-0.381 966 011 250 105 097 475 04 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 250 105 097 475 04(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.381 966 011 250 105 097 475 04(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.381 966 011 250 105 097 475 04| = 0.381 966 011 250 105 097 475 04


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: 0.381 966 011 250 105 097 475 04.

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) 0.381 966 011 250 105 097 475 04 × 2 = 0 + 0.763 932 022 500 210 194 950 08;
  • 2) 0.763 932 022 500 210 194 950 08 × 2 = 1 + 0.527 864 045 000 420 389 900 16;
  • 3) 0.527 864 045 000 420 389 900 16 × 2 = 1 + 0.055 728 090 000 840 779 800 32;
  • 4) 0.055 728 090 000 840 779 800 32 × 2 = 0 + 0.111 456 180 001 681 559 600 64;
  • 5) 0.111 456 180 001 681 559 600 64 × 2 = 0 + 0.222 912 360 003 363 119 201 28;
  • 6) 0.222 912 360 003 363 119 201 28 × 2 = 0 + 0.445 824 720 006 726 238 402 56;
  • 7) 0.445 824 720 006 726 238 402 56 × 2 = 0 + 0.891 649 440 013 452 476 805 12;
  • 8) 0.891 649 440 013 452 476 805 12 × 2 = 1 + 0.783 298 880 026 904 953 610 24;
  • 9) 0.783 298 880 026 904 953 610 24 × 2 = 1 + 0.566 597 760 053 809 907 220 48;
  • 10) 0.566 597 760 053 809 907 220 48 × 2 = 1 + 0.133 195 520 107 619 814 440 96;
  • 11) 0.133 195 520 107 619 814 440 96 × 2 = 0 + 0.266 391 040 215 239 628 881 92;
  • 12) 0.266 391 040 215 239 628 881 92 × 2 = 0 + 0.532 782 080 430 479 257 763 84;
  • 13) 0.532 782 080 430 479 257 763 84 × 2 = 1 + 0.065 564 160 860 958 515 527 68;
  • 14) 0.065 564 160 860 958 515 527 68 × 2 = 0 + 0.131 128 321 721 917 031 055 36;
  • 15) 0.131 128 321 721 917 031 055 36 × 2 = 0 + 0.262 256 643 443 834 062 110 72;
  • 16) 0.262 256 643 443 834 062 110 72 × 2 = 0 + 0.524 513 286 887 668 124 221 44;
  • 17) 0.524 513 286 887 668 124 221 44 × 2 = 1 + 0.049 026 573 775 336 248 442 88;
  • 18) 0.049 026 573 775 336 248 442 88 × 2 = 0 + 0.098 053 147 550 672 496 885 76;
  • 19) 0.098 053 147 550 672 496 885 76 × 2 = 0 + 0.196 106 295 101 344 993 771 52;
  • 20) 0.196 106 295 101 344 993 771 52 × 2 = 0 + 0.392 212 590 202 689 987 543 04;
  • 21) 0.392 212 590 202 689 987 543 04 × 2 = 0 + 0.784 425 180 405 379 975 086 08;
  • 22) 0.784 425 180 405 379 975 086 08 × 2 = 1 + 0.568 850 360 810 759 950 172 16;
  • 23) 0.568 850 360 810 759 950 172 16 × 2 = 1 + 0.137 700 721 621 519 900 344 32;
  • 24) 0.137 700 721 621 519 900 344 32 × 2 = 0 + 0.275 401 443 243 039 800 688 64;
  • 25) 0.275 401 443 243 039 800 688 64 × 2 = 0 + 0.550 802 886 486 079 601 377 28;
  • 26) 0.550 802 886 486 079 601 377 28 × 2 = 1 + 0.101 605 772 972 159 202 754 56;
  • 27) 0.101 605 772 972 159 202 754 56 × 2 = 0 + 0.203 211 545 944 318 405 509 12;
  • 28) 0.203 211 545 944 318 405 509 12 × 2 = 0 + 0.406 423 091 888 636 811 018 24;
  • 29) 0.406 423 091 888 636 811 018 24 × 2 = 0 + 0.812 846 183 777 273 622 036 48;
  • 30) 0.812 846 183 777 273 622 036 48 × 2 = 1 + 0.625 692 367 554 547 244 072 96;
  • 31) 0.625 692 367 554 547 244 072 96 × 2 = 1 + 0.251 384 735 109 094 488 145 92;
  • 32) 0.251 384 735 109 094 488 145 92 × 2 = 0 + 0.502 769 470 218 188 976 291 84;
  • 33) 0.502 769 470 218 188 976 291 84 × 2 = 1 + 0.005 538 940 436 377 952 583 68;
  • 34) 0.005 538 940 436 377 952 583 68 × 2 = 0 + 0.011 077 880 872 755 905 167 36;
  • 35) 0.011 077 880 872 755 905 167 36 × 2 = 0 + 0.022 155 761 745 511 810 334 72;
  • 36) 0.022 155 761 745 511 810 334 72 × 2 = 0 + 0.044 311 523 491 023 620 669 44;
  • 37) 0.044 311 523 491 023 620 669 44 × 2 = 0 + 0.088 623 046 982 047 241 338 88;
  • 38) 0.088 623 046 982 047 241 338 88 × 2 = 0 + 0.177 246 093 964 094 482 677 76;
  • 39) 0.177 246 093 964 094 482 677 76 × 2 = 0 + 0.354 492 187 928 188 965 355 52;
  • 40) 0.354 492 187 928 188 965 355 52 × 2 = 0 + 0.708 984 375 856 377 930 711 04;
  • 41) 0.708 984 375 856 377 930 711 04 × 2 = 1 + 0.417 968 751 712 755 861 422 08;
  • 42) 0.417 968 751 712 755 861 422 08 × 2 = 0 + 0.835 937 503 425 511 722 844 16;
  • 43) 0.835 937 503 425 511 722 844 16 × 2 = 1 + 0.671 875 006 851 023 445 688 32;
  • 44) 0.671 875 006 851 023 445 688 32 × 2 = 1 + 0.343 750 013 702 046 891 376 64;
  • 45) 0.343 750 013 702 046 891 376 64 × 2 = 0 + 0.687 500 027 404 093 782 753 28;
  • 46) 0.687 500 027 404 093 782 753 28 × 2 = 1 + 0.375 000 054 808 187 565 506 56;
  • 47) 0.375 000 054 808 187 565 506 56 × 2 = 0 + 0.750 000 109 616 375 131 013 12;
  • 48) 0.750 000 109 616 375 131 013 12 × 2 = 1 + 0.500 000 219 232 750 262 026 24;
  • 49) 0.500 000 219 232 750 262 026 24 × 2 = 1 + 0.000 000 438 465 500 524 052 48;
  • 50) 0.000 000 438 465 500 524 052 48 × 2 = 0 + 0.000 000 876 931 001 048 104 96;
  • 51) 0.000 000 876 931 001 048 104 96 × 2 = 0 + 0.000 001 753 862 002 096 209 92;
  • 52) 0.000 001 753 862 002 096 209 92 × 2 = 0 + 0.000 003 507 724 004 192 419 84;
  • 53) 0.000 003 507 724 004 192 419 84 × 2 = 0 + 0.000 007 015 448 008 384 839 68;
  • 54) 0.000 007 015 448 008 384 839 68 × 2 = 0 + 0.000 014 030 896 016 769 679 36;

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:


0.381 966 011 250 105 097 475 04(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 1000 00(2)

6. Positive number before normalization:

0.381 966 011 250 105 097 475 04(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 1000 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.381 966 011 250 105 097 475 04(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 1000 00(2) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 1000 00(2) × 20 =

1.1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0110 0000(2) × 2-2


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -2

Mantissa (not normalized):
1.1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0110 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-2 + 2(11-1) - 1 =

(-2 + 1 023)(10) =

1 021(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

1021(10) =

011 1111 1101(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 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0110 0000 =

1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0110 0000


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1101

Mantissa (52 bits) =
1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0110 0000


Decimal number -0.381 966 011 250 105 097 475 04 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0110 0000

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double 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 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 from the previous 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 multiplying operations, starting from the top of the list constructed 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, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' 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 -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 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:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. 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) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) 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:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100