × 2 = 0 + 0.000 000 001 483 998;
  • 2) 0.000 000 001 483 998 × 2 = 0 + 0.000 000 002 967 996;
  • 3) 0.000 000 002 967 996 × 2 = 0 + 0.000 000 005 935 992;
  • 4) 0.000 000 005 935 992 × 2 = 0 + 0.000 000 011 871 984;
  • 5) 0.000 000 011 871 984 × 2 = 0 + 0.000 000 023 743 968;
  • 6) 0.000 000 023 743 968 × 2 = 0 + 0.000 000 047 487 936;
  • 7) 0.000 000 047 487 936 × 2 = 0 + 0.000 000 094 975 872;
  • 8) 0.000 000 094 975 872 × 2 = 0 + 0.000 000 189 951 744;
  • 9) 0.000 000 189 951 744 × 2 = 0 + 0.000 000 379 903 488;
  • 10) 0.000 000 379 903 488 × 2 = 0 + 0.000 000 759 806 976;
  • 11) 0.000 000 759 806 976 × 2 = 0 + 0.000 001 519 613 952;
  • 12) 0.000 001 519 613 952 × 2 = 0 + 0.000 003 039 227 904;
  • 13) 0.000 003 039 227 904 × 2 = 0 + 0.000 006 078 455 808;
  • 14) 0.000 006 078 455 808 × 2 = 0 + 0.000 012 156 911 616;
  • 15) 0.000 012 156 911 616 × 2 = 0 + 0.000 024 313 823 232;
  • 16) 0.000 024 313 823 232 × 2 = 0 + 0.000 048 627 646 464;
  • 17) 0.000 048 627 646 464 × 2 = 0 + 0.000 097 255 292 928;
  • 18) 0.000 097 255 292 928 × 2 = 0 + 0.000 194 510 585 856;
  • 19) 0.000 194 510 585 856 × 2 = 0 + 0.000 389 021 171 712;
  • 20) 0.000 389 021 171 712 × 2 = 0 + 0.000 778 042 343 424;
  • 21) 0.000 778 042 343 424 × 2 = 0 + 0.001 556 084 686 848;
  • 22) 0.001 556 084 686 848 × 2 = 0 + 0.003 112 169 373 696;
  • 23) 0.003 112 169 373 696 × 2 = 0 + 0.006 224 338 747 392;
  • 24) 0.006 224 338 747 392 × 2 = 0 + 0.012 448 677 494 784;
  • 25) 0.012 448 677 494 784 × 2 = 0 + 0.024 897 354 989 568;
  • 26) 0.024 897 354 989 568 × 2 = 0 + 0.049 794 709 979 136;
  • 27) 0.049 794 709 979 136 × 2 = 0 + 0.099 589 419 958 272;
  • 28) 0.099 589 419 958 272 × 2 = 0 + 0.199 178 839 916 544;
  • 29) 0.199 178 839 916 544 × 2 = 0 + 0.398 357 679 833 088;
  • 30) 0.398 357 679 833 088 × 2 = 0 + 0.796 715 359 666 176;
  • 31) 0.796 715 359 666 176 × 2 = 1 + 0.593 430 719 332 352;
  • 32) 0.593 430 719 332 352 × 2 = 1 + 0.186 861 438 664 704;
  • 33) 0.186 861 438 664 704 × 2 = 0 + 0.373 722 877 329 408;
  • 34) 0.373 722 877 329 408 × 2 = 0 + 0.747 445 754 658 816;
  • 35) 0.747 445 754 658 816 × 2 = 1 + 0.494 891 509 317 632;
  • 36) 0.494 891 509 317 632 × 2 = 0 + 0.989 783 018 635 264;
  • 37) 0.989 783 018 635 264 × 2 = 1 + 0.979 566 037 270 528;
  • 38) 0.979 566 037 270 528 × 2 = 1 + 0.959 132 074 541 056;
  • 39) 0.959 132 074 541 056 × 2 = 1 + 0.918 264 149 082 112;
  • 40) 0.918 264 149 082 112 × 2 = 1 + 0.836 528 298 164 224;
  • 41) 0.836 528 298 164 224 × 2 = 1 + 0.673 056 596 328 448;
  • 42) 0.673 056 596 328 448 × 2 = 1 + 0.346 113 192 656 896;
  • 43) 0.346 113 192 656 896 × 2 = 0 + 0.692 226 385 313 792;
  • 44) 0.692 226 385 313 792 × 2 = 1 + 0.384 452 770 627 584;
  • 45) 0.384 452 770 627 584 × 2 = 0 + 0.768 905 541 255 168;
  • 46) 0.768 905 541 255 168 × 2 = 1 + 0.537 811 082 510 336;
  • 47) 0.537 811 082 510 336 × 2 = 1 + 0.075 622 165 020 672;
  • 48) 0.075 622 165 020 672 × 2 = 0 + 0.151 244 330 041 344;
  • 49) 0.151 244 330 041 344 × 2 = 0 + 0.302 488 660 082 688;
  • 50) 0.302 488 660 082 688 × 2 = 0 + 0.604 977 320 165 376;
  • 51) 0.604 977 320 165 376 × 2 = 1 + 0.209 954 640 330 752;
  • 52) 0.209 954 640 330 752 × 2 = 0 + 0.419 909 280 661 504;
  • 53) 0.419 909 280 661 504 × 2 = 0 + 0.839 818 561 323 008;
  • 54) 0.839 818 561 323 008 × 2 = 1 + 0.679 637 122 646 016;

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:


0.000 000 000 741 999(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0110 0010 01(2)

6. Positive number before normalization:

0.000 000 000 741 999(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0110 0010 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:


0.000 000 000 741 999(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0110 0010 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0110 0010 01(2) × 20 =

1.1001 0111 1110 1011 0001 001(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 0111 1110 1011 0001 001


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 1011 1111 0101 1000 1001 =

100 1011 1111 0101 1000 1001


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 1011 1111 0101 1000 1001


Decimal number -0.000 000 000 741 999 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 0101 1000 1001

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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