-0.000 007 820 338 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 007 820 338 3(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
-0.000 007 820 338 3(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:

|-0.000 007 820 338 3| = 0.000 007 820 338 3


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: 0.000 007 820 338 3.

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) 0.000 007 820 338 3 × 2 = 0 + 0.000 015 640 676 6;
  • 2) 0.000 015 640 676 6 × 2 = 0 + 0.000 031 281 353 2;
  • 3) 0.000 031 281 353 2 × 2 = 0 + 0.000 062 562 706 4;
  • 4) 0.000 062 562 706 4 × 2 = 0 + 0.000 125 125 412 8;
  • 5) 0.000 125 125 412 8 × 2 = 0 + 0.000 250 250 825 6;
  • 6) 0.000 250 250 825 6 × 2 = 0 + 0.000 500 501 651 2;
  • 7) 0.000 500 501 651 2 × 2 = 0 + 0.001 001 003 302 4;
  • 8) 0.001 001 003 302 4 × 2 = 0 + 0.002 002 006 604 8;
  • 9) 0.002 002 006 604 8 × 2 = 0 + 0.004 004 013 209 6;
  • 10) 0.004 004 013 209 6 × 2 = 0 + 0.008 008 026 419 2;
  • 11) 0.008 008 026 419 2 × 2 = 0 + 0.016 016 052 838 4;
  • 12) 0.016 016 052 838 4 × 2 = 0 + 0.032 032 105 676 8;
  • 13) 0.032 032 105 676 8 × 2 = 0 + 0.064 064 211 353 6;
  • 14) 0.064 064 211 353 6 × 2 = 0 + 0.128 128 422 707 2;
  • 15) 0.128 128 422 707 2 × 2 = 0 + 0.256 256 845 414 4;
  • 16) 0.256 256 845 414 4 × 2 = 0 + 0.512 513 690 828 8;
  • 17) 0.512 513 690 828 8 × 2 = 1 + 0.025 027 381 657 6;
  • 18) 0.025 027 381 657 6 × 2 = 0 + 0.050 054 763 315 2;
  • 19) 0.050 054 763 315 2 × 2 = 0 + 0.100 109 526 630 4;
  • 20) 0.100 109 526 630 4 × 2 = 0 + 0.200 219 053 260 8;
  • 21) 0.200 219 053 260 8 × 2 = 0 + 0.400 438 106 521 6;
  • 22) 0.400 438 106 521 6 × 2 = 0 + 0.800 876 213 043 2;
  • 23) 0.800 876 213 043 2 × 2 = 1 + 0.601 752 426 086 4;
  • 24) 0.601 752 426 086 4 × 2 = 1 + 0.203 504 852 172 8;
  • 25) 0.203 504 852 172 8 × 2 = 0 + 0.407 009 704 345 6;
  • 26) 0.407 009 704 345 6 × 2 = 0 + 0.814 019 408 691 2;
  • 27) 0.814 019 408 691 2 × 2 = 1 + 0.628 038 817 382 4;
  • 28) 0.628 038 817 382 4 × 2 = 1 + 0.256 077 634 764 8;
  • 29) 0.256 077 634 764 8 × 2 = 0 + 0.512 155 269 529 6;
  • 30) 0.512 155 269 529 6 × 2 = 1 + 0.024 310 539 059 2;
  • 31) 0.024 310 539 059 2 × 2 = 0 + 0.048 621 078 118 4;
  • 32) 0.048 621 078 118 4 × 2 = 0 + 0.097 242 156 236 8;
  • 33) 0.097 242 156 236 8 × 2 = 0 + 0.194 484 312 473 6;
  • 34) 0.194 484 312 473 6 × 2 = 0 + 0.388 968 624 947 2;
  • 35) 0.388 968 624 947 2 × 2 = 0 + 0.777 937 249 894 4;
  • 36) 0.777 937 249 894 4 × 2 = 1 + 0.555 874 499 788 8;
  • 37) 0.555 874 499 788 8 × 2 = 1 + 0.111 748 999 577 6;
  • 38) 0.111 748 999 577 6 × 2 = 0 + 0.223 497 999 155 2;
  • 39) 0.223 497 999 155 2 × 2 = 0 + 0.446 995 998 310 4;
  • 40) 0.446 995 998 310 4 × 2 = 0 + 0.893 991 996 620 8;
  • 41) 0.893 991 996 620 8 × 2 = 1 + 0.787 983 993 241 6;
  • 42) 0.787 983 993 241 6 × 2 = 1 + 0.575 967 986 483 2;
  • 43) 0.575 967 986 483 2 × 2 = 1 + 0.151 935 972 966 4;
  • 44) 0.151 935 972 966 4 × 2 = 0 + 0.303 871 945 932 8;
  • 45) 0.303 871 945 932 8 × 2 = 0 + 0.607 743 891 865 6;
  • 46) 0.607 743 891 865 6 × 2 = 1 + 0.215 487 783 731 2;
  • 47) 0.215 487 783 731 2 × 2 = 0 + 0.430 975 567 462 4;
  • 48) 0.430 975 567 462 4 × 2 = 0 + 0.861 951 134 924 8;
  • 49) 0.861 951 134 924 8 × 2 = 1 + 0.723 902 269 849 6;
  • 50) 0.723 902 269 849 6 × 2 = 1 + 0.447 804 539 699 2;
  • 51) 0.447 804 539 699 2 × 2 = 0 + 0.895 609 079 398 4;
  • 52) 0.895 609 079 398 4 × 2 = 1 + 0.791 218 158 796 8;
  • 53) 0.791 218 158 796 8 × 2 = 1 + 0.582 436 317 593 6;
  • 54) 0.582 436 317 593 6 × 2 = 1 + 0.164 872 635 187 2;
  • 55) 0.164 872 635 187 2 × 2 = 0 + 0.329 745 270 374 4;
  • 56) 0.329 745 270 374 4 × 2 = 0 + 0.659 490 540 748 8;
  • 57) 0.659 490 540 748 8 × 2 = 1 + 0.318 981 081 497 6;
  • 58) 0.318 981 081 497 6 × 2 = 0 + 0.637 962 162 995 2;
  • 59) 0.637 962 162 995 2 × 2 = 1 + 0.275 924 325 990 4;
  • 60) 0.275 924 325 990 4 × 2 = 0 + 0.551 848 651 980 8;
  • 61) 0.551 848 651 980 8 × 2 = 1 + 0.103 697 303 961 6;
  • 62) 0.103 697 303 961 6 × 2 = 0 + 0.207 394 607 923 2;
  • 63) 0.207 394 607 923 2 × 2 = 0 + 0.414 789 215 846 4;
  • 64) 0.414 789 215 846 4 × 2 = 0 + 0.829 578 431 692 8;
  • 65) 0.829 578 431 692 8 × 2 = 1 + 0.659 156 863 385 6;
  • 66) 0.659 156 863 385 6 × 2 = 1 + 0.318 313 726 771 2;
  • 67) 0.318 313 726 771 2 × 2 = 0 + 0.636 627 453 542 4;
  • 68) 0.636 627 453 542 4 × 2 = 1 + 0.273 254 907 084 8;
  • 69) 0.273 254 907 084 8 × 2 = 0 + 0.546 509 814 169 6;

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:


0.000 007 820 338 3(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0001 1000 1110 0100 1101 1100 1010 1000 1101 0(2)

6. Positive number before normalization:

0.000 007 820 338 3(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0001 1000 1110 0100 1101 1100 1010 1000 1101 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 007 820 338 3(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0001 1000 1110 0100 1101 1100 1010 1000 1101 0(2) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0001 1000 1110 0100 1101 1100 1010 1000 1101 0(2) × 20 =

1.0000 0110 0110 1000 0011 0001 1100 1001 1011 1001 0101 0001 1010(2) × 2-17


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): -17

Mantissa (not normalized):
1.0000 0110 0110 1000 0011 0001 1100 1001 1011 1001 0101 0001 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(11-1) - 1 =

(-17 + 1 023)(10) =

1 006(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;
  • 1 006 ÷ 2 = 503 + 0;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1006(10) =

011 1110 1110(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. 0000 0110 0110 1000 0011 0001 1100 1001 1011 1001 0101 0001 1010 =

0000 0110 0110 1000 0011 0001 1100 1001 1011 1001 0101 0001 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 1110 1110

Mantissa (52 bits) =
0000 0110 0110 1000 0011 0001 1100 1001 1011 1001 0101 0001 1010


Decimal number -0.000 007 820 338 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0000 0110 0110 1000 0011 0001 1100 1001 1011 1001 0101 0001 1010

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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