-0.000 000 000 742 147 676 905 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 905(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
-0.000 000 000 742 147 676 905(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:

|-0.000 000 000 742 147 676 905| = 0.000 000 000 742 147 676 905


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: 0.000 000 000 742 147 676 905.

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) 0.000 000 000 742 147 676 905 × 2 = 0 + 0.000 000 001 484 295 353 81;
  • 2) 0.000 000 001 484 295 353 81 × 2 = 0 + 0.000 000 002 968 590 707 62;
  • 3) 0.000 000 002 968 590 707 62 × 2 = 0 + 0.000 000 005 937 181 415 24;
  • 4) 0.000 000 005 937 181 415 24 × 2 = 0 + 0.000 000 011 874 362 830 48;
  • 5) 0.000 000 011 874 362 830 48 × 2 = 0 + 0.000 000 023 748 725 660 96;
  • 6) 0.000 000 023 748 725 660 96 × 2 = 0 + 0.000 000 047 497 451 321 92;
  • 7) 0.000 000 047 497 451 321 92 × 2 = 0 + 0.000 000 094 994 902 643 84;
  • 8) 0.000 000 094 994 902 643 84 × 2 = 0 + 0.000 000 189 989 805 287 68;
  • 9) 0.000 000 189 989 805 287 68 × 2 = 0 + 0.000 000 379 979 610 575 36;
  • 10) 0.000 000 379 979 610 575 36 × 2 = 0 + 0.000 000 759 959 221 150 72;
  • 11) 0.000 000 759 959 221 150 72 × 2 = 0 + 0.000 001 519 918 442 301 44;
  • 12) 0.000 001 519 918 442 301 44 × 2 = 0 + 0.000 003 039 836 884 602 88;
  • 13) 0.000 003 039 836 884 602 88 × 2 = 0 + 0.000 006 079 673 769 205 76;
  • 14) 0.000 006 079 673 769 205 76 × 2 = 0 + 0.000 012 159 347 538 411 52;
  • 15) 0.000 012 159 347 538 411 52 × 2 = 0 + 0.000 024 318 695 076 823 04;
  • 16) 0.000 024 318 695 076 823 04 × 2 = 0 + 0.000 048 637 390 153 646 08;
  • 17) 0.000 048 637 390 153 646 08 × 2 = 0 + 0.000 097 274 780 307 292 16;
  • 18) 0.000 097 274 780 307 292 16 × 2 = 0 + 0.000 194 549 560 614 584 32;
  • 19) 0.000 194 549 560 614 584 32 × 2 = 0 + 0.000 389 099 121 229 168 64;
  • 20) 0.000 389 099 121 229 168 64 × 2 = 0 + 0.000 778 198 242 458 337 28;
  • 21) 0.000 778 198 242 458 337 28 × 2 = 0 + 0.001 556 396 484 916 674 56;
  • 22) 0.001 556 396 484 916 674 56 × 2 = 0 + 0.003 112 792 969 833 349 12;
  • 23) 0.003 112 792 969 833 349 12 × 2 = 0 + 0.006 225 585 939 666 698 24;
  • 24) 0.006 225 585 939 666 698 24 × 2 = 0 + 0.012 451 171 879 333 396 48;
  • 25) 0.012 451 171 879 333 396 48 × 2 = 0 + 0.024 902 343 758 666 792 96;
  • 26) 0.024 902 343 758 666 792 96 × 2 = 0 + 0.049 804 687 517 333 585 92;
  • 27) 0.049 804 687 517 333 585 92 × 2 = 0 + 0.099 609 375 034 667 171 84;
  • 28) 0.099 609 375 034 667 171 84 × 2 = 0 + 0.199 218 750 069 334 343 68;
  • 29) 0.199 218 750 069 334 343 68 × 2 = 0 + 0.398 437 500 138 668 687 36;
  • 30) 0.398 437 500 138 668 687 36 × 2 = 0 + 0.796 875 000 277 337 374 72;
  • 31) 0.796 875 000 277 337 374 72 × 2 = 1 + 0.593 750 000 554 674 749 44;
  • 32) 0.593 750 000 554 674 749 44 × 2 = 1 + 0.187 500 001 109 349 498 88;
  • 33) 0.187 500 001 109 349 498 88 × 2 = 0 + 0.375 000 002 218 698 997 76;
  • 34) 0.375 000 002 218 698 997 76 × 2 = 0 + 0.750 000 004 437 397 995 52;
  • 35) 0.750 000 004 437 397 995 52 × 2 = 1 + 0.500 000 008 874 795 991 04;
  • 36) 0.500 000 008 874 795 991 04 × 2 = 1 + 0.000 000 017 749 591 982 08;
  • 37) 0.000 000 017 749 591 982 08 × 2 = 0 + 0.000 000 035 499 183 964 16;
  • 38) 0.000 000 035 499 183 964 16 × 2 = 0 + 0.000 000 070 998 367 928 32;
  • 39) 0.000 000 070 998 367 928 32 × 2 = 0 + 0.000 000 141 996 735 856 64;
  • 40) 0.000 000 141 996 735 856 64 × 2 = 0 + 0.000 000 283 993 471 713 28;
  • 41) 0.000 000 283 993 471 713 28 × 2 = 0 + 0.000 000 567 986 943 426 56;
  • 42) 0.000 000 567 986 943 426 56 × 2 = 0 + 0.000 001 135 973 886 853 12;
  • 43) 0.000 001 135 973 886 853 12 × 2 = 0 + 0.000 002 271 947 773 706 24;
  • 44) 0.000 002 271 947 773 706 24 × 2 = 0 + 0.000 004 543 895 547 412 48;
  • 45) 0.000 004 543 895 547 412 48 × 2 = 0 + 0.000 009 087 791 094 824 96;
  • 46) 0.000 009 087 791 094 824 96 × 2 = 0 + 0.000 018 175 582 189 649 92;
  • 47) 0.000 018 175 582 189 649 92 × 2 = 0 + 0.000 036 351 164 379 299 84;
  • 48) 0.000 036 351 164 379 299 84 × 2 = 0 + 0.000 072 702 328 758 599 68;
  • 49) 0.000 072 702 328 758 599 68 × 2 = 0 + 0.000 145 404 657 517 199 36;
  • 50) 0.000 145 404 657 517 199 36 × 2 = 0 + 0.000 290 809 315 034 398 72;
  • 51) 0.000 290 809 315 034 398 72 × 2 = 0 + 0.000 581 618 630 068 797 44;
  • 52) 0.000 581 618 630 068 797 44 × 2 = 0 + 0.001 163 237 260 137 594 88;
  • 53) 0.001 163 237 260 137 594 88 × 2 = 0 + 0.002 326 474 520 275 189 76;
  • 54) 0.002 326 474 520 275 189 76 × 2 = 0 + 0.004 652 949 040 550 379 52;

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:


0.000 000 000 742 147 676 905(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 905(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:


0.000 000 000 742 147 676 905(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 -0.000 000 000 742 147 676 905 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