-0.000 000 000 742 147 676 395 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 395(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
-0.000 000 000 742 147 676 395(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:

|-0.000 000 000 742 147 676 395| = 0.000 000 000 742 147 676 395


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: 0.000 000 000 742 147 676 395.

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) 0.000 000 000 742 147 676 395 × 2 = 0 + 0.000 000 001 484 295 352 79;
  • 2) 0.000 000 001 484 295 352 79 × 2 = 0 + 0.000 000 002 968 590 705 58;
  • 3) 0.000 000 002 968 590 705 58 × 2 = 0 + 0.000 000 005 937 181 411 16;
  • 4) 0.000 000 005 937 181 411 16 × 2 = 0 + 0.000 000 011 874 362 822 32;
  • 5) 0.000 000 011 874 362 822 32 × 2 = 0 + 0.000 000 023 748 725 644 64;
  • 6) 0.000 000 023 748 725 644 64 × 2 = 0 + 0.000 000 047 497 451 289 28;
  • 7) 0.000 000 047 497 451 289 28 × 2 = 0 + 0.000 000 094 994 902 578 56;
  • 8) 0.000 000 094 994 902 578 56 × 2 = 0 + 0.000 000 189 989 805 157 12;
  • 9) 0.000 000 189 989 805 157 12 × 2 = 0 + 0.000 000 379 979 610 314 24;
  • 10) 0.000 000 379 979 610 314 24 × 2 = 0 + 0.000 000 759 959 220 628 48;
  • 11) 0.000 000 759 959 220 628 48 × 2 = 0 + 0.000 001 519 918 441 256 96;
  • 12) 0.000 001 519 918 441 256 96 × 2 = 0 + 0.000 003 039 836 882 513 92;
  • 13) 0.000 003 039 836 882 513 92 × 2 = 0 + 0.000 006 079 673 765 027 84;
  • 14) 0.000 006 079 673 765 027 84 × 2 = 0 + 0.000 012 159 347 530 055 68;
  • 15) 0.000 012 159 347 530 055 68 × 2 = 0 + 0.000 024 318 695 060 111 36;
  • 16) 0.000 024 318 695 060 111 36 × 2 = 0 + 0.000 048 637 390 120 222 72;
  • 17) 0.000 048 637 390 120 222 72 × 2 = 0 + 0.000 097 274 780 240 445 44;
  • 18) 0.000 097 274 780 240 445 44 × 2 = 0 + 0.000 194 549 560 480 890 88;
  • 19) 0.000 194 549 560 480 890 88 × 2 = 0 + 0.000 389 099 120 961 781 76;
  • 20) 0.000 389 099 120 961 781 76 × 2 = 0 + 0.000 778 198 241 923 563 52;
  • 21) 0.000 778 198 241 923 563 52 × 2 = 0 + 0.001 556 396 483 847 127 04;
  • 22) 0.001 556 396 483 847 127 04 × 2 = 0 + 0.003 112 792 967 694 254 08;
  • 23) 0.003 112 792 967 694 254 08 × 2 = 0 + 0.006 225 585 935 388 508 16;
  • 24) 0.006 225 585 935 388 508 16 × 2 = 0 + 0.012 451 171 870 777 016 32;
  • 25) 0.012 451 171 870 777 016 32 × 2 = 0 + 0.024 902 343 741 554 032 64;
  • 26) 0.024 902 343 741 554 032 64 × 2 = 0 + 0.049 804 687 483 108 065 28;
  • 27) 0.049 804 687 483 108 065 28 × 2 = 0 + 0.099 609 374 966 216 130 56;
  • 28) 0.099 609 374 966 216 130 56 × 2 = 0 + 0.199 218 749 932 432 261 12;
  • 29) 0.199 218 749 932 432 261 12 × 2 = 0 + 0.398 437 499 864 864 522 24;
  • 30) 0.398 437 499 864 864 522 24 × 2 = 0 + 0.796 874 999 729 729 044 48;
  • 31) 0.796 874 999 729 729 044 48 × 2 = 1 + 0.593 749 999 459 458 088 96;
  • 32) 0.593 749 999 459 458 088 96 × 2 = 1 + 0.187 499 998 918 916 177 92;
  • 33) 0.187 499 998 918 916 177 92 × 2 = 0 + 0.374 999 997 837 832 355 84;
  • 34) 0.374 999 997 837 832 355 84 × 2 = 0 + 0.749 999 995 675 664 711 68;
  • 35) 0.749 999 995 675 664 711 68 × 2 = 1 + 0.499 999 991 351 329 423 36;
  • 36) 0.499 999 991 351 329 423 36 × 2 = 0 + 0.999 999 982 702 658 846 72;
  • 37) 0.999 999 982 702 658 846 72 × 2 = 1 + 0.999 999 965 405 317 693 44;
  • 38) 0.999 999 965 405 317 693 44 × 2 = 1 + 0.999 999 930 810 635 386 88;
  • 39) 0.999 999 930 810 635 386 88 × 2 = 1 + 0.999 999 861 621 270 773 76;
  • 40) 0.999 999 861 621 270 773 76 × 2 = 1 + 0.999 999 723 242 541 547 52;
  • 41) 0.999 999 723 242 541 547 52 × 2 = 1 + 0.999 999 446 485 083 095 04;
  • 42) 0.999 999 446 485 083 095 04 × 2 = 1 + 0.999 998 892 970 166 190 08;
  • 43) 0.999 998 892 970 166 190 08 × 2 = 1 + 0.999 997 785 940 332 380 16;
  • 44) 0.999 997 785 940 332 380 16 × 2 = 1 + 0.999 995 571 880 664 760 32;
  • 45) 0.999 995 571 880 664 760 32 × 2 = 1 + 0.999 991 143 761 329 520 64;
  • 46) 0.999 991 143 761 329 520 64 × 2 = 1 + 0.999 982 287 522 659 041 28;
  • 47) 0.999 982 287 522 659 041 28 × 2 = 1 + 0.999 964 575 045 318 082 56;
  • 48) 0.999 964 575 045 318 082 56 × 2 = 1 + 0.999 929 150 090 636 165 12;
  • 49) 0.999 929 150 090 636 165 12 × 2 = 1 + 0.999 858 300 181 272 330 24;
  • 50) 0.999 858 300 181 272 330 24 × 2 = 1 + 0.999 716 600 362 544 660 48;
  • 51) 0.999 716 600 362 544 660 48 × 2 = 1 + 0.999 433 200 725 089 320 96;
  • 52) 0.999 433 200 725 089 320 96 × 2 = 1 + 0.998 866 401 450 178 641 92;
  • 53) 0.998 866 401 450 178 641 92 × 2 = 1 + 0.997 732 802 900 357 283 84;
  • 54) 0.997 732 802 900 357 283 84 × 2 = 1 + 0.995 465 605 800 714 567 68;

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:


0.000 000 000 742 147 676 395(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 395(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:


0.000 000 000 742 147 676 395(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 -0.000 000 000 742 147 676 395 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 414 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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