-0.000 000 000 742 147 676 159 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 159(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
-0.000 000 000 742 147 676 159(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:

|-0.000 000 000 742 147 676 159| = 0.000 000 000 742 147 676 159


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: 0.000 000 000 742 147 676 159.

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) 0.000 000 000 742 147 676 159 × 2 = 0 + 0.000 000 001 484 295 352 318;
  • 2) 0.000 000 001 484 295 352 318 × 2 = 0 + 0.000 000 002 968 590 704 636;
  • 3) 0.000 000 002 968 590 704 636 × 2 = 0 + 0.000 000 005 937 181 409 272;
  • 4) 0.000 000 005 937 181 409 272 × 2 = 0 + 0.000 000 011 874 362 818 544;
  • 5) 0.000 000 011 874 362 818 544 × 2 = 0 + 0.000 000 023 748 725 637 088;
  • 6) 0.000 000 023 748 725 637 088 × 2 = 0 + 0.000 000 047 497 451 274 176;
  • 7) 0.000 000 047 497 451 274 176 × 2 = 0 + 0.000 000 094 994 902 548 352;
  • 8) 0.000 000 094 994 902 548 352 × 2 = 0 + 0.000 000 189 989 805 096 704;
  • 9) 0.000 000 189 989 805 096 704 × 2 = 0 + 0.000 000 379 979 610 193 408;
  • 10) 0.000 000 379 979 610 193 408 × 2 = 0 + 0.000 000 759 959 220 386 816;
  • 11) 0.000 000 759 959 220 386 816 × 2 = 0 + 0.000 001 519 918 440 773 632;
  • 12) 0.000 001 519 918 440 773 632 × 2 = 0 + 0.000 003 039 836 881 547 264;
  • 13) 0.000 003 039 836 881 547 264 × 2 = 0 + 0.000 006 079 673 763 094 528;
  • 14) 0.000 006 079 673 763 094 528 × 2 = 0 + 0.000 012 159 347 526 189 056;
  • 15) 0.000 012 159 347 526 189 056 × 2 = 0 + 0.000 024 318 695 052 378 112;
  • 16) 0.000 024 318 695 052 378 112 × 2 = 0 + 0.000 048 637 390 104 756 224;
  • 17) 0.000 048 637 390 104 756 224 × 2 = 0 + 0.000 097 274 780 209 512 448;
  • 18) 0.000 097 274 780 209 512 448 × 2 = 0 + 0.000 194 549 560 419 024 896;
  • 19) 0.000 194 549 560 419 024 896 × 2 = 0 + 0.000 389 099 120 838 049 792;
  • 20) 0.000 389 099 120 838 049 792 × 2 = 0 + 0.000 778 198 241 676 099 584;
  • 21) 0.000 778 198 241 676 099 584 × 2 = 0 + 0.001 556 396 483 352 199 168;
  • 22) 0.001 556 396 483 352 199 168 × 2 = 0 + 0.003 112 792 966 704 398 336;
  • 23) 0.003 112 792 966 704 398 336 × 2 = 0 + 0.006 225 585 933 408 796 672;
  • 24) 0.006 225 585 933 408 796 672 × 2 = 0 + 0.012 451 171 866 817 593 344;
  • 25) 0.012 451 171 866 817 593 344 × 2 = 0 + 0.024 902 343 733 635 186 688;
  • 26) 0.024 902 343 733 635 186 688 × 2 = 0 + 0.049 804 687 467 270 373 376;
  • 27) 0.049 804 687 467 270 373 376 × 2 = 0 + 0.099 609 374 934 540 746 752;
  • 28) 0.099 609 374 934 540 746 752 × 2 = 0 + 0.199 218 749 869 081 493 504;
  • 29) 0.199 218 749 869 081 493 504 × 2 = 0 + 0.398 437 499 738 162 987 008;
  • 30) 0.398 437 499 738 162 987 008 × 2 = 0 + 0.796 874 999 476 325 974 016;
  • 31) 0.796 874 999 476 325 974 016 × 2 = 1 + 0.593 749 998 952 651 948 032;
  • 32) 0.593 749 998 952 651 948 032 × 2 = 1 + 0.187 499 997 905 303 896 064;
  • 33) 0.187 499 997 905 303 896 064 × 2 = 0 + 0.374 999 995 810 607 792 128;
  • 34) 0.374 999 995 810 607 792 128 × 2 = 0 + 0.749 999 991 621 215 584 256;
  • 35) 0.749 999 991 621 215 584 256 × 2 = 1 + 0.499 999 983 242 431 168 512;
  • 36) 0.499 999 983 242 431 168 512 × 2 = 0 + 0.999 999 966 484 862 337 024;
  • 37) 0.999 999 966 484 862 337 024 × 2 = 1 + 0.999 999 932 969 724 674 048;
  • 38) 0.999 999 932 969 724 674 048 × 2 = 1 + 0.999 999 865 939 449 348 096;
  • 39) 0.999 999 865 939 449 348 096 × 2 = 1 + 0.999 999 731 878 898 696 192;
  • 40) 0.999 999 731 878 898 696 192 × 2 = 1 + 0.999 999 463 757 797 392 384;
  • 41) 0.999 999 463 757 797 392 384 × 2 = 1 + 0.999 998 927 515 594 784 768;
  • 42) 0.999 998 927 515 594 784 768 × 2 = 1 + 0.999 997 855 031 189 569 536;
  • 43) 0.999 997 855 031 189 569 536 × 2 = 1 + 0.999 995 710 062 379 139 072;
  • 44) 0.999 995 710 062 379 139 072 × 2 = 1 + 0.999 991 420 124 758 278 144;
  • 45) 0.999 991 420 124 758 278 144 × 2 = 1 + 0.999 982 840 249 516 556 288;
  • 46) 0.999 982 840 249 516 556 288 × 2 = 1 + 0.999 965 680 499 033 112 576;
  • 47) 0.999 965 680 499 033 112 576 × 2 = 1 + 0.999 931 360 998 066 225 152;
  • 48) 0.999 931 360 998 066 225 152 × 2 = 1 + 0.999 862 721 996 132 450 304;
  • 49) 0.999 862 721 996 132 450 304 × 2 = 1 + 0.999 725 443 992 264 900 608;
  • 50) 0.999 725 443 992 264 900 608 × 2 = 1 + 0.999 450 887 984 529 801 216;
  • 51) 0.999 450 887 984 529 801 216 × 2 = 1 + 0.998 901 775 969 059 602 432;
  • 52) 0.998 901 775 969 059 602 432 × 2 = 1 + 0.997 803 551 938 119 204 864;
  • 53) 0.997 803 551 938 119 204 864 × 2 = 1 + 0.995 607 103 876 238 409 728;
  • 54) 0.995 607 103 876 238 409 728 × 2 = 1 + 0.991 214 207 752 476 819 456;

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:


0.000 000 000 742 147 676 159(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 159(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:


0.000 000 000 742 147 676 159(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 -0.000 000 000 742 147 676 159 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