-0.000 000 000 742 147 676 646 695 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 695 7(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
-0.000 000 000 742 147 676 646 695 7(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:

|-0.000 000 000 742 147 676 646 695 7| = 0.000 000 000 742 147 676 646 695 7


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: 0.000 000 000 742 147 676 646 695 7.

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) 0.000 000 000 742 147 676 646 695 7 × 2 = 0 + 0.000 000 001 484 295 353 293 391 4;
  • 2) 0.000 000 001 484 295 353 293 391 4 × 2 = 0 + 0.000 000 002 968 590 706 586 782 8;
  • 3) 0.000 000 002 968 590 706 586 782 8 × 2 = 0 + 0.000 000 005 937 181 413 173 565 6;
  • 4) 0.000 000 005 937 181 413 173 565 6 × 2 = 0 + 0.000 000 011 874 362 826 347 131 2;
  • 5) 0.000 000 011 874 362 826 347 131 2 × 2 = 0 + 0.000 000 023 748 725 652 694 262 4;
  • 6) 0.000 000 023 748 725 652 694 262 4 × 2 = 0 + 0.000 000 047 497 451 305 388 524 8;
  • 7) 0.000 000 047 497 451 305 388 524 8 × 2 = 0 + 0.000 000 094 994 902 610 777 049 6;
  • 8) 0.000 000 094 994 902 610 777 049 6 × 2 = 0 + 0.000 000 189 989 805 221 554 099 2;
  • 9) 0.000 000 189 989 805 221 554 099 2 × 2 = 0 + 0.000 000 379 979 610 443 108 198 4;
  • 10) 0.000 000 379 979 610 443 108 198 4 × 2 = 0 + 0.000 000 759 959 220 886 216 396 8;
  • 11) 0.000 000 759 959 220 886 216 396 8 × 2 = 0 + 0.000 001 519 918 441 772 432 793 6;
  • 12) 0.000 001 519 918 441 772 432 793 6 × 2 = 0 + 0.000 003 039 836 883 544 865 587 2;
  • 13) 0.000 003 039 836 883 544 865 587 2 × 2 = 0 + 0.000 006 079 673 767 089 731 174 4;
  • 14) 0.000 006 079 673 767 089 731 174 4 × 2 = 0 + 0.000 012 159 347 534 179 462 348 8;
  • 15) 0.000 012 159 347 534 179 462 348 8 × 2 = 0 + 0.000 024 318 695 068 358 924 697 6;
  • 16) 0.000 024 318 695 068 358 924 697 6 × 2 = 0 + 0.000 048 637 390 136 717 849 395 2;
  • 17) 0.000 048 637 390 136 717 849 395 2 × 2 = 0 + 0.000 097 274 780 273 435 698 790 4;
  • 18) 0.000 097 274 780 273 435 698 790 4 × 2 = 0 + 0.000 194 549 560 546 871 397 580 8;
  • 19) 0.000 194 549 560 546 871 397 580 8 × 2 = 0 + 0.000 389 099 121 093 742 795 161 6;
  • 20) 0.000 389 099 121 093 742 795 161 6 × 2 = 0 + 0.000 778 198 242 187 485 590 323 2;
  • 21) 0.000 778 198 242 187 485 590 323 2 × 2 = 0 + 0.001 556 396 484 374 971 180 646 4;
  • 22) 0.001 556 396 484 374 971 180 646 4 × 2 = 0 + 0.003 112 792 968 749 942 361 292 8;
  • 23) 0.003 112 792 968 749 942 361 292 8 × 2 = 0 + 0.006 225 585 937 499 884 722 585 6;
  • 24) 0.006 225 585 937 499 884 722 585 6 × 2 = 0 + 0.012 451 171 874 999 769 445 171 2;
  • 25) 0.012 451 171 874 999 769 445 171 2 × 2 = 0 + 0.024 902 343 749 999 538 890 342 4;
  • 26) 0.024 902 343 749 999 538 890 342 4 × 2 = 0 + 0.049 804 687 499 999 077 780 684 8;
  • 27) 0.049 804 687 499 999 077 780 684 8 × 2 = 0 + 0.099 609 374 999 998 155 561 369 6;
  • 28) 0.099 609 374 999 998 155 561 369 6 × 2 = 0 + 0.199 218 749 999 996 311 122 739 2;
  • 29) 0.199 218 749 999 996 311 122 739 2 × 2 = 0 + 0.398 437 499 999 992 622 245 478 4;
  • 30) 0.398 437 499 999 992 622 245 478 4 × 2 = 0 + 0.796 874 999 999 985 244 490 956 8;
  • 31) 0.796 874 999 999 985 244 490 956 8 × 2 = 1 + 0.593 749 999 999 970 488 981 913 6;
  • 32) 0.593 749 999 999 970 488 981 913 6 × 2 = 1 + 0.187 499 999 999 940 977 963 827 2;
  • 33) 0.187 499 999 999 940 977 963 827 2 × 2 = 0 + 0.374 999 999 999 881 955 927 654 4;
  • 34) 0.374 999 999 999 881 955 927 654 4 × 2 = 0 + 0.749 999 999 999 763 911 855 308 8;
  • 35) 0.749 999 999 999 763 911 855 308 8 × 2 = 1 + 0.499 999 999 999 527 823 710 617 6;
  • 36) 0.499 999 999 999 527 823 710 617 6 × 2 = 0 + 0.999 999 999 999 055 647 421 235 2;
  • 37) 0.999 999 999 999 055 647 421 235 2 × 2 = 1 + 0.999 999 999 998 111 294 842 470 4;
  • 38) 0.999 999 999 998 111 294 842 470 4 × 2 = 1 + 0.999 999 999 996 222 589 684 940 8;
  • 39) 0.999 999 999 996 222 589 684 940 8 × 2 = 1 + 0.999 999 999 992 445 179 369 881 6;
  • 40) 0.999 999 999 992 445 179 369 881 6 × 2 = 1 + 0.999 999 999 984 890 358 739 763 2;
  • 41) 0.999 999 999 984 890 358 739 763 2 × 2 = 1 + 0.999 999 999 969 780 717 479 526 4;
  • 42) 0.999 999 999 969 780 717 479 526 4 × 2 = 1 + 0.999 999 999 939 561 434 959 052 8;
  • 43) 0.999 999 999 939 561 434 959 052 8 × 2 = 1 + 0.999 999 999 879 122 869 918 105 6;
  • 44) 0.999 999 999 879 122 869 918 105 6 × 2 = 1 + 0.999 999 999 758 245 739 836 211 2;
  • 45) 0.999 999 999 758 245 739 836 211 2 × 2 = 1 + 0.999 999 999 516 491 479 672 422 4;
  • 46) 0.999 999 999 516 491 479 672 422 4 × 2 = 1 + 0.999 999 999 032 982 959 344 844 8;
  • 47) 0.999 999 999 032 982 959 344 844 8 × 2 = 1 + 0.999 999 998 065 965 918 689 689 6;
  • 48) 0.999 999 998 065 965 918 689 689 6 × 2 = 1 + 0.999 999 996 131 931 837 379 379 2;
  • 49) 0.999 999 996 131 931 837 379 379 2 × 2 = 1 + 0.999 999 992 263 863 674 758 758 4;
  • 50) 0.999 999 992 263 863 674 758 758 4 × 2 = 1 + 0.999 999 984 527 727 349 517 516 8;
  • 51) 0.999 999 984 527 727 349 517 516 8 × 2 = 1 + 0.999 999 969 055 454 699 035 033 6;
  • 52) 0.999 999 969 055 454 699 035 033 6 × 2 = 1 + 0.999 999 938 110 909 398 070 067 2;
  • 53) 0.999 999 938 110 909 398 070 067 2 × 2 = 1 + 0.999 999 876 221 818 796 140 134 4;
  • 54) 0.999 999 876 221 818 796 140 134 4 × 2 = 1 + 0.999 999 752 443 637 592 280 268 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:


0.000 000 000 742 147 676 646 695 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 695 7(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:


0.000 000 000 742 147 676 646 695 7(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 -0.000 000 000 742 147 676 646 695 7 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