-0.000 000 000 742 147 676 645 85 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 645 85(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
-0.000 000 000 742 147 676 645 85(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:

|-0.000 000 000 742 147 676 645 85| = 0.000 000 000 742 147 676 645 85


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: 0.000 000 000 742 147 676 645 85.

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) 0.000 000 000 742 147 676 645 85 × 2 = 0 + 0.000 000 001 484 295 353 291 7;
  • 2) 0.000 000 001 484 295 353 291 7 × 2 = 0 + 0.000 000 002 968 590 706 583 4;
  • 3) 0.000 000 002 968 590 706 583 4 × 2 = 0 + 0.000 000 005 937 181 413 166 8;
  • 4) 0.000 000 005 937 181 413 166 8 × 2 = 0 + 0.000 000 011 874 362 826 333 6;
  • 5) 0.000 000 011 874 362 826 333 6 × 2 = 0 + 0.000 000 023 748 725 652 667 2;
  • 6) 0.000 000 023 748 725 652 667 2 × 2 = 0 + 0.000 000 047 497 451 305 334 4;
  • 7) 0.000 000 047 497 451 305 334 4 × 2 = 0 + 0.000 000 094 994 902 610 668 8;
  • 8) 0.000 000 094 994 902 610 668 8 × 2 = 0 + 0.000 000 189 989 805 221 337 6;
  • 9) 0.000 000 189 989 805 221 337 6 × 2 = 0 + 0.000 000 379 979 610 442 675 2;
  • 10) 0.000 000 379 979 610 442 675 2 × 2 = 0 + 0.000 000 759 959 220 885 350 4;
  • 11) 0.000 000 759 959 220 885 350 4 × 2 = 0 + 0.000 001 519 918 441 770 700 8;
  • 12) 0.000 001 519 918 441 770 700 8 × 2 = 0 + 0.000 003 039 836 883 541 401 6;
  • 13) 0.000 003 039 836 883 541 401 6 × 2 = 0 + 0.000 006 079 673 767 082 803 2;
  • 14) 0.000 006 079 673 767 082 803 2 × 2 = 0 + 0.000 012 159 347 534 165 606 4;
  • 15) 0.000 012 159 347 534 165 606 4 × 2 = 0 + 0.000 024 318 695 068 331 212 8;
  • 16) 0.000 024 318 695 068 331 212 8 × 2 = 0 + 0.000 048 637 390 136 662 425 6;
  • 17) 0.000 048 637 390 136 662 425 6 × 2 = 0 + 0.000 097 274 780 273 324 851 2;
  • 18) 0.000 097 274 780 273 324 851 2 × 2 = 0 + 0.000 194 549 560 546 649 702 4;
  • 19) 0.000 194 549 560 546 649 702 4 × 2 = 0 + 0.000 389 099 121 093 299 404 8;
  • 20) 0.000 389 099 121 093 299 404 8 × 2 = 0 + 0.000 778 198 242 186 598 809 6;
  • 21) 0.000 778 198 242 186 598 809 6 × 2 = 0 + 0.001 556 396 484 373 197 619 2;
  • 22) 0.001 556 396 484 373 197 619 2 × 2 = 0 + 0.003 112 792 968 746 395 238 4;
  • 23) 0.003 112 792 968 746 395 238 4 × 2 = 0 + 0.006 225 585 937 492 790 476 8;
  • 24) 0.006 225 585 937 492 790 476 8 × 2 = 0 + 0.012 451 171 874 985 580 953 6;
  • 25) 0.012 451 171 874 985 580 953 6 × 2 = 0 + 0.024 902 343 749 971 161 907 2;
  • 26) 0.024 902 343 749 971 161 907 2 × 2 = 0 + 0.049 804 687 499 942 323 814 4;
  • 27) 0.049 804 687 499 942 323 814 4 × 2 = 0 + 0.099 609 374 999 884 647 628 8;
  • 28) 0.099 609 374 999 884 647 628 8 × 2 = 0 + 0.199 218 749 999 769 295 257 6;
  • 29) 0.199 218 749 999 769 295 257 6 × 2 = 0 + 0.398 437 499 999 538 590 515 2;
  • 30) 0.398 437 499 999 538 590 515 2 × 2 = 0 + 0.796 874 999 999 077 181 030 4;
  • 31) 0.796 874 999 999 077 181 030 4 × 2 = 1 + 0.593 749 999 998 154 362 060 8;
  • 32) 0.593 749 999 998 154 362 060 8 × 2 = 1 + 0.187 499 999 996 308 724 121 6;
  • 33) 0.187 499 999 996 308 724 121 6 × 2 = 0 + 0.374 999 999 992 617 448 243 2;
  • 34) 0.374 999 999 992 617 448 243 2 × 2 = 0 + 0.749 999 999 985 234 896 486 4;
  • 35) 0.749 999 999 985 234 896 486 4 × 2 = 1 + 0.499 999 999 970 469 792 972 8;
  • 36) 0.499 999 999 970 469 792 972 8 × 2 = 0 + 0.999 999 999 940 939 585 945 6;
  • 37) 0.999 999 999 940 939 585 945 6 × 2 = 1 + 0.999 999 999 881 879 171 891 2;
  • 38) 0.999 999 999 881 879 171 891 2 × 2 = 1 + 0.999 999 999 763 758 343 782 4;
  • 39) 0.999 999 999 763 758 343 782 4 × 2 = 1 + 0.999 999 999 527 516 687 564 8;
  • 40) 0.999 999 999 527 516 687 564 8 × 2 = 1 + 0.999 999 999 055 033 375 129 6;
  • 41) 0.999 999 999 055 033 375 129 6 × 2 = 1 + 0.999 999 998 110 066 750 259 2;
  • 42) 0.999 999 998 110 066 750 259 2 × 2 = 1 + 0.999 999 996 220 133 500 518 4;
  • 43) 0.999 999 996 220 133 500 518 4 × 2 = 1 + 0.999 999 992 440 267 001 036 8;
  • 44) 0.999 999 992 440 267 001 036 8 × 2 = 1 + 0.999 999 984 880 534 002 073 6;
  • 45) 0.999 999 984 880 534 002 073 6 × 2 = 1 + 0.999 999 969 761 068 004 147 2;
  • 46) 0.999 999 969 761 068 004 147 2 × 2 = 1 + 0.999 999 939 522 136 008 294 4;
  • 47) 0.999 999 939 522 136 008 294 4 × 2 = 1 + 0.999 999 879 044 272 016 588 8;
  • 48) 0.999 999 879 044 272 016 588 8 × 2 = 1 + 0.999 999 758 088 544 033 177 6;
  • 49) 0.999 999 758 088 544 033 177 6 × 2 = 1 + 0.999 999 516 177 088 066 355 2;
  • 50) 0.999 999 516 177 088 066 355 2 × 2 = 1 + 0.999 999 032 354 176 132 710 4;
  • 51) 0.999 999 032 354 176 132 710 4 × 2 = 1 + 0.999 998 064 708 352 265 420 8;
  • 52) 0.999 998 064 708 352 265 420 8 × 2 = 1 + 0.999 996 129 416 704 530 841 6;
  • 53) 0.999 996 129 416 704 530 841 6 × 2 = 1 + 0.999 992 258 833 409 061 683 2;
  • 54) 0.999 992 258 833 409 061 683 2 × 2 = 1 + 0.999 984 517 666 818 123 366 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:


0.000 000 000 742 147 676 645 85(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 645 85(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 645 85(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 0111 1111 1111 1111 111


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 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 645 85 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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