-0.000 000 000 742 147 676 646 684 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 684 8(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
-0.000 000 000 742 147 676 646 684 8(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:

|-0.000 000 000 742 147 676 646 684 8| = 0.000 000 000 742 147 676 646 684 8


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: 0.000 000 000 742 147 676 646 684 8.

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) 0.000 000 000 742 147 676 646 684 8 × 2 = 0 + 0.000 000 001 484 295 353 293 369 6;
  • 2) 0.000 000 001 484 295 353 293 369 6 × 2 = 0 + 0.000 000 002 968 590 706 586 739 2;
  • 3) 0.000 000 002 968 590 706 586 739 2 × 2 = 0 + 0.000 000 005 937 181 413 173 478 4;
  • 4) 0.000 000 005 937 181 413 173 478 4 × 2 = 0 + 0.000 000 011 874 362 826 346 956 8;
  • 5) 0.000 000 011 874 362 826 346 956 8 × 2 = 0 + 0.000 000 023 748 725 652 693 913 6;
  • 6) 0.000 000 023 748 725 652 693 913 6 × 2 = 0 + 0.000 000 047 497 451 305 387 827 2;
  • 7) 0.000 000 047 497 451 305 387 827 2 × 2 = 0 + 0.000 000 094 994 902 610 775 654 4;
  • 8) 0.000 000 094 994 902 610 775 654 4 × 2 = 0 + 0.000 000 189 989 805 221 551 308 8;
  • 9) 0.000 000 189 989 805 221 551 308 8 × 2 = 0 + 0.000 000 379 979 610 443 102 617 6;
  • 10) 0.000 000 379 979 610 443 102 617 6 × 2 = 0 + 0.000 000 759 959 220 886 205 235 2;
  • 11) 0.000 000 759 959 220 886 205 235 2 × 2 = 0 + 0.000 001 519 918 441 772 410 470 4;
  • 12) 0.000 001 519 918 441 772 410 470 4 × 2 = 0 + 0.000 003 039 836 883 544 820 940 8;
  • 13) 0.000 003 039 836 883 544 820 940 8 × 2 = 0 + 0.000 006 079 673 767 089 641 881 6;
  • 14) 0.000 006 079 673 767 089 641 881 6 × 2 = 0 + 0.000 012 159 347 534 179 283 763 2;
  • 15) 0.000 012 159 347 534 179 283 763 2 × 2 = 0 + 0.000 024 318 695 068 358 567 526 4;
  • 16) 0.000 024 318 695 068 358 567 526 4 × 2 = 0 + 0.000 048 637 390 136 717 135 052 8;
  • 17) 0.000 048 637 390 136 717 135 052 8 × 2 = 0 + 0.000 097 274 780 273 434 270 105 6;
  • 18) 0.000 097 274 780 273 434 270 105 6 × 2 = 0 + 0.000 194 549 560 546 868 540 211 2;
  • 19) 0.000 194 549 560 546 868 540 211 2 × 2 = 0 + 0.000 389 099 121 093 737 080 422 4;
  • 20) 0.000 389 099 121 093 737 080 422 4 × 2 = 0 + 0.000 778 198 242 187 474 160 844 8;
  • 21) 0.000 778 198 242 187 474 160 844 8 × 2 = 0 + 0.001 556 396 484 374 948 321 689 6;
  • 22) 0.001 556 396 484 374 948 321 689 6 × 2 = 0 + 0.003 112 792 968 749 896 643 379 2;
  • 23) 0.003 112 792 968 749 896 643 379 2 × 2 = 0 + 0.006 225 585 937 499 793 286 758 4;
  • 24) 0.006 225 585 937 499 793 286 758 4 × 2 = 0 + 0.012 451 171 874 999 586 573 516 8;
  • 25) 0.012 451 171 874 999 586 573 516 8 × 2 = 0 + 0.024 902 343 749 999 173 147 033 6;
  • 26) 0.024 902 343 749 999 173 147 033 6 × 2 = 0 + 0.049 804 687 499 998 346 294 067 2;
  • 27) 0.049 804 687 499 998 346 294 067 2 × 2 = 0 + 0.099 609 374 999 996 692 588 134 4;
  • 28) 0.099 609 374 999 996 692 588 134 4 × 2 = 0 + 0.199 218 749 999 993 385 176 268 8;
  • 29) 0.199 218 749 999 993 385 176 268 8 × 2 = 0 + 0.398 437 499 999 986 770 352 537 6;
  • 30) 0.398 437 499 999 986 770 352 537 6 × 2 = 0 + 0.796 874 999 999 973 540 705 075 2;
  • 31) 0.796 874 999 999 973 540 705 075 2 × 2 = 1 + 0.593 749 999 999 947 081 410 150 4;
  • 32) 0.593 749 999 999 947 081 410 150 4 × 2 = 1 + 0.187 499 999 999 894 162 820 300 8;
  • 33) 0.187 499 999 999 894 162 820 300 8 × 2 = 0 + 0.374 999 999 999 788 325 640 601 6;
  • 34) 0.374 999 999 999 788 325 640 601 6 × 2 = 0 + 0.749 999 999 999 576 651 281 203 2;
  • 35) 0.749 999 999 999 576 651 281 203 2 × 2 = 1 + 0.499 999 999 999 153 302 562 406 4;
  • 36) 0.499 999 999 999 153 302 562 406 4 × 2 = 0 + 0.999 999 999 998 306 605 124 812 8;
  • 37) 0.999 999 999 998 306 605 124 812 8 × 2 = 1 + 0.999 999 999 996 613 210 249 625 6;
  • 38) 0.999 999 999 996 613 210 249 625 6 × 2 = 1 + 0.999 999 999 993 226 420 499 251 2;
  • 39) 0.999 999 999 993 226 420 499 251 2 × 2 = 1 + 0.999 999 999 986 452 840 998 502 4;
  • 40) 0.999 999 999 986 452 840 998 502 4 × 2 = 1 + 0.999 999 999 972 905 681 997 004 8;
  • 41) 0.999 999 999 972 905 681 997 004 8 × 2 = 1 + 0.999 999 999 945 811 363 994 009 6;
  • 42) 0.999 999 999 945 811 363 994 009 6 × 2 = 1 + 0.999 999 999 891 622 727 988 019 2;
  • 43) 0.999 999 999 891 622 727 988 019 2 × 2 = 1 + 0.999 999 999 783 245 455 976 038 4;
  • 44) 0.999 999 999 783 245 455 976 038 4 × 2 = 1 + 0.999 999 999 566 490 911 952 076 8;
  • 45) 0.999 999 999 566 490 911 952 076 8 × 2 = 1 + 0.999 999 999 132 981 823 904 153 6;
  • 46) 0.999 999 999 132 981 823 904 153 6 × 2 = 1 + 0.999 999 998 265 963 647 808 307 2;
  • 47) 0.999 999 998 265 963 647 808 307 2 × 2 = 1 + 0.999 999 996 531 927 295 616 614 4;
  • 48) 0.999 999 996 531 927 295 616 614 4 × 2 = 1 + 0.999 999 993 063 854 591 233 228 8;
  • 49) 0.999 999 993 063 854 591 233 228 8 × 2 = 1 + 0.999 999 986 127 709 182 466 457 6;
  • 50) 0.999 999 986 127 709 182 466 457 6 × 2 = 1 + 0.999 999 972 255 418 364 932 915 2;
  • 51) 0.999 999 972 255 418 364 932 915 2 × 2 = 1 + 0.999 999 944 510 836 729 865 830 4;
  • 52) 0.999 999 944 510 836 729 865 830 4 × 2 = 1 + 0.999 999 889 021 673 459 731 660 8;
  • 53) 0.999 999 889 021 673 459 731 660 8 × 2 = 1 + 0.999 999 778 043 346 919 463 321 6;
  • 54) 0.999 999 778 043 346 919 463 321 6 × 2 = 1 + 0.999 999 556 086 693 838 926 643 2;

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:


0.000 000 000 742 147 676 646 684 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 684 8(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:


0.000 000 000 742 147 676 646 684 8(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 -0.000 000 000 742 147 676 646 684 8 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