-0.000 000 000 742 147 676 181 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 181(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
-0.000 000 000 742 147 676 181(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:

|-0.000 000 000 742 147 676 181| = 0.000 000 000 742 147 676 181


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: 0.000 000 000 742 147 676 181.

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) 0.000 000 000 742 147 676 181 × 2 = 0 + 0.000 000 001 484 295 352 362;
  • 2) 0.000 000 001 484 295 352 362 × 2 = 0 + 0.000 000 002 968 590 704 724;
  • 3) 0.000 000 002 968 590 704 724 × 2 = 0 + 0.000 000 005 937 181 409 448;
  • 4) 0.000 000 005 937 181 409 448 × 2 = 0 + 0.000 000 011 874 362 818 896;
  • 5) 0.000 000 011 874 362 818 896 × 2 = 0 + 0.000 000 023 748 725 637 792;
  • 6) 0.000 000 023 748 725 637 792 × 2 = 0 + 0.000 000 047 497 451 275 584;
  • 7) 0.000 000 047 497 451 275 584 × 2 = 0 + 0.000 000 094 994 902 551 168;
  • 8) 0.000 000 094 994 902 551 168 × 2 = 0 + 0.000 000 189 989 805 102 336;
  • 9) 0.000 000 189 989 805 102 336 × 2 = 0 + 0.000 000 379 979 610 204 672;
  • 10) 0.000 000 379 979 610 204 672 × 2 = 0 + 0.000 000 759 959 220 409 344;
  • 11) 0.000 000 759 959 220 409 344 × 2 = 0 + 0.000 001 519 918 440 818 688;
  • 12) 0.000 001 519 918 440 818 688 × 2 = 0 + 0.000 003 039 836 881 637 376;
  • 13) 0.000 003 039 836 881 637 376 × 2 = 0 + 0.000 006 079 673 763 274 752;
  • 14) 0.000 006 079 673 763 274 752 × 2 = 0 + 0.000 012 159 347 526 549 504;
  • 15) 0.000 012 159 347 526 549 504 × 2 = 0 + 0.000 024 318 695 053 099 008;
  • 16) 0.000 024 318 695 053 099 008 × 2 = 0 + 0.000 048 637 390 106 198 016;
  • 17) 0.000 048 637 390 106 198 016 × 2 = 0 + 0.000 097 274 780 212 396 032;
  • 18) 0.000 097 274 780 212 396 032 × 2 = 0 + 0.000 194 549 560 424 792 064;
  • 19) 0.000 194 549 560 424 792 064 × 2 = 0 + 0.000 389 099 120 849 584 128;
  • 20) 0.000 389 099 120 849 584 128 × 2 = 0 + 0.000 778 198 241 699 168 256;
  • 21) 0.000 778 198 241 699 168 256 × 2 = 0 + 0.001 556 396 483 398 336 512;
  • 22) 0.001 556 396 483 398 336 512 × 2 = 0 + 0.003 112 792 966 796 673 024;
  • 23) 0.003 112 792 966 796 673 024 × 2 = 0 + 0.006 225 585 933 593 346 048;
  • 24) 0.006 225 585 933 593 346 048 × 2 = 0 + 0.012 451 171 867 186 692 096;
  • 25) 0.012 451 171 867 186 692 096 × 2 = 0 + 0.024 902 343 734 373 384 192;
  • 26) 0.024 902 343 734 373 384 192 × 2 = 0 + 0.049 804 687 468 746 768 384;
  • 27) 0.049 804 687 468 746 768 384 × 2 = 0 + 0.099 609 374 937 493 536 768;
  • 28) 0.099 609 374 937 493 536 768 × 2 = 0 + 0.199 218 749 874 987 073 536;
  • 29) 0.199 218 749 874 987 073 536 × 2 = 0 + 0.398 437 499 749 974 147 072;
  • 30) 0.398 437 499 749 974 147 072 × 2 = 0 + 0.796 874 999 499 948 294 144;
  • 31) 0.796 874 999 499 948 294 144 × 2 = 1 + 0.593 749 998 999 896 588 288;
  • 32) 0.593 749 998 999 896 588 288 × 2 = 1 + 0.187 499 997 999 793 176 576;
  • 33) 0.187 499 997 999 793 176 576 × 2 = 0 + 0.374 999 995 999 586 353 152;
  • 34) 0.374 999 995 999 586 353 152 × 2 = 0 + 0.749 999 991 999 172 706 304;
  • 35) 0.749 999 991 999 172 706 304 × 2 = 1 + 0.499 999 983 998 345 412 608;
  • 36) 0.499 999 983 998 345 412 608 × 2 = 0 + 0.999 999 967 996 690 825 216;
  • 37) 0.999 999 967 996 690 825 216 × 2 = 1 + 0.999 999 935 993 381 650 432;
  • 38) 0.999 999 935 993 381 650 432 × 2 = 1 + 0.999 999 871 986 763 300 864;
  • 39) 0.999 999 871 986 763 300 864 × 2 = 1 + 0.999 999 743 973 526 601 728;
  • 40) 0.999 999 743 973 526 601 728 × 2 = 1 + 0.999 999 487 947 053 203 456;
  • 41) 0.999 999 487 947 053 203 456 × 2 = 1 + 0.999 998 975 894 106 406 912;
  • 42) 0.999 998 975 894 106 406 912 × 2 = 1 + 0.999 997 951 788 212 813 824;
  • 43) 0.999 997 951 788 212 813 824 × 2 = 1 + 0.999 995 903 576 425 627 648;
  • 44) 0.999 995 903 576 425 627 648 × 2 = 1 + 0.999 991 807 152 851 255 296;
  • 45) 0.999 991 807 152 851 255 296 × 2 = 1 + 0.999 983 614 305 702 510 592;
  • 46) 0.999 983 614 305 702 510 592 × 2 = 1 + 0.999 967 228 611 405 021 184;
  • 47) 0.999 967 228 611 405 021 184 × 2 = 1 + 0.999 934 457 222 810 042 368;
  • 48) 0.999 934 457 222 810 042 368 × 2 = 1 + 0.999 868 914 445 620 084 736;
  • 49) 0.999 868 914 445 620 084 736 × 2 = 1 + 0.999 737 828 891 240 169 472;
  • 50) 0.999 737 828 891 240 169 472 × 2 = 1 + 0.999 475 657 782 480 338 944;
  • 51) 0.999 475 657 782 480 338 944 × 2 = 1 + 0.998 951 315 564 960 677 888;
  • 52) 0.998 951 315 564 960 677 888 × 2 = 1 + 0.997 902 631 129 921 355 776;
  • 53) 0.997 902 631 129 921 355 776 × 2 = 1 + 0.995 805 262 259 842 711 552;
  • 54) 0.995 805 262 259 842 711 552 × 2 = 1 + 0.991 610 524 519 685 423 104;

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:


0.000 000 000 742 147 676 181(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 181(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:


0.000 000 000 742 147 676 181(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 -0.000 000 000 742 147 676 181 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