-0.000 000 000 742 147 676 646 733 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 733 3(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
-0.000 000 000 742 147 676 646 733 3(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:

|-0.000 000 000 742 147 676 646 733 3| = 0.000 000 000 742 147 676 646 733 3


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: 0.000 000 000 742 147 676 646 733 3.

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) 0.000 000 000 742 147 676 646 733 3 × 2 = 0 + 0.000 000 001 484 295 353 293 466 6;
  • 2) 0.000 000 001 484 295 353 293 466 6 × 2 = 0 + 0.000 000 002 968 590 706 586 933 2;
  • 3) 0.000 000 002 968 590 706 586 933 2 × 2 = 0 + 0.000 000 005 937 181 413 173 866 4;
  • 4) 0.000 000 005 937 181 413 173 866 4 × 2 = 0 + 0.000 000 011 874 362 826 347 732 8;
  • 5) 0.000 000 011 874 362 826 347 732 8 × 2 = 0 + 0.000 000 023 748 725 652 695 465 6;
  • 6) 0.000 000 023 748 725 652 695 465 6 × 2 = 0 + 0.000 000 047 497 451 305 390 931 2;
  • 7) 0.000 000 047 497 451 305 390 931 2 × 2 = 0 + 0.000 000 094 994 902 610 781 862 4;
  • 8) 0.000 000 094 994 902 610 781 862 4 × 2 = 0 + 0.000 000 189 989 805 221 563 724 8;
  • 9) 0.000 000 189 989 805 221 563 724 8 × 2 = 0 + 0.000 000 379 979 610 443 127 449 6;
  • 10) 0.000 000 379 979 610 443 127 449 6 × 2 = 0 + 0.000 000 759 959 220 886 254 899 2;
  • 11) 0.000 000 759 959 220 886 254 899 2 × 2 = 0 + 0.000 001 519 918 441 772 509 798 4;
  • 12) 0.000 001 519 918 441 772 509 798 4 × 2 = 0 + 0.000 003 039 836 883 545 019 596 8;
  • 13) 0.000 003 039 836 883 545 019 596 8 × 2 = 0 + 0.000 006 079 673 767 090 039 193 6;
  • 14) 0.000 006 079 673 767 090 039 193 6 × 2 = 0 + 0.000 012 159 347 534 180 078 387 2;
  • 15) 0.000 012 159 347 534 180 078 387 2 × 2 = 0 + 0.000 024 318 695 068 360 156 774 4;
  • 16) 0.000 024 318 695 068 360 156 774 4 × 2 = 0 + 0.000 048 637 390 136 720 313 548 8;
  • 17) 0.000 048 637 390 136 720 313 548 8 × 2 = 0 + 0.000 097 274 780 273 440 627 097 6;
  • 18) 0.000 097 274 780 273 440 627 097 6 × 2 = 0 + 0.000 194 549 560 546 881 254 195 2;
  • 19) 0.000 194 549 560 546 881 254 195 2 × 2 = 0 + 0.000 389 099 121 093 762 508 390 4;
  • 20) 0.000 389 099 121 093 762 508 390 4 × 2 = 0 + 0.000 778 198 242 187 525 016 780 8;
  • 21) 0.000 778 198 242 187 525 016 780 8 × 2 = 0 + 0.001 556 396 484 375 050 033 561 6;
  • 22) 0.001 556 396 484 375 050 033 561 6 × 2 = 0 + 0.003 112 792 968 750 100 067 123 2;
  • 23) 0.003 112 792 968 750 100 067 123 2 × 2 = 0 + 0.006 225 585 937 500 200 134 246 4;
  • 24) 0.006 225 585 937 500 200 134 246 4 × 2 = 0 + 0.012 451 171 875 000 400 268 492 8;
  • 25) 0.012 451 171 875 000 400 268 492 8 × 2 = 0 + 0.024 902 343 750 000 800 536 985 6;
  • 26) 0.024 902 343 750 000 800 536 985 6 × 2 = 0 + 0.049 804 687 500 001 601 073 971 2;
  • 27) 0.049 804 687 500 001 601 073 971 2 × 2 = 0 + 0.099 609 375 000 003 202 147 942 4;
  • 28) 0.099 609 375 000 003 202 147 942 4 × 2 = 0 + 0.199 218 750 000 006 404 295 884 8;
  • 29) 0.199 218 750 000 006 404 295 884 8 × 2 = 0 + 0.398 437 500 000 012 808 591 769 6;
  • 30) 0.398 437 500 000 012 808 591 769 6 × 2 = 0 + 0.796 875 000 000 025 617 183 539 2;
  • 31) 0.796 875 000 000 025 617 183 539 2 × 2 = 1 + 0.593 750 000 000 051 234 367 078 4;
  • 32) 0.593 750 000 000 051 234 367 078 4 × 2 = 1 + 0.187 500 000 000 102 468 734 156 8;
  • 33) 0.187 500 000 000 102 468 734 156 8 × 2 = 0 + 0.375 000 000 000 204 937 468 313 6;
  • 34) 0.375 000 000 000 204 937 468 313 6 × 2 = 0 + 0.750 000 000 000 409 874 936 627 2;
  • 35) 0.750 000 000 000 409 874 936 627 2 × 2 = 1 + 0.500 000 000 000 819 749 873 254 4;
  • 36) 0.500 000 000 000 819 749 873 254 4 × 2 = 1 + 0.000 000 000 001 639 499 746 508 8;
  • 37) 0.000 000 000 001 639 499 746 508 8 × 2 = 0 + 0.000 000 000 003 278 999 493 017 6;
  • 38) 0.000 000 000 003 278 999 493 017 6 × 2 = 0 + 0.000 000 000 006 557 998 986 035 2;
  • 39) 0.000 000 000 006 557 998 986 035 2 × 2 = 0 + 0.000 000 000 013 115 997 972 070 4;
  • 40) 0.000 000 000 013 115 997 972 070 4 × 2 = 0 + 0.000 000 000 026 231 995 944 140 8;
  • 41) 0.000 000 000 026 231 995 944 140 8 × 2 = 0 + 0.000 000 000 052 463 991 888 281 6;
  • 42) 0.000 000 000 052 463 991 888 281 6 × 2 = 0 + 0.000 000 000 104 927 983 776 563 2;
  • 43) 0.000 000 000 104 927 983 776 563 2 × 2 = 0 + 0.000 000 000 209 855 967 553 126 4;
  • 44) 0.000 000 000 209 855 967 553 126 4 × 2 = 0 + 0.000 000 000 419 711 935 106 252 8;
  • 45) 0.000 000 000 419 711 935 106 252 8 × 2 = 0 + 0.000 000 000 839 423 870 212 505 6;
  • 46) 0.000 000 000 839 423 870 212 505 6 × 2 = 0 + 0.000 000 001 678 847 740 425 011 2;
  • 47) 0.000 000 001 678 847 740 425 011 2 × 2 = 0 + 0.000 000 003 357 695 480 850 022 4;
  • 48) 0.000 000 003 357 695 480 850 022 4 × 2 = 0 + 0.000 000 006 715 390 961 700 044 8;
  • 49) 0.000 000 006 715 390 961 700 044 8 × 2 = 0 + 0.000 000 013 430 781 923 400 089 6;
  • 50) 0.000 000 013 430 781 923 400 089 6 × 2 = 0 + 0.000 000 026 861 563 846 800 179 2;
  • 51) 0.000 000 026 861 563 846 800 179 2 × 2 = 0 + 0.000 000 053 723 127 693 600 358 4;
  • 52) 0.000 000 053 723 127 693 600 358 4 × 2 = 0 + 0.000 000 107 446 255 387 200 716 8;
  • 53) 0.000 000 107 446 255 387 200 716 8 × 2 = 0 + 0.000 000 214 892 510 774 401 433 6;
  • 54) 0.000 000 214 892 510 774 401 433 6 × 2 = 0 + 0.000 000 429 785 021 548 802 867 2;

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:


0.000 000 000 742 147 676 646 733 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 733 3(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:


0.000 000 000 742 147 676 646 733 3(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 -0.000 000 000 742 147 676 646 733 3 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