-0.000 000 000 742 147 676 648 36 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 648 36(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 36(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 36| = 0.000 000 000 742 147 676 648 36


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 36.

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 36 × 2 = 0 + 0.000 000 001 484 295 353 296 72;
  • 2) 0.000 000 001 484 295 353 296 72 × 2 = 0 + 0.000 000 002 968 590 706 593 44;
  • 3) 0.000 000 002 968 590 706 593 44 × 2 = 0 + 0.000 000 005 937 181 413 186 88;
  • 4) 0.000 000 005 937 181 413 186 88 × 2 = 0 + 0.000 000 011 874 362 826 373 76;
  • 5) 0.000 000 011 874 362 826 373 76 × 2 = 0 + 0.000 000 023 748 725 652 747 52;
  • 6) 0.000 000 023 748 725 652 747 52 × 2 = 0 + 0.000 000 047 497 451 305 495 04;
  • 7) 0.000 000 047 497 451 305 495 04 × 2 = 0 + 0.000 000 094 994 902 610 990 08;
  • 8) 0.000 000 094 994 902 610 990 08 × 2 = 0 + 0.000 000 189 989 805 221 980 16;
  • 9) 0.000 000 189 989 805 221 980 16 × 2 = 0 + 0.000 000 379 979 610 443 960 32;
  • 10) 0.000 000 379 979 610 443 960 32 × 2 = 0 + 0.000 000 759 959 220 887 920 64;
  • 11) 0.000 000 759 959 220 887 920 64 × 2 = 0 + 0.000 001 519 918 441 775 841 28;
  • 12) 0.000 001 519 918 441 775 841 28 × 2 = 0 + 0.000 003 039 836 883 551 682 56;
  • 13) 0.000 003 039 836 883 551 682 56 × 2 = 0 + 0.000 006 079 673 767 103 365 12;
  • 14) 0.000 006 079 673 767 103 365 12 × 2 = 0 + 0.000 012 159 347 534 206 730 24;
  • 15) 0.000 012 159 347 534 206 730 24 × 2 = 0 + 0.000 024 318 695 068 413 460 48;
  • 16) 0.000 024 318 695 068 413 460 48 × 2 = 0 + 0.000 048 637 390 136 826 920 96;
  • 17) 0.000 048 637 390 136 826 920 96 × 2 = 0 + 0.000 097 274 780 273 653 841 92;
  • 18) 0.000 097 274 780 273 653 841 92 × 2 = 0 + 0.000 194 549 560 547 307 683 84;
  • 19) 0.000 194 549 560 547 307 683 84 × 2 = 0 + 0.000 389 099 121 094 615 367 68;
  • 20) 0.000 389 099 121 094 615 367 68 × 2 = 0 + 0.000 778 198 242 189 230 735 36;
  • 21) 0.000 778 198 242 189 230 735 36 × 2 = 0 + 0.001 556 396 484 378 461 470 72;
  • 22) 0.001 556 396 484 378 461 470 72 × 2 = 0 + 0.003 112 792 968 756 922 941 44;
  • 23) 0.003 112 792 968 756 922 941 44 × 2 = 0 + 0.006 225 585 937 513 845 882 88;
  • 24) 0.006 225 585 937 513 845 882 88 × 2 = 0 + 0.012 451 171 875 027 691 765 76;
  • 25) 0.012 451 171 875 027 691 765 76 × 2 = 0 + 0.024 902 343 750 055 383 531 52;
  • 26) 0.024 902 343 750 055 383 531 52 × 2 = 0 + 0.049 804 687 500 110 767 063 04;
  • 27) 0.049 804 687 500 110 767 063 04 × 2 = 0 + 0.099 609 375 000 221 534 126 08;
  • 28) 0.099 609 375 000 221 534 126 08 × 2 = 0 + 0.199 218 750 000 443 068 252 16;
  • 29) 0.199 218 750 000 443 068 252 16 × 2 = 0 + 0.398 437 500 000 886 136 504 32;
  • 30) 0.398 437 500 000 886 136 504 32 × 2 = 0 + 0.796 875 000 001 772 273 008 64;
  • 31) 0.796 875 000 001 772 273 008 64 × 2 = 1 + 0.593 750 000 003 544 546 017 28;
  • 32) 0.593 750 000 003 544 546 017 28 × 2 = 1 + 0.187 500 000 007 089 092 034 56;
  • 33) 0.187 500 000 007 089 092 034 56 × 2 = 0 + 0.375 000 000 014 178 184 069 12;
  • 34) 0.375 000 000 014 178 184 069 12 × 2 = 0 + 0.750 000 000 028 356 368 138 24;
  • 35) 0.750 000 000 028 356 368 138 24 × 2 = 1 + 0.500 000 000 056 712 736 276 48;
  • 36) 0.500 000 000 056 712 736 276 48 × 2 = 1 + 0.000 000 000 113 425 472 552 96;
  • 37) 0.000 000 000 113 425 472 552 96 × 2 = 0 + 0.000 000 000 226 850 945 105 92;
  • 38) 0.000 000 000 226 850 945 105 92 × 2 = 0 + 0.000 000 000 453 701 890 211 84;
  • 39) 0.000 000 000 453 701 890 211 84 × 2 = 0 + 0.000 000 000 907 403 780 423 68;
  • 40) 0.000 000 000 907 403 780 423 68 × 2 = 0 + 0.000 000 001 814 807 560 847 36;
  • 41) 0.000 000 001 814 807 560 847 36 × 2 = 0 + 0.000 000 003 629 615 121 694 72;
  • 42) 0.000 000 003 629 615 121 694 72 × 2 = 0 + 0.000 000 007 259 230 243 389 44;
  • 43) 0.000 000 007 259 230 243 389 44 × 2 = 0 + 0.000 000 014 518 460 486 778 88;
  • 44) 0.000 000 014 518 460 486 778 88 × 2 = 0 + 0.000 000 029 036 920 973 557 76;
  • 45) 0.000 000 029 036 920 973 557 76 × 2 = 0 + 0.000 000 058 073 841 947 115 52;
  • 46) 0.000 000 058 073 841 947 115 52 × 2 = 0 + 0.000 000 116 147 683 894 231 04;
  • 47) 0.000 000 116 147 683 894 231 04 × 2 = 0 + 0.000 000 232 295 367 788 462 08;
  • 48) 0.000 000 232 295 367 788 462 08 × 2 = 0 + 0.000 000 464 590 735 576 924 16;
  • 49) 0.000 000 464 590 735 576 924 16 × 2 = 0 + 0.000 000 929 181 471 153 848 32;
  • 50) 0.000 000 929 181 471 153 848 32 × 2 = 0 + 0.000 001 858 362 942 307 696 64;
  • 51) 0.000 001 858 362 942 307 696 64 × 2 = 0 + 0.000 003 716 725 884 615 393 28;
  • 52) 0.000 003 716 725 884 615 393 28 × 2 = 0 + 0.000 007 433 451 769 230 786 56;
  • 53) 0.000 007 433 451 769 230 786 56 × 2 = 0 + 0.000 014 866 903 538 461 573 12;
  • 54) 0.000 014 866 903 538 461 573 12 × 2 = 0 + 0.000 029 733 807 076 923 146 24;

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 36(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 36(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 36(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 36 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