-0.000 000 000 742 147 676 648 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 648(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
-0.000 000 000 742 147 676 648(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:

|-0.000 000 000 742 147 676 648| = 0.000 000 000 742 147 676 648


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: 0.000 000 000 742 147 676 648.

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) 0.000 000 000 742 147 676 648 × 2 = 0 + 0.000 000 001 484 295 353 296;
  • 2) 0.000 000 001 484 295 353 296 × 2 = 0 + 0.000 000 002 968 590 706 592;
  • 3) 0.000 000 002 968 590 706 592 × 2 = 0 + 0.000 000 005 937 181 413 184;
  • 4) 0.000 000 005 937 181 413 184 × 2 = 0 + 0.000 000 011 874 362 826 368;
  • 5) 0.000 000 011 874 362 826 368 × 2 = 0 + 0.000 000 023 748 725 652 736;
  • 6) 0.000 000 023 748 725 652 736 × 2 = 0 + 0.000 000 047 497 451 305 472;
  • 7) 0.000 000 047 497 451 305 472 × 2 = 0 + 0.000 000 094 994 902 610 944;
  • 8) 0.000 000 094 994 902 610 944 × 2 = 0 + 0.000 000 189 989 805 221 888;
  • 9) 0.000 000 189 989 805 221 888 × 2 = 0 + 0.000 000 379 979 610 443 776;
  • 10) 0.000 000 379 979 610 443 776 × 2 = 0 + 0.000 000 759 959 220 887 552;
  • 11) 0.000 000 759 959 220 887 552 × 2 = 0 + 0.000 001 519 918 441 775 104;
  • 12) 0.000 001 519 918 441 775 104 × 2 = 0 + 0.000 003 039 836 883 550 208;
  • 13) 0.000 003 039 836 883 550 208 × 2 = 0 + 0.000 006 079 673 767 100 416;
  • 14) 0.000 006 079 673 767 100 416 × 2 = 0 + 0.000 012 159 347 534 200 832;
  • 15) 0.000 012 159 347 534 200 832 × 2 = 0 + 0.000 024 318 695 068 401 664;
  • 16) 0.000 024 318 695 068 401 664 × 2 = 0 + 0.000 048 637 390 136 803 328;
  • 17) 0.000 048 637 390 136 803 328 × 2 = 0 + 0.000 097 274 780 273 606 656;
  • 18) 0.000 097 274 780 273 606 656 × 2 = 0 + 0.000 194 549 560 547 213 312;
  • 19) 0.000 194 549 560 547 213 312 × 2 = 0 + 0.000 389 099 121 094 426 624;
  • 20) 0.000 389 099 121 094 426 624 × 2 = 0 + 0.000 778 198 242 188 853 248;
  • 21) 0.000 778 198 242 188 853 248 × 2 = 0 + 0.001 556 396 484 377 706 496;
  • 22) 0.001 556 396 484 377 706 496 × 2 = 0 + 0.003 112 792 968 755 412 992;
  • 23) 0.003 112 792 968 755 412 992 × 2 = 0 + 0.006 225 585 937 510 825 984;
  • 24) 0.006 225 585 937 510 825 984 × 2 = 0 + 0.012 451 171 875 021 651 968;
  • 25) 0.012 451 171 875 021 651 968 × 2 = 0 + 0.024 902 343 750 043 303 936;
  • 26) 0.024 902 343 750 043 303 936 × 2 = 0 + 0.049 804 687 500 086 607 872;
  • 27) 0.049 804 687 500 086 607 872 × 2 = 0 + 0.099 609 375 000 173 215 744;
  • 28) 0.099 609 375 000 173 215 744 × 2 = 0 + 0.199 218 750 000 346 431 488;
  • 29) 0.199 218 750 000 346 431 488 × 2 = 0 + 0.398 437 500 000 692 862 976;
  • 30) 0.398 437 500 000 692 862 976 × 2 = 0 + 0.796 875 000 001 385 725 952;
  • 31) 0.796 875 000 001 385 725 952 × 2 = 1 + 0.593 750 000 002 771 451 904;
  • 32) 0.593 750 000 002 771 451 904 × 2 = 1 + 0.187 500 000 005 542 903 808;
  • 33) 0.187 500 000 005 542 903 808 × 2 = 0 + 0.375 000 000 011 085 807 616;
  • 34) 0.375 000 000 011 085 807 616 × 2 = 0 + 0.750 000 000 022 171 615 232;
  • 35) 0.750 000 000 022 171 615 232 × 2 = 1 + 0.500 000 000 044 343 230 464;
  • 36) 0.500 000 000 044 343 230 464 × 2 = 1 + 0.000 000 000 088 686 460 928;
  • 37) 0.000 000 000 088 686 460 928 × 2 = 0 + 0.000 000 000 177 372 921 856;
  • 38) 0.000 000 000 177 372 921 856 × 2 = 0 + 0.000 000 000 354 745 843 712;
  • 39) 0.000 000 000 354 745 843 712 × 2 = 0 + 0.000 000 000 709 491 687 424;
  • 40) 0.000 000 000 709 491 687 424 × 2 = 0 + 0.000 000 001 418 983 374 848;
  • 41) 0.000 000 001 418 983 374 848 × 2 = 0 + 0.000 000 002 837 966 749 696;
  • 42) 0.000 000 002 837 966 749 696 × 2 = 0 + 0.000 000 005 675 933 499 392;
  • 43) 0.000 000 005 675 933 499 392 × 2 = 0 + 0.000 000 011 351 866 998 784;
  • 44) 0.000 000 011 351 866 998 784 × 2 = 0 + 0.000 000 022 703 733 997 568;
  • 45) 0.000 000 022 703 733 997 568 × 2 = 0 + 0.000 000 045 407 467 995 136;
  • 46) 0.000 000 045 407 467 995 136 × 2 = 0 + 0.000 000 090 814 935 990 272;
  • 47) 0.000 000 090 814 935 990 272 × 2 = 0 + 0.000 000 181 629 871 980 544;
  • 48) 0.000 000 181 629 871 980 544 × 2 = 0 + 0.000 000 363 259 743 961 088;
  • 49) 0.000 000 363 259 743 961 088 × 2 = 0 + 0.000 000 726 519 487 922 176;
  • 50) 0.000 000 726 519 487 922 176 × 2 = 0 + 0.000 001 453 038 975 844 352;
  • 51) 0.000 001 453 038 975 844 352 × 2 = 0 + 0.000 002 906 077 951 688 704;
  • 52) 0.000 002 906 077 951 688 704 × 2 = 0 + 0.000 005 812 155 903 377 408;
  • 53) 0.000 005 812 155 903 377 408 × 2 = 0 + 0.000 011 624 311 806 754 816;
  • 54) 0.000 011 624 311 806 754 816 × 2 = 0 + 0.000 023 248 623 613 509 632;

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:


0.000 000 000 742 147 676 648(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 648(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:


0.000 000 000 742 147 676 648(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 -0.000 000 000 742 147 676 648 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