-0.381 966 011 250 103 02 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 250 103 02(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
-0.381 966 011 250 103 02(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:

|-0.381 966 011 250 103 02| = 0.381 966 011 250 103 02


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: 0.381 966 011 250 103 02.

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) 0.381 966 011 250 103 02 × 2 = 0 + 0.763 932 022 500 206 04;
  • 2) 0.763 932 022 500 206 04 × 2 = 1 + 0.527 864 045 000 412 08;
  • 3) 0.527 864 045 000 412 08 × 2 = 1 + 0.055 728 090 000 824 16;
  • 4) 0.055 728 090 000 824 16 × 2 = 0 + 0.111 456 180 001 648 32;
  • 5) 0.111 456 180 001 648 32 × 2 = 0 + 0.222 912 360 003 296 64;
  • 6) 0.222 912 360 003 296 64 × 2 = 0 + 0.445 824 720 006 593 28;
  • 7) 0.445 824 720 006 593 28 × 2 = 0 + 0.891 649 440 013 186 56;
  • 8) 0.891 649 440 013 186 56 × 2 = 1 + 0.783 298 880 026 373 12;
  • 9) 0.783 298 880 026 373 12 × 2 = 1 + 0.566 597 760 052 746 24;
  • 10) 0.566 597 760 052 746 24 × 2 = 1 + 0.133 195 520 105 492 48;
  • 11) 0.133 195 520 105 492 48 × 2 = 0 + 0.266 391 040 210 984 96;
  • 12) 0.266 391 040 210 984 96 × 2 = 0 + 0.532 782 080 421 969 92;
  • 13) 0.532 782 080 421 969 92 × 2 = 1 + 0.065 564 160 843 939 84;
  • 14) 0.065 564 160 843 939 84 × 2 = 0 + 0.131 128 321 687 879 68;
  • 15) 0.131 128 321 687 879 68 × 2 = 0 + 0.262 256 643 375 759 36;
  • 16) 0.262 256 643 375 759 36 × 2 = 0 + 0.524 513 286 751 518 72;
  • 17) 0.524 513 286 751 518 72 × 2 = 1 + 0.049 026 573 503 037 44;
  • 18) 0.049 026 573 503 037 44 × 2 = 0 + 0.098 053 147 006 074 88;
  • 19) 0.098 053 147 006 074 88 × 2 = 0 + 0.196 106 294 012 149 76;
  • 20) 0.196 106 294 012 149 76 × 2 = 0 + 0.392 212 588 024 299 52;
  • 21) 0.392 212 588 024 299 52 × 2 = 0 + 0.784 425 176 048 599 04;
  • 22) 0.784 425 176 048 599 04 × 2 = 1 + 0.568 850 352 097 198 08;
  • 23) 0.568 850 352 097 198 08 × 2 = 1 + 0.137 700 704 194 396 16;
  • 24) 0.137 700 704 194 396 16 × 2 = 0 + 0.275 401 408 388 792 32;
  • 25) 0.275 401 408 388 792 32 × 2 = 0 + 0.550 802 816 777 584 64;
  • 26) 0.550 802 816 777 584 64 × 2 = 1 + 0.101 605 633 555 169 28;
  • 27) 0.101 605 633 555 169 28 × 2 = 0 + 0.203 211 267 110 338 56;
  • 28) 0.203 211 267 110 338 56 × 2 = 0 + 0.406 422 534 220 677 12;
  • 29) 0.406 422 534 220 677 12 × 2 = 0 + 0.812 845 068 441 354 24;
  • 30) 0.812 845 068 441 354 24 × 2 = 1 + 0.625 690 136 882 708 48;
  • 31) 0.625 690 136 882 708 48 × 2 = 1 + 0.251 380 273 765 416 96;
  • 32) 0.251 380 273 765 416 96 × 2 = 0 + 0.502 760 547 530 833 92;
  • 33) 0.502 760 547 530 833 92 × 2 = 1 + 0.005 521 095 061 667 84;
  • 34) 0.005 521 095 061 667 84 × 2 = 0 + 0.011 042 190 123 335 68;
  • 35) 0.011 042 190 123 335 68 × 2 = 0 + 0.022 084 380 246 671 36;
  • 36) 0.022 084 380 246 671 36 × 2 = 0 + 0.044 168 760 493 342 72;
  • 37) 0.044 168 760 493 342 72 × 2 = 0 + 0.088 337 520 986 685 44;
  • 38) 0.088 337 520 986 685 44 × 2 = 0 + 0.176 675 041 973 370 88;
  • 39) 0.176 675 041 973 370 88 × 2 = 0 + 0.353 350 083 946 741 76;
  • 40) 0.353 350 083 946 741 76 × 2 = 0 + 0.706 700 167 893 483 52;
  • 41) 0.706 700 167 893 483 52 × 2 = 1 + 0.413 400 335 786 967 04;
  • 42) 0.413 400 335 786 967 04 × 2 = 0 + 0.826 800 671 573 934 08;
  • 43) 0.826 800 671 573 934 08 × 2 = 1 + 0.653 601 343 147 868 16;
  • 44) 0.653 601 343 147 868 16 × 2 = 1 + 0.307 202 686 295 736 32;
  • 45) 0.307 202 686 295 736 32 × 2 = 0 + 0.614 405 372 591 472 64;
  • 46) 0.614 405 372 591 472 64 × 2 = 1 + 0.228 810 745 182 945 28;
  • 47) 0.228 810 745 182 945 28 × 2 = 0 + 0.457 621 490 365 890 56;
  • 48) 0.457 621 490 365 890 56 × 2 = 0 + 0.915 242 980 731 781 12;
  • 49) 0.915 242 980 731 781 12 × 2 = 1 + 0.830 485 961 463 562 24;
  • 50) 0.830 485 961 463 562 24 × 2 = 1 + 0.660 971 922 927 124 48;
  • 51) 0.660 971 922 927 124 48 × 2 = 1 + 0.321 943 845 854 248 96;
  • 52) 0.321 943 845 854 248 96 × 2 = 0 + 0.643 887 691 708 497 92;
  • 53) 0.643 887 691 708 497 92 × 2 = 1 + 0.287 775 383 416 995 84;
  • 54) 0.287 775 383 416 995 84 × 2 = 0 + 0.575 550 766 833 991 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:


0.381 966 011 250 103 02(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0100 1110 10(2)

6. Positive number before normalization:

0.381 966 011 250 103 02(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0100 1110 10(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:


0.381 966 011 250 103 02(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0100 1110 10(2) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0100 1110 10(2) × 20 =

1.1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0011 1010(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 0011 1010


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 0011 1010 =

1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0011 1010


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 0011 1010


Decimal number -0.381 966 011 250 103 02 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 0011 1010

» Convert decimal -0.381 966 011 250 103 71 to 64 bit double precision IEEE 754 binary floating point representation standard

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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