-0.000 000 000 742 147 676 803 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 803(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
-0.000 000 000 742 147 676 803(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:

|-0.000 000 000 742 147 676 803| = 0.000 000 000 742 147 676 803


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: 0.000 000 000 742 147 676 803.

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) 0.000 000 000 742 147 676 803 × 2 = 0 + 0.000 000 001 484 295 353 606;
  • 2) 0.000 000 001 484 295 353 606 × 2 = 0 + 0.000 000 002 968 590 707 212;
  • 3) 0.000 000 002 968 590 707 212 × 2 = 0 + 0.000 000 005 937 181 414 424;
  • 4) 0.000 000 005 937 181 414 424 × 2 = 0 + 0.000 000 011 874 362 828 848;
  • 5) 0.000 000 011 874 362 828 848 × 2 = 0 + 0.000 000 023 748 725 657 696;
  • 6) 0.000 000 023 748 725 657 696 × 2 = 0 + 0.000 000 047 497 451 315 392;
  • 7) 0.000 000 047 497 451 315 392 × 2 = 0 + 0.000 000 094 994 902 630 784;
  • 8) 0.000 000 094 994 902 630 784 × 2 = 0 + 0.000 000 189 989 805 261 568;
  • 9) 0.000 000 189 989 805 261 568 × 2 = 0 + 0.000 000 379 979 610 523 136;
  • 10) 0.000 000 379 979 610 523 136 × 2 = 0 + 0.000 000 759 959 221 046 272;
  • 11) 0.000 000 759 959 221 046 272 × 2 = 0 + 0.000 001 519 918 442 092 544;
  • 12) 0.000 001 519 918 442 092 544 × 2 = 0 + 0.000 003 039 836 884 185 088;
  • 13) 0.000 003 039 836 884 185 088 × 2 = 0 + 0.000 006 079 673 768 370 176;
  • 14) 0.000 006 079 673 768 370 176 × 2 = 0 + 0.000 012 159 347 536 740 352;
  • 15) 0.000 012 159 347 536 740 352 × 2 = 0 + 0.000 024 318 695 073 480 704;
  • 16) 0.000 024 318 695 073 480 704 × 2 = 0 + 0.000 048 637 390 146 961 408;
  • 17) 0.000 048 637 390 146 961 408 × 2 = 0 + 0.000 097 274 780 293 922 816;
  • 18) 0.000 097 274 780 293 922 816 × 2 = 0 + 0.000 194 549 560 587 845 632;
  • 19) 0.000 194 549 560 587 845 632 × 2 = 0 + 0.000 389 099 121 175 691 264;
  • 20) 0.000 389 099 121 175 691 264 × 2 = 0 + 0.000 778 198 242 351 382 528;
  • 21) 0.000 778 198 242 351 382 528 × 2 = 0 + 0.001 556 396 484 702 765 056;
  • 22) 0.001 556 396 484 702 765 056 × 2 = 0 + 0.003 112 792 969 405 530 112;
  • 23) 0.003 112 792 969 405 530 112 × 2 = 0 + 0.006 225 585 938 811 060 224;
  • 24) 0.006 225 585 938 811 060 224 × 2 = 0 + 0.012 451 171 877 622 120 448;
  • 25) 0.012 451 171 877 622 120 448 × 2 = 0 + 0.024 902 343 755 244 240 896;
  • 26) 0.024 902 343 755 244 240 896 × 2 = 0 + 0.049 804 687 510 488 481 792;
  • 27) 0.049 804 687 510 488 481 792 × 2 = 0 + 0.099 609 375 020 976 963 584;
  • 28) 0.099 609 375 020 976 963 584 × 2 = 0 + 0.199 218 750 041 953 927 168;
  • 29) 0.199 218 750 041 953 927 168 × 2 = 0 + 0.398 437 500 083 907 854 336;
  • 30) 0.398 437 500 083 907 854 336 × 2 = 0 + 0.796 875 000 167 815 708 672;
  • 31) 0.796 875 000 167 815 708 672 × 2 = 1 + 0.593 750 000 335 631 417 344;
  • 32) 0.593 750 000 335 631 417 344 × 2 = 1 + 0.187 500 000 671 262 834 688;
  • 33) 0.187 500 000 671 262 834 688 × 2 = 0 + 0.375 000 001 342 525 669 376;
  • 34) 0.375 000 001 342 525 669 376 × 2 = 0 + 0.750 000 002 685 051 338 752;
  • 35) 0.750 000 002 685 051 338 752 × 2 = 1 + 0.500 000 005 370 102 677 504;
  • 36) 0.500 000 005 370 102 677 504 × 2 = 1 + 0.000 000 010 740 205 355 008;
  • 37) 0.000 000 010 740 205 355 008 × 2 = 0 + 0.000 000 021 480 410 710 016;
  • 38) 0.000 000 021 480 410 710 016 × 2 = 0 + 0.000 000 042 960 821 420 032;
  • 39) 0.000 000 042 960 821 420 032 × 2 = 0 + 0.000 000 085 921 642 840 064;
  • 40) 0.000 000 085 921 642 840 064 × 2 = 0 + 0.000 000 171 843 285 680 128;
  • 41) 0.000 000 171 843 285 680 128 × 2 = 0 + 0.000 000 343 686 571 360 256;
  • 42) 0.000 000 343 686 571 360 256 × 2 = 0 + 0.000 000 687 373 142 720 512;
  • 43) 0.000 000 687 373 142 720 512 × 2 = 0 + 0.000 001 374 746 285 441 024;
  • 44) 0.000 001 374 746 285 441 024 × 2 = 0 + 0.000 002 749 492 570 882 048;
  • 45) 0.000 002 749 492 570 882 048 × 2 = 0 + 0.000 005 498 985 141 764 096;
  • 46) 0.000 005 498 985 141 764 096 × 2 = 0 + 0.000 010 997 970 283 528 192;
  • 47) 0.000 010 997 970 283 528 192 × 2 = 0 + 0.000 021 995 940 567 056 384;
  • 48) 0.000 021 995 940 567 056 384 × 2 = 0 + 0.000 043 991 881 134 112 768;
  • 49) 0.000 043 991 881 134 112 768 × 2 = 0 + 0.000 087 983 762 268 225 536;
  • 50) 0.000 087 983 762 268 225 536 × 2 = 0 + 0.000 175 967 524 536 451 072;
  • 51) 0.000 175 967 524 536 451 072 × 2 = 0 + 0.000 351 935 049 072 902 144;
  • 52) 0.000 351 935 049 072 902 144 × 2 = 0 + 0.000 703 870 098 145 804 288;
  • 53) 0.000 703 870 098 145 804 288 × 2 = 0 + 0.001 407 740 196 291 608 576;
  • 54) 0.001 407 740 196 291 608 576 × 2 = 0 + 0.002 815 480 392 583 217 152;

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:


0.000 000 000 742 147 676 803(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 803(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 803(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 803 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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