-0.000 375 658 489 56 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 375 658 489 56(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
-0.000 375 658 489 56(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:

|-0.000 375 658 489 56| = 0.000 375 658 489 56


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: 0.000 375 658 489 56.

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) 0.000 375 658 489 56 × 2 = 0 + 0.000 751 316 979 12;
  • 2) 0.000 751 316 979 12 × 2 = 0 + 0.001 502 633 958 24;
  • 3) 0.001 502 633 958 24 × 2 = 0 + 0.003 005 267 916 48;
  • 4) 0.003 005 267 916 48 × 2 = 0 + 0.006 010 535 832 96;
  • 5) 0.006 010 535 832 96 × 2 = 0 + 0.012 021 071 665 92;
  • 6) 0.012 021 071 665 92 × 2 = 0 + 0.024 042 143 331 84;
  • 7) 0.024 042 143 331 84 × 2 = 0 + 0.048 084 286 663 68;
  • 8) 0.048 084 286 663 68 × 2 = 0 + 0.096 168 573 327 36;
  • 9) 0.096 168 573 327 36 × 2 = 0 + 0.192 337 146 654 72;
  • 10) 0.192 337 146 654 72 × 2 = 0 + 0.384 674 293 309 44;
  • 11) 0.384 674 293 309 44 × 2 = 0 + 0.769 348 586 618 88;
  • 12) 0.769 348 586 618 88 × 2 = 1 + 0.538 697 173 237 76;
  • 13) 0.538 697 173 237 76 × 2 = 1 + 0.077 394 346 475 52;
  • 14) 0.077 394 346 475 52 × 2 = 0 + 0.154 788 692 951 04;
  • 15) 0.154 788 692 951 04 × 2 = 0 + 0.309 577 385 902 08;
  • 16) 0.309 577 385 902 08 × 2 = 0 + 0.619 154 771 804 16;
  • 17) 0.619 154 771 804 16 × 2 = 1 + 0.238 309 543 608 32;
  • 18) 0.238 309 543 608 32 × 2 = 0 + 0.476 619 087 216 64;
  • 19) 0.476 619 087 216 64 × 2 = 0 + 0.953 238 174 433 28;
  • 20) 0.953 238 174 433 28 × 2 = 1 + 0.906 476 348 866 56;
  • 21) 0.906 476 348 866 56 × 2 = 1 + 0.812 952 697 733 12;
  • 22) 0.812 952 697 733 12 × 2 = 1 + 0.625 905 395 466 24;
  • 23) 0.625 905 395 466 24 × 2 = 1 + 0.251 810 790 932 48;
  • 24) 0.251 810 790 932 48 × 2 = 0 + 0.503 621 581 864 96;
  • 25) 0.503 621 581 864 96 × 2 = 1 + 0.007 243 163 729 92;
  • 26) 0.007 243 163 729 92 × 2 = 0 + 0.014 486 327 459 84;
  • 27) 0.014 486 327 459 84 × 2 = 0 + 0.028 972 654 919 68;
  • 28) 0.028 972 654 919 68 × 2 = 0 + 0.057 945 309 839 36;
  • 29) 0.057 945 309 839 36 × 2 = 0 + 0.115 890 619 678 72;
  • 30) 0.115 890 619 678 72 × 2 = 0 + 0.231 781 239 357 44;
  • 31) 0.231 781 239 357 44 × 2 = 0 + 0.463 562 478 714 88;
  • 32) 0.463 562 478 714 88 × 2 = 0 + 0.927 124 957 429 76;
  • 33) 0.927 124 957 429 76 × 2 = 1 + 0.854 249 914 859 52;
  • 34) 0.854 249 914 859 52 × 2 = 1 + 0.708 499 829 719 04;
  • 35) 0.708 499 829 719 04 × 2 = 1 + 0.416 999 659 438 08;
  • 36) 0.416 999 659 438 08 × 2 = 0 + 0.833 999 318 876 16;
  • 37) 0.833 999 318 876 16 × 2 = 1 + 0.667 998 637 752 32;
  • 38) 0.667 998 637 752 32 × 2 = 1 + 0.335 997 275 504 64;
  • 39) 0.335 997 275 504 64 × 2 = 0 + 0.671 994 551 009 28;
  • 40) 0.671 994 551 009 28 × 2 = 1 + 0.343 989 102 018 56;
  • 41) 0.343 989 102 018 56 × 2 = 0 + 0.687 978 204 037 12;
  • 42) 0.687 978 204 037 12 × 2 = 1 + 0.375 956 408 074 24;
  • 43) 0.375 956 408 074 24 × 2 = 0 + 0.751 912 816 148 48;
  • 44) 0.751 912 816 148 48 × 2 = 1 + 0.503 825 632 296 96;
  • 45) 0.503 825 632 296 96 × 2 = 1 + 0.007 651 264 593 92;
  • 46) 0.007 651 264 593 92 × 2 = 0 + 0.015 302 529 187 84;
  • 47) 0.015 302 529 187 84 × 2 = 0 + 0.030 605 058 375 68;
  • 48) 0.030 605 058 375 68 × 2 = 0 + 0.061 210 116 751 36;
  • 49) 0.061 210 116 751 36 × 2 = 0 + 0.122 420 233 502 72;
  • 50) 0.122 420 233 502 72 × 2 = 0 + 0.244 840 467 005 44;
  • 51) 0.244 840 467 005 44 × 2 = 0 + 0.489 680 934 010 88;
  • 52) 0.489 680 934 010 88 × 2 = 0 + 0.979 361 868 021 76;
  • 53) 0.979 361 868 021 76 × 2 = 1 + 0.958 723 736 043 52;
  • 54) 0.958 723 736 043 52 × 2 = 1 + 0.917 447 472 087 04;
  • 55) 0.917 447 472 087 04 × 2 = 1 + 0.834 894 944 174 08;
  • 56) 0.834 894 944 174 08 × 2 = 1 + 0.669 789 888 348 16;
  • 57) 0.669 789 888 348 16 × 2 = 1 + 0.339 579 776 696 32;
  • 58) 0.339 579 776 696 32 × 2 = 0 + 0.679 159 553 392 64;
  • 59) 0.679 159 553 392 64 × 2 = 1 + 0.358 319 106 785 28;
  • 60) 0.358 319 106 785 28 × 2 = 0 + 0.716 638 213 570 56;
  • 61) 0.716 638 213 570 56 × 2 = 1 + 0.433 276 427 141 12;
  • 62) 0.433 276 427 141 12 × 2 = 0 + 0.866 552 854 282 24;
  • 63) 0.866 552 854 282 24 × 2 = 1 + 0.733 105 708 564 48;
  • 64) 0.733 105 708 564 48 × 2 = 1 + 0.466 211 417 128 96;

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:


0.000 375 658 489 56(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0101 1000 0000 1111 1010 1011(2)

6. Positive number before normalization:

0.000 375 658 489 56(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0101 1000 0000 1111 1010 1011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 375 658 489 56(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0101 1000 0000 1111 1010 1011(2) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0101 1000 0000 1111 1010 1011(2) × 20 =

1.1000 1001 1110 1000 0000 1110 1101 0101 1000 0000 1111 1010 1011(2) × 2-12


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): -12

Mantissa (not normalized):
1.1000 1001 1110 1000 0000 1110 1101 0101 1000 0000 1111 1010 1011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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 1001 1110 1000 0000 1110 1101 0101 1000 0000 1111 1010 1011 =

1000 1001 1110 1000 0000 1110 1101 0101 1000 0000 1111 1010 1011


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 0011

Mantissa (52 bits) =
1000 1001 1110 1000 0000 1110 1101 0101 1000 0000 1111 1010 1011


Decimal number -0.000 375 658 489 56 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 1000 1001 1110 1000 0000 1110 1101 0101 1000 0000 1111 1010 1011

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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