-0.000 000 000 742 147 676 646 700 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 700 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
-0.000 000 000 742 147 676 646 700 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:

|-0.000 000 000 742 147 676 646 700 8| = 0.000 000 000 742 147 676 646 700 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: 0.000 000 000 742 147 676 646 700 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) 0.000 000 000 742 147 676 646 700 8 × 2 = 0 + 0.000 000 001 484 295 353 293 401 6;
  • 2) 0.000 000 001 484 295 353 293 401 6 × 2 = 0 + 0.000 000 002 968 590 706 586 803 2;
  • 3) 0.000 000 002 968 590 706 586 803 2 × 2 = 0 + 0.000 000 005 937 181 413 173 606 4;
  • 4) 0.000 000 005 937 181 413 173 606 4 × 2 = 0 + 0.000 000 011 874 362 826 347 212 8;
  • 5) 0.000 000 011 874 362 826 347 212 8 × 2 = 0 + 0.000 000 023 748 725 652 694 425 6;
  • 6) 0.000 000 023 748 725 652 694 425 6 × 2 = 0 + 0.000 000 047 497 451 305 388 851 2;
  • 7) 0.000 000 047 497 451 305 388 851 2 × 2 = 0 + 0.000 000 094 994 902 610 777 702 4;
  • 8) 0.000 000 094 994 902 610 777 702 4 × 2 = 0 + 0.000 000 189 989 805 221 555 404 8;
  • 9) 0.000 000 189 989 805 221 555 404 8 × 2 = 0 + 0.000 000 379 979 610 443 110 809 6;
  • 10) 0.000 000 379 979 610 443 110 809 6 × 2 = 0 + 0.000 000 759 959 220 886 221 619 2;
  • 11) 0.000 000 759 959 220 886 221 619 2 × 2 = 0 + 0.000 001 519 918 441 772 443 238 4;
  • 12) 0.000 001 519 918 441 772 443 238 4 × 2 = 0 + 0.000 003 039 836 883 544 886 476 8;
  • 13) 0.000 003 039 836 883 544 886 476 8 × 2 = 0 + 0.000 006 079 673 767 089 772 953 6;
  • 14) 0.000 006 079 673 767 089 772 953 6 × 2 = 0 + 0.000 012 159 347 534 179 545 907 2;
  • 15) 0.000 012 159 347 534 179 545 907 2 × 2 = 0 + 0.000 024 318 695 068 359 091 814 4;
  • 16) 0.000 024 318 695 068 359 091 814 4 × 2 = 0 + 0.000 048 637 390 136 718 183 628 8;
  • 17) 0.000 048 637 390 136 718 183 628 8 × 2 = 0 + 0.000 097 274 780 273 436 367 257 6;
  • 18) 0.000 097 274 780 273 436 367 257 6 × 2 = 0 + 0.000 194 549 560 546 872 734 515 2;
  • 19) 0.000 194 549 560 546 872 734 515 2 × 2 = 0 + 0.000 389 099 121 093 745 469 030 4;
  • 20) 0.000 389 099 121 093 745 469 030 4 × 2 = 0 + 0.000 778 198 242 187 490 938 060 8;
  • 21) 0.000 778 198 242 187 490 938 060 8 × 2 = 0 + 0.001 556 396 484 374 981 876 121 6;
  • 22) 0.001 556 396 484 374 981 876 121 6 × 2 = 0 + 0.003 112 792 968 749 963 752 243 2;
  • 23) 0.003 112 792 968 749 963 752 243 2 × 2 = 0 + 0.006 225 585 937 499 927 504 486 4;
  • 24) 0.006 225 585 937 499 927 504 486 4 × 2 = 0 + 0.012 451 171 874 999 855 008 972 8;
  • 25) 0.012 451 171 874 999 855 008 972 8 × 2 = 0 + 0.024 902 343 749 999 710 017 945 6;
  • 26) 0.024 902 343 749 999 710 017 945 6 × 2 = 0 + 0.049 804 687 499 999 420 035 891 2;
  • 27) 0.049 804 687 499 999 420 035 891 2 × 2 = 0 + 0.099 609 374 999 998 840 071 782 4;
  • 28) 0.099 609 374 999 998 840 071 782 4 × 2 = 0 + 0.199 218 749 999 997 680 143 564 8;
  • 29) 0.199 218 749 999 997 680 143 564 8 × 2 = 0 + 0.398 437 499 999 995 360 287 129 6;
  • 30) 0.398 437 499 999 995 360 287 129 6 × 2 = 0 + 0.796 874 999 999 990 720 574 259 2;
  • 31) 0.796 874 999 999 990 720 574 259 2 × 2 = 1 + 0.593 749 999 999 981 441 148 518 4;
  • 32) 0.593 749 999 999 981 441 148 518 4 × 2 = 1 + 0.187 499 999 999 962 882 297 036 8;
  • 33) 0.187 499 999 999 962 882 297 036 8 × 2 = 0 + 0.374 999 999 999 925 764 594 073 6;
  • 34) 0.374 999 999 999 925 764 594 073 6 × 2 = 0 + 0.749 999 999 999 851 529 188 147 2;
  • 35) 0.749 999 999 999 851 529 188 147 2 × 2 = 1 + 0.499 999 999 999 703 058 376 294 4;
  • 36) 0.499 999 999 999 703 058 376 294 4 × 2 = 0 + 0.999 999 999 999 406 116 752 588 8;
  • 37) 0.999 999 999 999 406 116 752 588 8 × 2 = 1 + 0.999 999 999 998 812 233 505 177 6;
  • 38) 0.999 999 999 998 812 233 505 177 6 × 2 = 1 + 0.999 999 999 997 624 467 010 355 2;
  • 39) 0.999 999 999 997 624 467 010 355 2 × 2 = 1 + 0.999 999 999 995 248 934 020 710 4;
  • 40) 0.999 999 999 995 248 934 020 710 4 × 2 = 1 + 0.999 999 999 990 497 868 041 420 8;
  • 41) 0.999 999 999 990 497 868 041 420 8 × 2 = 1 + 0.999 999 999 980 995 736 082 841 6;
  • 42) 0.999 999 999 980 995 736 082 841 6 × 2 = 1 + 0.999 999 999 961 991 472 165 683 2;
  • 43) 0.999 999 999 961 991 472 165 683 2 × 2 = 1 + 0.999 999 999 923 982 944 331 366 4;
  • 44) 0.999 999 999 923 982 944 331 366 4 × 2 = 1 + 0.999 999 999 847 965 888 662 732 8;
  • 45) 0.999 999 999 847 965 888 662 732 8 × 2 = 1 + 0.999 999 999 695 931 777 325 465 6;
  • 46) 0.999 999 999 695 931 777 325 465 6 × 2 = 1 + 0.999 999 999 391 863 554 650 931 2;
  • 47) 0.999 999 999 391 863 554 650 931 2 × 2 = 1 + 0.999 999 998 783 727 109 301 862 4;
  • 48) 0.999 999 998 783 727 109 301 862 4 × 2 = 1 + 0.999 999 997 567 454 218 603 724 8;
  • 49) 0.999 999 997 567 454 218 603 724 8 × 2 = 1 + 0.999 999 995 134 908 437 207 449 6;
  • 50) 0.999 999 995 134 908 437 207 449 6 × 2 = 1 + 0.999 999 990 269 816 874 414 899 2;
  • 51) 0.999 999 990 269 816 874 414 899 2 × 2 = 1 + 0.999 999 980 539 633 748 829 798 4;
  • 52) 0.999 999 980 539 633 748 829 798 4 × 2 = 1 + 0.999 999 961 079 267 497 659 596 8;
  • 53) 0.999 999 961 079 267 497 659 596 8 × 2 = 1 + 0.999 999 922 158 534 995 319 193 6;
  • 54) 0.999 999 922 158 534 995 319 193 6 × 2 = 1 + 0.999 999 844 317 069 990 638 387 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:


0.000 000 000 742 147 676 646 700 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 700 8(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:


0.000 000 000 742 147 676 646 700 8(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 -0.000 000 000 742 147 676 646 700 8 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