-0.000 035 666 846 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 846 9(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
-0.000 035 666 846 9(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:

|-0.000 035 666 846 9| = 0.000 035 666 846 9


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: 0.000 035 666 846 9.

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) 0.000 035 666 846 9 × 2 = 0 + 0.000 071 333 693 8;
  • 2) 0.000 071 333 693 8 × 2 = 0 + 0.000 142 667 387 6;
  • 3) 0.000 142 667 387 6 × 2 = 0 + 0.000 285 334 775 2;
  • 4) 0.000 285 334 775 2 × 2 = 0 + 0.000 570 669 550 4;
  • 5) 0.000 570 669 550 4 × 2 = 0 + 0.001 141 339 100 8;
  • 6) 0.001 141 339 100 8 × 2 = 0 + 0.002 282 678 201 6;
  • 7) 0.002 282 678 201 6 × 2 = 0 + 0.004 565 356 403 2;
  • 8) 0.004 565 356 403 2 × 2 = 0 + 0.009 130 712 806 4;
  • 9) 0.009 130 712 806 4 × 2 = 0 + 0.018 261 425 612 8;
  • 10) 0.018 261 425 612 8 × 2 = 0 + 0.036 522 851 225 6;
  • 11) 0.036 522 851 225 6 × 2 = 0 + 0.073 045 702 451 2;
  • 12) 0.073 045 702 451 2 × 2 = 0 + 0.146 091 404 902 4;
  • 13) 0.146 091 404 902 4 × 2 = 0 + 0.292 182 809 804 8;
  • 14) 0.292 182 809 804 8 × 2 = 0 + 0.584 365 619 609 6;
  • 15) 0.584 365 619 609 6 × 2 = 1 + 0.168 731 239 219 2;
  • 16) 0.168 731 239 219 2 × 2 = 0 + 0.337 462 478 438 4;
  • 17) 0.337 462 478 438 4 × 2 = 0 + 0.674 924 956 876 8;
  • 18) 0.674 924 956 876 8 × 2 = 1 + 0.349 849 913 753 6;
  • 19) 0.349 849 913 753 6 × 2 = 0 + 0.699 699 827 507 2;
  • 20) 0.699 699 827 507 2 × 2 = 1 + 0.399 399 655 014 4;
  • 21) 0.399 399 655 014 4 × 2 = 0 + 0.798 799 310 028 8;
  • 22) 0.798 799 310 028 8 × 2 = 1 + 0.597 598 620 057 6;
  • 23) 0.597 598 620 057 6 × 2 = 1 + 0.195 197 240 115 2;
  • 24) 0.195 197 240 115 2 × 2 = 0 + 0.390 394 480 230 4;
  • 25) 0.390 394 480 230 4 × 2 = 0 + 0.780 788 960 460 8;
  • 26) 0.780 788 960 460 8 × 2 = 1 + 0.561 577 920 921 6;
  • 27) 0.561 577 920 921 6 × 2 = 1 + 0.123 155 841 843 2;
  • 28) 0.123 155 841 843 2 × 2 = 0 + 0.246 311 683 686 4;
  • 29) 0.246 311 683 686 4 × 2 = 0 + 0.492 623 367 372 8;
  • 30) 0.492 623 367 372 8 × 2 = 0 + 0.985 246 734 745 6;
  • 31) 0.985 246 734 745 6 × 2 = 1 + 0.970 493 469 491 2;
  • 32) 0.970 493 469 491 2 × 2 = 1 + 0.940 986 938 982 4;
  • 33) 0.940 986 938 982 4 × 2 = 1 + 0.881 973 877 964 8;
  • 34) 0.881 973 877 964 8 × 2 = 1 + 0.763 947 755 929 6;
  • 35) 0.763 947 755 929 6 × 2 = 1 + 0.527 895 511 859 2;
  • 36) 0.527 895 511 859 2 × 2 = 1 + 0.055 791 023 718 4;
  • 37) 0.055 791 023 718 4 × 2 = 0 + 0.111 582 047 436 8;
  • 38) 0.111 582 047 436 8 × 2 = 0 + 0.223 164 094 873 6;
  • 39) 0.223 164 094 873 6 × 2 = 0 + 0.446 328 189 747 2;
  • 40) 0.446 328 189 747 2 × 2 = 0 + 0.892 656 379 494 4;
  • 41) 0.892 656 379 494 4 × 2 = 1 + 0.785 312 758 988 8;
  • 42) 0.785 312 758 988 8 × 2 = 1 + 0.570 625 517 977 6;
  • 43) 0.570 625 517 977 6 × 2 = 1 + 0.141 251 035 955 2;
  • 44) 0.141 251 035 955 2 × 2 = 0 + 0.282 502 071 910 4;
  • 45) 0.282 502 071 910 4 × 2 = 0 + 0.565 004 143 820 8;
  • 46) 0.565 004 143 820 8 × 2 = 1 + 0.130 008 287 641 6;
  • 47) 0.130 008 287 641 6 × 2 = 0 + 0.260 016 575 283 2;
  • 48) 0.260 016 575 283 2 × 2 = 0 + 0.520 033 150 566 4;
  • 49) 0.520 033 150 566 4 × 2 = 1 + 0.040 066 301 132 8;
  • 50) 0.040 066 301 132 8 × 2 = 0 + 0.080 132 602 265 6;
  • 51) 0.080 132 602 265 6 × 2 = 0 + 0.160 265 204 531 2;
  • 52) 0.160 265 204 531 2 × 2 = 0 + 0.320 530 409 062 4;
  • 53) 0.320 530 409 062 4 × 2 = 0 + 0.641 060 818 124 8;
  • 54) 0.641 060 818 124 8 × 2 = 1 + 0.282 121 636 249 6;
  • 55) 0.282 121 636 249 6 × 2 = 0 + 0.564 243 272 499 2;
  • 56) 0.564 243 272 499 2 × 2 = 1 + 0.128 486 544 998 4;
  • 57) 0.128 486 544 998 4 × 2 = 0 + 0.256 973 089 996 8;
  • 58) 0.256 973 089 996 8 × 2 = 0 + 0.513 946 179 993 6;
  • 59) 0.513 946 179 993 6 × 2 = 1 + 0.027 892 359 987 2;
  • 60) 0.027 892 359 987 2 × 2 = 0 + 0.055 784 719 974 4;
  • 61) 0.055 784 719 974 4 × 2 = 0 + 0.111 569 439 948 8;
  • 62) 0.111 569 439 948 8 × 2 = 0 + 0.223 138 879 897 6;
  • 63) 0.223 138 879 897 6 × 2 = 0 + 0.446 277 759 795 2;
  • 64) 0.446 277 759 795 2 × 2 = 0 + 0.892 555 519 590 4;
  • 65) 0.892 555 519 590 4 × 2 = 1 + 0.785 111 039 180 8;
  • 66) 0.785 111 039 180 8 × 2 = 1 + 0.570 222 078 361 6;
  • 67) 0.570 222 078 361 6 × 2 = 1 + 0.140 444 156 723 2;

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:


0.000 035 666 846 9(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1111 0000 1110 0100 1000 0101 0010 0000 111(2)

6. Positive number before normalization:

0.000 035 666 846 9(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1111 0000 1110 0100 1000 0101 0010 0000 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 035 666 846 9(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1111 0000 1110 0100 1000 0101 0010 0000 111(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1111 0000 1110 0100 1000 0101 0010 0000 111(2) × 20 =

1.0010 1011 0011 0001 1111 1000 0111 0010 0100 0010 1001 0000 0111(2) × 2-15


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): -15

Mantissa (not normalized):
1.0010 1011 0011 0001 1111 1000 0111 0010 0100 0010 1001 0000 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 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) =

1008(10) =

011 1111 0000(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. 0010 1011 0011 0001 1111 1000 0111 0010 0100 0010 1001 0000 0111 =

0010 1011 0011 0001 1111 1000 0111 0010 0100 0010 1001 0000 0111


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 0000

Mantissa (52 bits) =
0010 1011 0011 0001 1111 1000 0111 0010 0100 0010 1001 0000 0111


Decimal number -0.000 035 666 846 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 1000 0111 0010 0100 0010 1001 0000 0111

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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