-0.000 000 000 742 147 676 588 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 588(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
-0.000 000 000 742 147 676 588(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:

|-0.000 000 000 742 147 676 588| = 0.000 000 000 742 147 676 588


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: 0.000 000 000 742 147 676 588.

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) 0.000 000 000 742 147 676 588 × 2 = 0 + 0.000 000 001 484 295 353 176;
  • 2) 0.000 000 001 484 295 353 176 × 2 = 0 + 0.000 000 002 968 590 706 352;
  • 3) 0.000 000 002 968 590 706 352 × 2 = 0 + 0.000 000 005 937 181 412 704;
  • 4) 0.000 000 005 937 181 412 704 × 2 = 0 + 0.000 000 011 874 362 825 408;
  • 5) 0.000 000 011 874 362 825 408 × 2 = 0 + 0.000 000 023 748 725 650 816;
  • 6) 0.000 000 023 748 725 650 816 × 2 = 0 + 0.000 000 047 497 451 301 632;
  • 7) 0.000 000 047 497 451 301 632 × 2 = 0 + 0.000 000 094 994 902 603 264;
  • 8) 0.000 000 094 994 902 603 264 × 2 = 0 + 0.000 000 189 989 805 206 528;
  • 9) 0.000 000 189 989 805 206 528 × 2 = 0 + 0.000 000 379 979 610 413 056;
  • 10) 0.000 000 379 979 610 413 056 × 2 = 0 + 0.000 000 759 959 220 826 112;
  • 11) 0.000 000 759 959 220 826 112 × 2 = 0 + 0.000 001 519 918 441 652 224;
  • 12) 0.000 001 519 918 441 652 224 × 2 = 0 + 0.000 003 039 836 883 304 448;
  • 13) 0.000 003 039 836 883 304 448 × 2 = 0 + 0.000 006 079 673 766 608 896;
  • 14) 0.000 006 079 673 766 608 896 × 2 = 0 + 0.000 012 159 347 533 217 792;
  • 15) 0.000 012 159 347 533 217 792 × 2 = 0 + 0.000 024 318 695 066 435 584;
  • 16) 0.000 024 318 695 066 435 584 × 2 = 0 + 0.000 048 637 390 132 871 168;
  • 17) 0.000 048 637 390 132 871 168 × 2 = 0 + 0.000 097 274 780 265 742 336;
  • 18) 0.000 097 274 780 265 742 336 × 2 = 0 + 0.000 194 549 560 531 484 672;
  • 19) 0.000 194 549 560 531 484 672 × 2 = 0 + 0.000 389 099 121 062 969 344;
  • 20) 0.000 389 099 121 062 969 344 × 2 = 0 + 0.000 778 198 242 125 938 688;
  • 21) 0.000 778 198 242 125 938 688 × 2 = 0 + 0.001 556 396 484 251 877 376;
  • 22) 0.001 556 396 484 251 877 376 × 2 = 0 + 0.003 112 792 968 503 754 752;
  • 23) 0.003 112 792 968 503 754 752 × 2 = 0 + 0.006 225 585 937 007 509 504;
  • 24) 0.006 225 585 937 007 509 504 × 2 = 0 + 0.012 451 171 874 015 019 008;
  • 25) 0.012 451 171 874 015 019 008 × 2 = 0 + 0.024 902 343 748 030 038 016;
  • 26) 0.024 902 343 748 030 038 016 × 2 = 0 + 0.049 804 687 496 060 076 032;
  • 27) 0.049 804 687 496 060 076 032 × 2 = 0 + 0.099 609 374 992 120 152 064;
  • 28) 0.099 609 374 992 120 152 064 × 2 = 0 + 0.199 218 749 984 240 304 128;
  • 29) 0.199 218 749 984 240 304 128 × 2 = 0 + 0.398 437 499 968 480 608 256;
  • 30) 0.398 437 499 968 480 608 256 × 2 = 0 + 0.796 874 999 936 961 216 512;
  • 31) 0.796 874 999 936 961 216 512 × 2 = 1 + 0.593 749 999 873 922 433 024;
  • 32) 0.593 749 999 873 922 433 024 × 2 = 1 + 0.187 499 999 747 844 866 048;
  • 33) 0.187 499 999 747 844 866 048 × 2 = 0 + 0.374 999 999 495 689 732 096;
  • 34) 0.374 999 999 495 689 732 096 × 2 = 0 + 0.749 999 998 991 379 464 192;
  • 35) 0.749 999 998 991 379 464 192 × 2 = 1 + 0.499 999 997 982 758 928 384;
  • 36) 0.499 999 997 982 758 928 384 × 2 = 0 + 0.999 999 995 965 517 856 768;
  • 37) 0.999 999 995 965 517 856 768 × 2 = 1 + 0.999 999 991 931 035 713 536;
  • 38) 0.999 999 991 931 035 713 536 × 2 = 1 + 0.999 999 983 862 071 427 072;
  • 39) 0.999 999 983 862 071 427 072 × 2 = 1 + 0.999 999 967 724 142 854 144;
  • 40) 0.999 999 967 724 142 854 144 × 2 = 1 + 0.999 999 935 448 285 708 288;
  • 41) 0.999 999 935 448 285 708 288 × 2 = 1 + 0.999 999 870 896 571 416 576;
  • 42) 0.999 999 870 896 571 416 576 × 2 = 1 + 0.999 999 741 793 142 833 152;
  • 43) 0.999 999 741 793 142 833 152 × 2 = 1 + 0.999 999 483 586 285 666 304;
  • 44) 0.999 999 483 586 285 666 304 × 2 = 1 + 0.999 998 967 172 571 332 608;
  • 45) 0.999 998 967 172 571 332 608 × 2 = 1 + 0.999 997 934 345 142 665 216;
  • 46) 0.999 997 934 345 142 665 216 × 2 = 1 + 0.999 995 868 690 285 330 432;
  • 47) 0.999 995 868 690 285 330 432 × 2 = 1 + 0.999 991 737 380 570 660 864;
  • 48) 0.999 991 737 380 570 660 864 × 2 = 1 + 0.999 983 474 761 141 321 728;
  • 49) 0.999 983 474 761 141 321 728 × 2 = 1 + 0.999 966 949 522 282 643 456;
  • 50) 0.999 966 949 522 282 643 456 × 2 = 1 + 0.999 933 899 044 565 286 912;
  • 51) 0.999 933 899 044 565 286 912 × 2 = 1 + 0.999 867 798 089 130 573 824;
  • 52) 0.999 867 798 089 130 573 824 × 2 = 1 + 0.999 735 596 178 261 147 648;
  • 53) 0.999 735 596 178 261 147 648 × 2 = 1 + 0.999 471 192 356 522 295 296;
  • 54) 0.999 471 192 356 522 295 296 × 2 = 1 + 0.998 942 384 713 044 590 592;

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:


0.000 000 000 742 147 676 588(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 588(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 588(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 1111 1111 1111 111


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 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 588 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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