-0.381 966 011 250 105 097 474 67 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 250 105 097 474 67(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
-0.381 966 011 250 105 097 474 67(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:

|-0.381 966 011 250 105 097 474 67| = 0.381 966 011 250 105 097 474 67


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: 0.381 966 011 250 105 097 474 67.

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) 0.381 966 011 250 105 097 474 67 × 2 = 0 + 0.763 932 022 500 210 194 949 34;
  • 2) 0.763 932 022 500 210 194 949 34 × 2 = 1 + 0.527 864 045 000 420 389 898 68;
  • 3) 0.527 864 045 000 420 389 898 68 × 2 = 1 + 0.055 728 090 000 840 779 797 36;
  • 4) 0.055 728 090 000 840 779 797 36 × 2 = 0 + 0.111 456 180 001 681 559 594 72;
  • 5) 0.111 456 180 001 681 559 594 72 × 2 = 0 + 0.222 912 360 003 363 119 189 44;
  • 6) 0.222 912 360 003 363 119 189 44 × 2 = 0 + 0.445 824 720 006 726 238 378 88;
  • 7) 0.445 824 720 006 726 238 378 88 × 2 = 0 + 0.891 649 440 013 452 476 757 76;
  • 8) 0.891 649 440 013 452 476 757 76 × 2 = 1 + 0.783 298 880 026 904 953 515 52;
  • 9) 0.783 298 880 026 904 953 515 52 × 2 = 1 + 0.566 597 760 053 809 907 031 04;
  • 10) 0.566 597 760 053 809 907 031 04 × 2 = 1 + 0.133 195 520 107 619 814 062 08;
  • 11) 0.133 195 520 107 619 814 062 08 × 2 = 0 + 0.266 391 040 215 239 628 124 16;
  • 12) 0.266 391 040 215 239 628 124 16 × 2 = 0 + 0.532 782 080 430 479 256 248 32;
  • 13) 0.532 782 080 430 479 256 248 32 × 2 = 1 + 0.065 564 160 860 958 512 496 64;
  • 14) 0.065 564 160 860 958 512 496 64 × 2 = 0 + 0.131 128 321 721 917 024 993 28;
  • 15) 0.131 128 321 721 917 024 993 28 × 2 = 0 + 0.262 256 643 443 834 049 986 56;
  • 16) 0.262 256 643 443 834 049 986 56 × 2 = 0 + 0.524 513 286 887 668 099 973 12;
  • 17) 0.524 513 286 887 668 099 973 12 × 2 = 1 + 0.049 026 573 775 336 199 946 24;
  • 18) 0.049 026 573 775 336 199 946 24 × 2 = 0 + 0.098 053 147 550 672 399 892 48;
  • 19) 0.098 053 147 550 672 399 892 48 × 2 = 0 + 0.196 106 295 101 344 799 784 96;
  • 20) 0.196 106 295 101 344 799 784 96 × 2 = 0 + 0.392 212 590 202 689 599 569 92;
  • 21) 0.392 212 590 202 689 599 569 92 × 2 = 0 + 0.784 425 180 405 379 199 139 84;
  • 22) 0.784 425 180 405 379 199 139 84 × 2 = 1 + 0.568 850 360 810 758 398 279 68;
  • 23) 0.568 850 360 810 758 398 279 68 × 2 = 1 + 0.137 700 721 621 516 796 559 36;
  • 24) 0.137 700 721 621 516 796 559 36 × 2 = 0 + 0.275 401 443 243 033 593 118 72;
  • 25) 0.275 401 443 243 033 593 118 72 × 2 = 0 + 0.550 802 886 486 067 186 237 44;
  • 26) 0.550 802 886 486 067 186 237 44 × 2 = 1 + 0.101 605 772 972 134 372 474 88;
  • 27) 0.101 605 772 972 134 372 474 88 × 2 = 0 + 0.203 211 545 944 268 744 949 76;
  • 28) 0.203 211 545 944 268 744 949 76 × 2 = 0 + 0.406 423 091 888 537 489 899 52;
  • 29) 0.406 423 091 888 537 489 899 52 × 2 = 0 + 0.812 846 183 777 074 979 799 04;
  • 30) 0.812 846 183 777 074 979 799 04 × 2 = 1 + 0.625 692 367 554 149 959 598 08;
  • 31) 0.625 692 367 554 149 959 598 08 × 2 = 1 + 0.251 384 735 108 299 919 196 16;
  • 32) 0.251 384 735 108 299 919 196 16 × 2 = 0 + 0.502 769 470 216 599 838 392 32;
  • 33) 0.502 769 470 216 599 838 392 32 × 2 = 1 + 0.005 538 940 433 199 676 784 64;
  • 34) 0.005 538 940 433 199 676 784 64 × 2 = 0 + 0.011 077 880 866 399 353 569 28;
  • 35) 0.011 077 880 866 399 353 569 28 × 2 = 0 + 0.022 155 761 732 798 707 138 56;
  • 36) 0.022 155 761 732 798 707 138 56 × 2 = 0 + 0.044 311 523 465 597 414 277 12;
  • 37) 0.044 311 523 465 597 414 277 12 × 2 = 0 + 0.088 623 046 931 194 828 554 24;
  • 38) 0.088 623 046 931 194 828 554 24 × 2 = 0 + 0.177 246 093 862 389 657 108 48;
  • 39) 0.177 246 093 862 389 657 108 48 × 2 = 0 + 0.354 492 187 724 779 314 216 96;
  • 40) 0.354 492 187 724 779 314 216 96 × 2 = 0 + 0.708 984 375 449 558 628 433 92;
  • 41) 0.708 984 375 449 558 628 433 92 × 2 = 1 + 0.417 968 750 899 117 256 867 84;
  • 42) 0.417 968 750 899 117 256 867 84 × 2 = 0 + 0.835 937 501 798 234 513 735 68;
  • 43) 0.835 937 501 798 234 513 735 68 × 2 = 1 + 0.671 875 003 596 469 027 471 36;
  • 44) 0.671 875 003 596 469 027 471 36 × 2 = 1 + 0.343 750 007 192 938 054 942 72;
  • 45) 0.343 750 007 192 938 054 942 72 × 2 = 0 + 0.687 500 014 385 876 109 885 44;
  • 46) 0.687 500 014 385 876 109 885 44 × 2 = 1 + 0.375 000 028 771 752 219 770 88;
  • 47) 0.375 000 028 771 752 219 770 88 × 2 = 0 + 0.750 000 057 543 504 439 541 76;
  • 48) 0.750 000 057 543 504 439 541 76 × 2 = 1 + 0.500 000 115 087 008 879 083 52;
  • 49) 0.500 000 115 087 008 879 083 52 × 2 = 1 + 0.000 000 230 174 017 758 167 04;
  • 50) 0.000 000 230 174 017 758 167 04 × 2 = 0 + 0.000 000 460 348 035 516 334 08;
  • 51) 0.000 000 460 348 035 516 334 08 × 2 = 0 + 0.000 000 920 696 071 032 668 16;
  • 52) 0.000 000 920 696 071 032 668 16 × 2 = 0 + 0.000 001 841 392 142 065 336 32;
  • 53) 0.000 001 841 392 142 065 336 32 × 2 = 0 + 0.000 003 682 784 284 130 672 64;
  • 54) 0.000 003 682 784 284 130 672 64 × 2 = 0 + 0.000 007 365 568 568 261 345 28;

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:


0.381 966 011 250 105 097 474 67(10) =

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

6. Positive number before normalization:

0.381 966 011 250 105 097 474 67(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:


0.381 966 011 250 105 097 474 67(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 -0.381 966 011 250 105 097 474 67 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