-0.000 000 000 742 147 676 202 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 202(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
-0.000 000 000 742 147 676 202(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:

|-0.000 000 000 742 147 676 202| = 0.000 000 000 742 147 676 202


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: 0.000 000 000 742 147 676 202.

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) 0.000 000 000 742 147 676 202 × 2 = 0 + 0.000 000 001 484 295 352 404;
  • 2) 0.000 000 001 484 295 352 404 × 2 = 0 + 0.000 000 002 968 590 704 808;
  • 3) 0.000 000 002 968 590 704 808 × 2 = 0 + 0.000 000 005 937 181 409 616;
  • 4) 0.000 000 005 937 181 409 616 × 2 = 0 + 0.000 000 011 874 362 819 232;
  • 5) 0.000 000 011 874 362 819 232 × 2 = 0 + 0.000 000 023 748 725 638 464;
  • 6) 0.000 000 023 748 725 638 464 × 2 = 0 + 0.000 000 047 497 451 276 928;
  • 7) 0.000 000 047 497 451 276 928 × 2 = 0 + 0.000 000 094 994 902 553 856;
  • 8) 0.000 000 094 994 902 553 856 × 2 = 0 + 0.000 000 189 989 805 107 712;
  • 9) 0.000 000 189 989 805 107 712 × 2 = 0 + 0.000 000 379 979 610 215 424;
  • 10) 0.000 000 379 979 610 215 424 × 2 = 0 + 0.000 000 759 959 220 430 848;
  • 11) 0.000 000 759 959 220 430 848 × 2 = 0 + 0.000 001 519 918 440 861 696;
  • 12) 0.000 001 519 918 440 861 696 × 2 = 0 + 0.000 003 039 836 881 723 392;
  • 13) 0.000 003 039 836 881 723 392 × 2 = 0 + 0.000 006 079 673 763 446 784;
  • 14) 0.000 006 079 673 763 446 784 × 2 = 0 + 0.000 012 159 347 526 893 568;
  • 15) 0.000 012 159 347 526 893 568 × 2 = 0 + 0.000 024 318 695 053 787 136;
  • 16) 0.000 024 318 695 053 787 136 × 2 = 0 + 0.000 048 637 390 107 574 272;
  • 17) 0.000 048 637 390 107 574 272 × 2 = 0 + 0.000 097 274 780 215 148 544;
  • 18) 0.000 097 274 780 215 148 544 × 2 = 0 + 0.000 194 549 560 430 297 088;
  • 19) 0.000 194 549 560 430 297 088 × 2 = 0 + 0.000 389 099 120 860 594 176;
  • 20) 0.000 389 099 120 860 594 176 × 2 = 0 + 0.000 778 198 241 721 188 352;
  • 21) 0.000 778 198 241 721 188 352 × 2 = 0 + 0.001 556 396 483 442 376 704;
  • 22) 0.001 556 396 483 442 376 704 × 2 = 0 + 0.003 112 792 966 884 753 408;
  • 23) 0.003 112 792 966 884 753 408 × 2 = 0 + 0.006 225 585 933 769 506 816;
  • 24) 0.006 225 585 933 769 506 816 × 2 = 0 + 0.012 451 171 867 539 013 632;
  • 25) 0.012 451 171 867 539 013 632 × 2 = 0 + 0.024 902 343 735 078 027 264;
  • 26) 0.024 902 343 735 078 027 264 × 2 = 0 + 0.049 804 687 470 156 054 528;
  • 27) 0.049 804 687 470 156 054 528 × 2 = 0 + 0.099 609 374 940 312 109 056;
  • 28) 0.099 609 374 940 312 109 056 × 2 = 0 + 0.199 218 749 880 624 218 112;
  • 29) 0.199 218 749 880 624 218 112 × 2 = 0 + 0.398 437 499 761 248 436 224;
  • 30) 0.398 437 499 761 248 436 224 × 2 = 0 + 0.796 874 999 522 496 872 448;
  • 31) 0.796 874 999 522 496 872 448 × 2 = 1 + 0.593 749 999 044 993 744 896;
  • 32) 0.593 749 999 044 993 744 896 × 2 = 1 + 0.187 499 998 089 987 489 792;
  • 33) 0.187 499 998 089 987 489 792 × 2 = 0 + 0.374 999 996 179 974 979 584;
  • 34) 0.374 999 996 179 974 979 584 × 2 = 0 + 0.749 999 992 359 949 959 168;
  • 35) 0.749 999 992 359 949 959 168 × 2 = 1 + 0.499 999 984 719 899 918 336;
  • 36) 0.499 999 984 719 899 918 336 × 2 = 0 + 0.999 999 969 439 799 836 672;
  • 37) 0.999 999 969 439 799 836 672 × 2 = 1 + 0.999 999 938 879 599 673 344;
  • 38) 0.999 999 938 879 599 673 344 × 2 = 1 + 0.999 999 877 759 199 346 688;
  • 39) 0.999 999 877 759 199 346 688 × 2 = 1 + 0.999 999 755 518 398 693 376;
  • 40) 0.999 999 755 518 398 693 376 × 2 = 1 + 0.999 999 511 036 797 386 752;
  • 41) 0.999 999 511 036 797 386 752 × 2 = 1 + 0.999 999 022 073 594 773 504;
  • 42) 0.999 999 022 073 594 773 504 × 2 = 1 + 0.999 998 044 147 189 547 008;
  • 43) 0.999 998 044 147 189 547 008 × 2 = 1 + 0.999 996 088 294 379 094 016;
  • 44) 0.999 996 088 294 379 094 016 × 2 = 1 + 0.999 992 176 588 758 188 032;
  • 45) 0.999 992 176 588 758 188 032 × 2 = 1 + 0.999 984 353 177 516 376 064;
  • 46) 0.999 984 353 177 516 376 064 × 2 = 1 + 0.999 968 706 355 032 752 128;
  • 47) 0.999 968 706 355 032 752 128 × 2 = 1 + 0.999 937 412 710 065 504 256;
  • 48) 0.999 937 412 710 065 504 256 × 2 = 1 + 0.999 874 825 420 131 008 512;
  • 49) 0.999 874 825 420 131 008 512 × 2 = 1 + 0.999 749 650 840 262 017 024;
  • 50) 0.999 749 650 840 262 017 024 × 2 = 1 + 0.999 499 301 680 524 034 048;
  • 51) 0.999 499 301 680 524 034 048 × 2 = 1 + 0.998 998 603 361 048 068 096;
  • 52) 0.998 998 603 361 048 068 096 × 2 = 1 + 0.997 997 206 722 096 136 192;
  • 53) 0.997 997 206 722 096 136 192 × 2 = 1 + 0.995 994 413 444 192 272 384;
  • 54) 0.995 994 413 444 192 272 384 × 2 = 1 + 0.991 988 826 888 384 544 768;

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:


0.000 000 000 742 147 676 202(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 202(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:


0.000 000 000 742 147 676 202(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 -0.000 000 000 742 147 676 202 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