-0.000 000 000 742 147 676 646 722 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 722 8(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
-0.000 000 000 742 147 676 646 722 8(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:

|-0.000 000 000 742 147 676 646 722 8| = 0.000 000 000 742 147 676 646 722 8


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: 0.000 000 000 742 147 676 646 722 8.

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) 0.000 000 000 742 147 676 646 722 8 × 2 = 0 + 0.000 000 001 484 295 353 293 445 6;
  • 2) 0.000 000 001 484 295 353 293 445 6 × 2 = 0 + 0.000 000 002 968 590 706 586 891 2;
  • 3) 0.000 000 002 968 590 706 586 891 2 × 2 = 0 + 0.000 000 005 937 181 413 173 782 4;
  • 4) 0.000 000 005 937 181 413 173 782 4 × 2 = 0 + 0.000 000 011 874 362 826 347 564 8;
  • 5) 0.000 000 011 874 362 826 347 564 8 × 2 = 0 + 0.000 000 023 748 725 652 695 129 6;
  • 6) 0.000 000 023 748 725 652 695 129 6 × 2 = 0 + 0.000 000 047 497 451 305 390 259 2;
  • 7) 0.000 000 047 497 451 305 390 259 2 × 2 = 0 + 0.000 000 094 994 902 610 780 518 4;
  • 8) 0.000 000 094 994 902 610 780 518 4 × 2 = 0 + 0.000 000 189 989 805 221 561 036 8;
  • 9) 0.000 000 189 989 805 221 561 036 8 × 2 = 0 + 0.000 000 379 979 610 443 122 073 6;
  • 10) 0.000 000 379 979 610 443 122 073 6 × 2 = 0 + 0.000 000 759 959 220 886 244 147 2;
  • 11) 0.000 000 759 959 220 886 244 147 2 × 2 = 0 + 0.000 001 519 918 441 772 488 294 4;
  • 12) 0.000 001 519 918 441 772 488 294 4 × 2 = 0 + 0.000 003 039 836 883 544 976 588 8;
  • 13) 0.000 003 039 836 883 544 976 588 8 × 2 = 0 + 0.000 006 079 673 767 089 953 177 6;
  • 14) 0.000 006 079 673 767 089 953 177 6 × 2 = 0 + 0.000 012 159 347 534 179 906 355 2;
  • 15) 0.000 012 159 347 534 179 906 355 2 × 2 = 0 + 0.000 024 318 695 068 359 812 710 4;
  • 16) 0.000 024 318 695 068 359 812 710 4 × 2 = 0 + 0.000 048 637 390 136 719 625 420 8;
  • 17) 0.000 048 637 390 136 719 625 420 8 × 2 = 0 + 0.000 097 274 780 273 439 250 841 6;
  • 18) 0.000 097 274 780 273 439 250 841 6 × 2 = 0 + 0.000 194 549 560 546 878 501 683 2;
  • 19) 0.000 194 549 560 546 878 501 683 2 × 2 = 0 + 0.000 389 099 121 093 757 003 366 4;
  • 20) 0.000 389 099 121 093 757 003 366 4 × 2 = 0 + 0.000 778 198 242 187 514 006 732 8;
  • 21) 0.000 778 198 242 187 514 006 732 8 × 2 = 0 + 0.001 556 396 484 375 028 013 465 6;
  • 22) 0.001 556 396 484 375 028 013 465 6 × 2 = 0 + 0.003 112 792 968 750 056 026 931 2;
  • 23) 0.003 112 792 968 750 056 026 931 2 × 2 = 0 + 0.006 225 585 937 500 112 053 862 4;
  • 24) 0.006 225 585 937 500 112 053 862 4 × 2 = 0 + 0.012 451 171 875 000 224 107 724 8;
  • 25) 0.012 451 171 875 000 224 107 724 8 × 2 = 0 + 0.024 902 343 750 000 448 215 449 6;
  • 26) 0.024 902 343 750 000 448 215 449 6 × 2 = 0 + 0.049 804 687 500 000 896 430 899 2;
  • 27) 0.049 804 687 500 000 896 430 899 2 × 2 = 0 + 0.099 609 375 000 001 792 861 798 4;
  • 28) 0.099 609 375 000 001 792 861 798 4 × 2 = 0 + 0.199 218 750 000 003 585 723 596 8;
  • 29) 0.199 218 750 000 003 585 723 596 8 × 2 = 0 + 0.398 437 500 000 007 171 447 193 6;
  • 30) 0.398 437 500 000 007 171 447 193 6 × 2 = 0 + 0.796 875 000 000 014 342 894 387 2;
  • 31) 0.796 875 000 000 014 342 894 387 2 × 2 = 1 + 0.593 750 000 000 028 685 788 774 4;
  • 32) 0.593 750 000 000 028 685 788 774 4 × 2 = 1 + 0.187 500 000 000 057 371 577 548 8;
  • 33) 0.187 500 000 000 057 371 577 548 8 × 2 = 0 + 0.375 000 000 000 114 743 155 097 6;
  • 34) 0.375 000 000 000 114 743 155 097 6 × 2 = 0 + 0.750 000 000 000 229 486 310 195 2;
  • 35) 0.750 000 000 000 229 486 310 195 2 × 2 = 1 + 0.500 000 000 000 458 972 620 390 4;
  • 36) 0.500 000 000 000 458 972 620 390 4 × 2 = 1 + 0.000 000 000 000 917 945 240 780 8;
  • 37) 0.000 000 000 000 917 945 240 780 8 × 2 = 0 + 0.000 000 000 001 835 890 481 561 6;
  • 38) 0.000 000 000 001 835 890 481 561 6 × 2 = 0 + 0.000 000 000 003 671 780 963 123 2;
  • 39) 0.000 000 000 003 671 780 963 123 2 × 2 = 0 + 0.000 000 000 007 343 561 926 246 4;
  • 40) 0.000 000 000 007 343 561 926 246 4 × 2 = 0 + 0.000 000 000 014 687 123 852 492 8;
  • 41) 0.000 000 000 014 687 123 852 492 8 × 2 = 0 + 0.000 000 000 029 374 247 704 985 6;
  • 42) 0.000 000 000 029 374 247 704 985 6 × 2 = 0 + 0.000 000 000 058 748 495 409 971 2;
  • 43) 0.000 000 000 058 748 495 409 971 2 × 2 = 0 + 0.000 000 000 117 496 990 819 942 4;
  • 44) 0.000 000 000 117 496 990 819 942 4 × 2 = 0 + 0.000 000 000 234 993 981 639 884 8;
  • 45) 0.000 000 000 234 993 981 639 884 8 × 2 = 0 + 0.000 000 000 469 987 963 279 769 6;
  • 46) 0.000 000 000 469 987 963 279 769 6 × 2 = 0 + 0.000 000 000 939 975 926 559 539 2;
  • 47) 0.000 000 000 939 975 926 559 539 2 × 2 = 0 + 0.000 000 001 879 951 853 119 078 4;
  • 48) 0.000 000 001 879 951 853 119 078 4 × 2 = 0 + 0.000 000 003 759 903 706 238 156 8;
  • 49) 0.000 000 003 759 903 706 238 156 8 × 2 = 0 + 0.000 000 007 519 807 412 476 313 6;
  • 50) 0.000 000 007 519 807 412 476 313 6 × 2 = 0 + 0.000 000 015 039 614 824 952 627 2;
  • 51) 0.000 000 015 039 614 824 952 627 2 × 2 = 0 + 0.000 000 030 079 229 649 905 254 4;
  • 52) 0.000 000 030 079 229 649 905 254 4 × 2 = 0 + 0.000 000 060 158 459 299 810 508 8;
  • 53) 0.000 000 060 158 459 299 810 508 8 × 2 = 0 + 0.000 000 120 316 918 599 621 017 6;
  • 54) 0.000 000 120 316 918 599 621 017 6 × 2 = 0 + 0.000 000 240 633 837 199 242 035 2;

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:


0.000 000 000 742 147 676 646 722 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 722 8(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:


0.000 000 000 742 147 676 646 722 8(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 -0.000 000 000 742 147 676 646 722 8 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