-0.000 000 000 742 147 676 73 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 73(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 73(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 73| = 0.000 000 000 742 147 676 73


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 73.

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 73 × 2 = 0 + 0.000 000 001 484 295 353 46;
  • 2) 0.000 000 001 484 295 353 46 × 2 = 0 + 0.000 000 002 968 590 706 92;
  • 3) 0.000 000 002 968 590 706 92 × 2 = 0 + 0.000 000 005 937 181 413 84;
  • 4) 0.000 000 005 937 181 413 84 × 2 = 0 + 0.000 000 011 874 362 827 68;
  • 5) 0.000 000 011 874 362 827 68 × 2 = 0 + 0.000 000 023 748 725 655 36;
  • 6) 0.000 000 023 748 725 655 36 × 2 = 0 + 0.000 000 047 497 451 310 72;
  • 7) 0.000 000 047 497 451 310 72 × 2 = 0 + 0.000 000 094 994 902 621 44;
  • 8) 0.000 000 094 994 902 621 44 × 2 = 0 + 0.000 000 189 989 805 242 88;
  • 9) 0.000 000 189 989 805 242 88 × 2 = 0 + 0.000 000 379 979 610 485 76;
  • 10) 0.000 000 379 979 610 485 76 × 2 = 0 + 0.000 000 759 959 220 971 52;
  • 11) 0.000 000 759 959 220 971 52 × 2 = 0 + 0.000 001 519 918 441 943 04;
  • 12) 0.000 001 519 918 441 943 04 × 2 = 0 + 0.000 003 039 836 883 886 08;
  • 13) 0.000 003 039 836 883 886 08 × 2 = 0 + 0.000 006 079 673 767 772 16;
  • 14) 0.000 006 079 673 767 772 16 × 2 = 0 + 0.000 012 159 347 535 544 32;
  • 15) 0.000 012 159 347 535 544 32 × 2 = 0 + 0.000 024 318 695 071 088 64;
  • 16) 0.000 024 318 695 071 088 64 × 2 = 0 + 0.000 048 637 390 142 177 28;
  • 17) 0.000 048 637 390 142 177 28 × 2 = 0 + 0.000 097 274 780 284 354 56;
  • 18) 0.000 097 274 780 284 354 56 × 2 = 0 + 0.000 194 549 560 568 709 12;
  • 19) 0.000 194 549 560 568 709 12 × 2 = 0 + 0.000 389 099 121 137 418 24;
  • 20) 0.000 389 099 121 137 418 24 × 2 = 0 + 0.000 778 198 242 274 836 48;
  • 21) 0.000 778 198 242 274 836 48 × 2 = 0 + 0.001 556 396 484 549 672 96;
  • 22) 0.001 556 396 484 549 672 96 × 2 = 0 + 0.003 112 792 969 099 345 92;
  • 23) 0.003 112 792 969 099 345 92 × 2 = 0 + 0.006 225 585 938 198 691 84;
  • 24) 0.006 225 585 938 198 691 84 × 2 = 0 + 0.012 451 171 876 397 383 68;
  • 25) 0.012 451 171 876 397 383 68 × 2 = 0 + 0.024 902 343 752 794 767 36;
  • 26) 0.024 902 343 752 794 767 36 × 2 = 0 + 0.049 804 687 505 589 534 72;
  • 27) 0.049 804 687 505 589 534 72 × 2 = 0 + 0.099 609 375 011 179 069 44;
  • 28) 0.099 609 375 011 179 069 44 × 2 = 0 + 0.199 218 750 022 358 138 88;
  • 29) 0.199 218 750 022 358 138 88 × 2 = 0 + 0.398 437 500 044 716 277 76;
  • 30) 0.398 437 500 044 716 277 76 × 2 = 0 + 0.796 875 000 089 432 555 52;
  • 31) 0.796 875 000 089 432 555 52 × 2 = 1 + 0.593 750 000 178 865 111 04;
  • 32) 0.593 750 000 178 865 111 04 × 2 = 1 + 0.187 500 000 357 730 222 08;
  • 33) 0.187 500 000 357 730 222 08 × 2 = 0 + 0.375 000 000 715 460 444 16;
  • 34) 0.375 000 000 715 460 444 16 × 2 = 0 + 0.750 000 001 430 920 888 32;
  • 35) 0.750 000 001 430 920 888 32 × 2 = 1 + 0.500 000 002 861 841 776 64;
  • 36) 0.500 000 002 861 841 776 64 × 2 = 1 + 0.000 000 005 723 683 553 28;
  • 37) 0.000 000 005 723 683 553 28 × 2 = 0 + 0.000 000 011 447 367 106 56;
  • 38) 0.000 000 011 447 367 106 56 × 2 = 0 + 0.000 000 022 894 734 213 12;
  • 39) 0.000 000 022 894 734 213 12 × 2 = 0 + 0.000 000 045 789 468 426 24;
  • 40) 0.000 000 045 789 468 426 24 × 2 = 0 + 0.000 000 091 578 936 852 48;
  • 41) 0.000 000 091 578 936 852 48 × 2 = 0 + 0.000 000 183 157 873 704 96;
  • 42) 0.000 000 183 157 873 704 96 × 2 = 0 + 0.000 000 366 315 747 409 92;
  • 43) 0.000 000 366 315 747 409 92 × 2 = 0 + 0.000 000 732 631 494 819 84;
  • 44) 0.000 000 732 631 494 819 84 × 2 = 0 + 0.000 001 465 262 989 639 68;
  • 45) 0.000 001 465 262 989 639 68 × 2 = 0 + 0.000 002 930 525 979 279 36;
  • 46) 0.000 002 930 525 979 279 36 × 2 = 0 + 0.000 005 861 051 958 558 72;
  • 47) 0.000 005 861 051 958 558 72 × 2 = 0 + 0.000 011 722 103 917 117 44;
  • 48) 0.000 011 722 103 917 117 44 × 2 = 0 + 0.000 023 444 207 834 234 88;
  • 49) 0.000 023 444 207 834 234 88 × 2 = 0 + 0.000 046 888 415 668 469 76;
  • 50) 0.000 046 888 415 668 469 76 × 2 = 0 + 0.000 093 776 831 336 939 52;
  • 51) 0.000 093 776 831 336 939 52 × 2 = 0 + 0.000 187 553 662 673 879 04;
  • 52) 0.000 187 553 662 673 879 04 × 2 = 0 + 0.000 375 107 325 347 758 08;
  • 53) 0.000 375 107 325 347 758 08 × 2 = 0 + 0.000 750 214 650 695 516 16;
  • 54) 0.000 750 214 650 695 516 16 × 2 = 0 + 0.001 500 429 301 391 032 32;

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 73(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 73(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 73(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 73 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