-0.000 000 000 742 147 676 646 95 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 95(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 95(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 95| = 0.000 000 000 742 147 676 646 95


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 95.

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 95 × 2 = 0 + 0.000 000 001 484 295 353 293 9;
  • 2) 0.000 000 001 484 295 353 293 9 × 2 = 0 + 0.000 000 002 968 590 706 587 8;
  • 3) 0.000 000 002 968 590 706 587 8 × 2 = 0 + 0.000 000 005 937 181 413 175 6;
  • 4) 0.000 000 005 937 181 413 175 6 × 2 = 0 + 0.000 000 011 874 362 826 351 2;
  • 5) 0.000 000 011 874 362 826 351 2 × 2 = 0 + 0.000 000 023 748 725 652 702 4;
  • 6) 0.000 000 023 748 725 652 702 4 × 2 = 0 + 0.000 000 047 497 451 305 404 8;
  • 7) 0.000 000 047 497 451 305 404 8 × 2 = 0 + 0.000 000 094 994 902 610 809 6;
  • 8) 0.000 000 094 994 902 610 809 6 × 2 = 0 + 0.000 000 189 989 805 221 619 2;
  • 9) 0.000 000 189 989 805 221 619 2 × 2 = 0 + 0.000 000 379 979 610 443 238 4;
  • 10) 0.000 000 379 979 610 443 238 4 × 2 = 0 + 0.000 000 759 959 220 886 476 8;
  • 11) 0.000 000 759 959 220 886 476 8 × 2 = 0 + 0.000 001 519 918 441 772 953 6;
  • 12) 0.000 001 519 918 441 772 953 6 × 2 = 0 + 0.000 003 039 836 883 545 907 2;
  • 13) 0.000 003 039 836 883 545 907 2 × 2 = 0 + 0.000 006 079 673 767 091 814 4;
  • 14) 0.000 006 079 673 767 091 814 4 × 2 = 0 + 0.000 012 159 347 534 183 628 8;
  • 15) 0.000 012 159 347 534 183 628 8 × 2 = 0 + 0.000 024 318 695 068 367 257 6;
  • 16) 0.000 024 318 695 068 367 257 6 × 2 = 0 + 0.000 048 637 390 136 734 515 2;
  • 17) 0.000 048 637 390 136 734 515 2 × 2 = 0 + 0.000 097 274 780 273 469 030 4;
  • 18) 0.000 097 274 780 273 469 030 4 × 2 = 0 + 0.000 194 549 560 546 938 060 8;
  • 19) 0.000 194 549 560 546 938 060 8 × 2 = 0 + 0.000 389 099 121 093 876 121 6;
  • 20) 0.000 389 099 121 093 876 121 6 × 2 = 0 + 0.000 778 198 242 187 752 243 2;
  • 21) 0.000 778 198 242 187 752 243 2 × 2 = 0 + 0.001 556 396 484 375 504 486 4;
  • 22) 0.001 556 396 484 375 504 486 4 × 2 = 0 + 0.003 112 792 968 751 008 972 8;
  • 23) 0.003 112 792 968 751 008 972 8 × 2 = 0 + 0.006 225 585 937 502 017 945 6;
  • 24) 0.006 225 585 937 502 017 945 6 × 2 = 0 + 0.012 451 171 875 004 035 891 2;
  • 25) 0.012 451 171 875 004 035 891 2 × 2 = 0 + 0.024 902 343 750 008 071 782 4;
  • 26) 0.024 902 343 750 008 071 782 4 × 2 = 0 + 0.049 804 687 500 016 143 564 8;
  • 27) 0.049 804 687 500 016 143 564 8 × 2 = 0 + 0.099 609 375 000 032 287 129 6;
  • 28) 0.099 609 375 000 032 287 129 6 × 2 = 0 + 0.199 218 750 000 064 574 259 2;
  • 29) 0.199 218 750 000 064 574 259 2 × 2 = 0 + 0.398 437 500 000 129 148 518 4;
  • 30) 0.398 437 500 000 129 148 518 4 × 2 = 0 + 0.796 875 000 000 258 297 036 8;
  • 31) 0.796 875 000 000 258 297 036 8 × 2 = 1 + 0.593 750 000 000 516 594 073 6;
  • 32) 0.593 750 000 000 516 594 073 6 × 2 = 1 + 0.187 500 000 001 033 188 147 2;
  • 33) 0.187 500 000 001 033 188 147 2 × 2 = 0 + 0.375 000 000 002 066 376 294 4;
  • 34) 0.375 000 000 002 066 376 294 4 × 2 = 0 + 0.750 000 000 004 132 752 588 8;
  • 35) 0.750 000 000 004 132 752 588 8 × 2 = 1 + 0.500 000 000 008 265 505 177 6;
  • 36) 0.500 000 000 008 265 505 177 6 × 2 = 1 + 0.000 000 000 016 531 010 355 2;
  • 37) 0.000 000 000 016 531 010 355 2 × 2 = 0 + 0.000 000 000 033 062 020 710 4;
  • 38) 0.000 000 000 033 062 020 710 4 × 2 = 0 + 0.000 000 000 066 124 041 420 8;
  • 39) 0.000 000 000 066 124 041 420 8 × 2 = 0 + 0.000 000 000 132 248 082 841 6;
  • 40) 0.000 000 000 132 248 082 841 6 × 2 = 0 + 0.000 000 000 264 496 165 683 2;
  • 41) 0.000 000 000 264 496 165 683 2 × 2 = 0 + 0.000 000 000 528 992 331 366 4;
  • 42) 0.000 000 000 528 992 331 366 4 × 2 = 0 + 0.000 000 001 057 984 662 732 8;
  • 43) 0.000 000 001 057 984 662 732 8 × 2 = 0 + 0.000 000 002 115 969 325 465 6;
  • 44) 0.000 000 002 115 969 325 465 6 × 2 = 0 + 0.000 000 004 231 938 650 931 2;
  • 45) 0.000 000 004 231 938 650 931 2 × 2 = 0 + 0.000 000 008 463 877 301 862 4;
  • 46) 0.000 000 008 463 877 301 862 4 × 2 = 0 + 0.000 000 016 927 754 603 724 8;
  • 47) 0.000 000 016 927 754 603 724 8 × 2 = 0 + 0.000 000 033 855 509 207 449 6;
  • 48) 0.000 000 033 855 509 207 449 6 × 2 = 0 + 0.000 000 067 711 018 414 899 2;
  • 49) 0.000 000 067 711 018 414 899 2 × 2 = 0 + 0.000 000 135 422 036 829 798 4;
  • 50) 0.000 000 135 422 036 829 798 4 × 2 = 0 + 0.000 000 270 844 073 659 596 8;
  • 51) 0.000 000 270 844 073 659 596 8 × 2 = 0 + 0.000 000 541 688 147 319 193 6;
  • 52) 0.000 000 541 688 147 319 193 6 × 2 = 0 + 0.000 001 083 376 294 638 387 2;
  • 53) 0.000 001 083 376 294 638 387 2 × 2 = 0 + 0.000 002 166 752 589 276 774 4;
  • 54) 0.000 002 166 752 589 276 774 4 × 2 = 0 + 0.000 004 333 505 178 553 548 8;

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 95(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 95(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 95(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 95 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