-0.000 000 000 742 147 683 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 683 1(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
-0.000 000 000 742 147 683 1(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:

|-0.000 000 000 742 147 683 1| = 0.000 000 000 742 147 683 1


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: 0.000 000 000 742 147 683 1.

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) 0.000 000 000 742 147 683 1 × 2 = 0 + 0.000 000 001 484 295 366 2;
  • 2) 0.000 000 001 484 295 366 2 × 2 = 0 + 0.000 000 002 968 590 732 4;
  • 3) 0.000 000 002 968 590 732 4 × 2 = 0 + 0.000 000 005 937 181 464 8;
  • 4) 0.000 000 005 937 181 464 8 × 2 = 0 + 0.000 000 011 874 362 929 6;
  • 5) 0.000 000 011 874 362 929 6 × 2 = 0 + 0.000 000 023 748 725 859 2;
  • 6) 0.000 000 023 748 725 859 2 × 2 = 0 + 0.000 000 047 497 451 718 4;
  • 7) 0.000 000 047 497 451 718 4 × 2 = 0 + 0.000 000 094 994 903 436 8;
  • 8) 0.000 000 094 994 903 436 8 × 2 = 0 + 0.000 000 189 989 806 873 6;
  • 9) 0.000 000 189 989 806 873 6 × 2 = 0 + 0.000 000 379 979 613 747 2;
  • 10) 0.000 000 379 979 613 747 2 × 2 = 0 + 0.000 000 759 959 227 494 4;
  • 11) 0.000 000 759 959 227 494 4 × 2 = 0 + 0.000 001 519 918 454 988 8;
  • 12) 0.000 001 519 918 454 988 8 × 2 = 0 + 0.000 003 039 836 909 977 6;
  • 13) 0.000 003 039 836 909 977 6 × 2 = 0 + 0.000 006 079 673 819 955 2;
  • 14) 0.000 006 079 673 819 955 2 × 2 = 0 + 0.000 012 159 347 639 910 4;
  • 15) 0.000 012 159 347 639 910 4 × 2 = 0 + 0.000 024 318 695 279 820 8;
  • 16) 0.000 024 318 695 279 820 8 × 2 = 0 + 0.000 048 637 390 559 641 6;
  • 17) 0.000 048 637 390 559 641 6 × 2 = 0 + 0.000 097 274 781 119 283 2;
  • 18) 0.000 097 274 781 119 283 2 × 2 = 0 + 0.000 194 549 562 238 566 4;
  • 19) 0.000 194 549 562 238 566 4 × 2 = 0 + 0.000 389 099 124 477 132 8;
  • 20) 0.000 389 099 124 477 132 8 × 2 = 0 + 0.000 778 198 248 954 265 6;
  • 21) 0.000 778 198 248 954 265 6 × 2 = 0 + 0.001 556 396 497 908 531 2;
  • 22) 0.001 556 396 497 908 531 2 × 2 = 0 + 0.003 112 792 995 817 062 4;
  • 23) 0.003 112 792 995 817 062 4 × 2 = 0 + 0.006 225 585 991 634 124 8;
  • 24) 0.006 225 585 991 634 124 8 × 2 = 0 + 0.012 451 171 983 268 249 6;
  • 25) 0.012 451 171 983 268 249 6 × 2 = 0 + 0.024 902 343 966 536 499 2;
  • 26) 0.024 902 343 966 536 499 2 × 2 = 0 + 0.049 804 687 933 072 998 4;
  • 27) 0.049 804 687 933 072 998 4 × 2 = 0 + 0.099 609 375 866 145 996 8;
  • 28) 0.099 609 375 866 145 996 8 × 2 = 0 + 0.199 218 751 732 291 993 6;
  • 29) 0.199 218 751 732 291 993 6 × 2 = 0 + 0.398 437 503 464 583 987 2;
  • 30) 0.398 437 503 464 583 987 2 × 2 = 0 + 0.796 875 006 929 167 974 4;
  • 31) 0.796 875 006 929 167 974 4 × 2 = 1 + 0.593 750 013 858 335 948 8;
  • 32) 0.593 750 013 858 335 948 8 × 2 = 1 + 0.187 500 027 716 671 897 6;
  • 33) 0.187 500 027 716 671 897 6 × 2 = 0 + 0.375 000 055 433 343 795 2;
  • 34) 0.375 000 055 433 343 795 2 × 2 = 0 + 0.750 000 110 866 687 590 4;
  • 35) 0.750 000 110 866 687 590 4 × 2 = 1 + 0.500 000 221 733 375 180 8;
  • 36) 0.500 000 221 733 375 180 8 × 2 = 1 + 0.000 000 443 466 750 361 6;
  • 37) 0.000 000 443 466 750 361 6 × 2 = 0 + 0.000 000 886 933 500 723 2;
  • 38) 0.000 000 886 933 500 723 2 × 2 = 0 + 0.000 001 773 867 001 446 4;
  • 39) 0.000 001 773 867 001 446 4 × 2 = 0 + 0.000 003 547 734 002 892 8;
  • 40) 0.000 003 547 734 002 892 8 × 2 = 0 + 0.000 007 095 468 005 785 6;
  • 41) 0.000 007 095 468 005 785 6 × 2 = 0 + 0.000 014 190 936 011 571 2;
  • 42) 0.000 014 190 936 011 571 2 × 2 = 0 + 0.000 028 381 872 023 142 4;
  • 43) 0.000 028 381 872 023 142 4 × 2 = 0 + 0.000 056 763 744 046 284 8;
  • 44) 0.000 056 763 744 046 284 8 × 2 = 0 + 0.000 113 527 488 092 569 6;
  • 45) 0.000 113 527 488 092 569 6 × 2 = 0 + 0.000 227 054 976 185 139 2;
  • 46) 0.000 227 054 976 185 139 2 × 2 = 0 + 0.000 454 109 952 370 278 4;
  • 47) 0.000 454 109 952 370 278 4 × 2 = 0 + 0.000 908 219 904 740 556 8;
  • 48) 0.000 908 219 904 740 556 8 × 2 = 0 + 0.001 816 439 809 481 113 6;
  • 49) 0.001 816 439 809 481 113 6 × 2 = 0 + 0.003 632 879 618 962 227 2;
  • 50) 0.003 632 879 618 962 227 2 × 2 = 0 + 0.007 265 759 237 924 454 4;
  • 51) 0.007 265 759 237 924 454 4 × 2 = 0 + 0.014 531 518 475 848 908 8;
  • 52) 0.014 531 518 475 848 908 8 × 2 = 0 + 0.029 063 036 951 697 817 6;
  • 53) 0.029 063 036 951 697 817 6 × 2 = 0 + 0.058 126 073 903 395 635 2;
  • 54) 0.058 126 073 903 395 635 2 × 2 = 0 + 0.116 252 147 806 791 270 4;

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:


0.000 000 000 742 147 683 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 683 1(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:


0.000 000 000 742 147 683 1(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 -0.000 000 000 742 147 683 1 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