-0.000 035 666 865 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 865(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
-0.000 035 666 865(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:

|-0.000 035 666 865| = 0.000 035 666 865


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: 0.000 035 666 865.

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) 0.000 035 666 865 × 2 = 0 + 0.000 071 333 73;
  • 2) 0.000 071 333 73 × 2 = 0 + 0.000 142 667 46;
  • 3) 0.000 142 667 46 × 2 = 0 + 0.000 285 334 92;
  • 4) 0.000 285 334 92 × 2 = 0 + 0.000 570 669 84;
  • 5) 0.000 570 669 84 × 2 = 0 + 0.001 141 339 68;
  • 6) 0.001 141 339 68 × 2 = 0 + 0.002 282 679 36;
  • 7) 0.002 282 679 36 × 2 = 0 + 0.004 565 358 72;
  • 8) 0.004 565 358 72 × 2 = 0 + 0.009 130 717 44;
  • 9) 0.009 130 717 44 × 2 = 0 + 0.018 261 434 88;
  • 10) 0.018 261 434 88 × 2 = 0 + 0.036 522 869 76;
  • 11) 0.036 522 869 76 × 2 = 0 + 0.073 045 739 52;
  • 12) 0.073 045 739 52 × 2 = 0 + 0.146 091 479 04;
  • 13) 0.146 091 479 04 × 2 = 0 + 0.292 182 958 08;
  • 14) 0.292 182 958 08 × 2 = 0 + 0.584 365 916 16;
  • 15) 0.584 365 916 16 × 2 = 1 + 0.168 731 832 32;
  • 16) 0.168 731 832 32 × 2 = 0 + 0.337 463 664 64;
  • 17) 0.337 463 664 64 × 2 = 0 + 0.674 927 329 28;
  • 18) 0.674 927 329 28 × 2 = 1 + 0.349 854 658 56;
  • 19) 0.349 854 658 56 × 2 = 0 + 0.699 709 317 12;
  • 20) 0.699 709 317 12 × 2 = 1 + 0.399 418 634 24;
  • 21) 0.399 418 634 24 × 2 = 0 + 0.798 837 268 48;
  • 22) 0.798 837 268 48 × 2 = 1 + 0.597 674 536 96;
  • 23) 0.597 674 536 96 × 2 = 1 + 0.195 349 073 92;
  • 24) 0.195 349 073 92 × 2 = 0 + 0.390 698 147 84;
  • 25) 0.390 698 147 84 × 2 = 0 + 0.781 396 295 68;
  • 26) 0.781 396 295 68 × 2 = 1 + 0.562 792 591 36;
  • 27) 0.562 792 591 36 × 2 = 1 + 0.125 585 182 72;
  • 28) 0.125 585 182 72 × 2 = 0 + 0.251 170 365 44;
  • 29) 0.251 170 365 44 × 2 = 0 + 0.502 340 730 88;
  • 30) 0.502 340 730 88 × 2 = 1 + 0.004 681 461 76;
  • 31) 0.004 681 461 76 × 2 = 0 + 0.009 362 923 52;
  • 32) 0.009 362 923 52 × 2 = 0 + 0.018 725 847 04;
  • 33) 0.018 725 847 04 × 2 = 0 + 0.037 451 694 08;
  • 34) 0.037 451 694 08 × 2 = 0 + 0.074 903 388 16;
  • 35) 0.074 903 388 16 × 2 = 0 + 0.149 806 776 32;
  • 36) 0.149 806 776 32 × 2 = 0 + 0.299 613 552 64;
  • 37) 0.299 613 552 64 × 2 = 0 + 0.599 227 105 28;
  • 38) 0.599 227 105 28 × 2 = 1 + 0.198 454 210 56;
  • 39) 0.198 454 210 56 × 2 = 0 + 0.396 908 421 12;
  • 40) 0.396 908 421 12 × 2 = 0 + 0.793 816 842 24;
  • 41) 0.793 816 842 24 × 2 = 1 + 0.587 633 684 48;
  • 42) 0.587 633 684 48 × 2 = 1 + 0.175 267 368 96;
  • 43) 0.175 267 368 96 × 2 = 0 + 0.350 534 737 92;
  • 44) 0.350 534 737 92 × 2 = 0 + 0.701 069 475 84;
  • 45) 0.701 069 475 84 × 2 = 1 + 0.402 138 951 68;
  • 46) 0.402 138 951 68 × 2 = 0 + 0.804 277 903 36;
  • 47) 0.804 277 903 36 × 2 = 1 + 0.608 555 806 72;
  • 48) 0.608 555 806 72 × 2 = 1 + 0.217 111 613 44;
  • 49) 0.217 111 613 44 × 2 = 0 + 0.434 223 226 88;
  • 50) 0.434 223 226 88 × 2 = 0 + 0.868 446 453 76;
  • 51) 0.868 446 453 76 × 2 = 1 + 0.736 892 907 52;
  • 52) 0.736 892 907 52 × 2 = 1 + 0.473 785 815 04;
  • 53) 0.473 785 815 04 × 2 = 0 + 0.947 571 630 08;
  • 54) 0.947 571 630 08 × 2 = 1 + 0.895 143 260 16;
  • 55) 0.895 143 260 16 × 2 = 1 + 0.790 286 520 32;
  • 56) 0.790 286 520 32 × 2 = 1 + 0.580 573 040 64;
  • 57) 0.580 573 040 64 × 2 = 1 + 0.161 146 081 28;
  • 58) 0.161 146 081 28 × 2 = 0 + 0.322 292 162 56;
  • 59) 0.322 292 162 56 × 2 = 0 + 0.644 584 325 12;
  • 60) 0.644 584 325 12 × 2 = 1 + 0.289 168 650 24;
  • 61) 0.289 168 650 24 × 2 = 0 + 0.578 337 300 48;
  • 62) 0.578 337 300 48 × 2 = 1 + 0.156 674 600 96;
  • 63) 0.156 674 600 96 × 2 = 0 + 0.313 349 201 92;
  • 64) 0.313 349 201 92 × 2 = 0 + 0.626 698 403 84;
  • 65) 0.626 698 403 84 × 2 = 1 + 0.253 396 807 68;
  • 66) 0.253 396 807 68 × 2 = 0 + 0.506 793 615 36;
  • 67) 0.506 793 615 36 × 2 = 1 + 0.013 587 230 72;

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:


0.000 035 666 865(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0000 0100 1100 1011 0011 0111 1001 0100 101(2)

6. Positive number before normalization:

0.000 035 666 865(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0000 0100 1100 1011 0011 0111 1001 0100 101(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:


0.000 035 666 865(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0000 0100 1100 1011 0011 0111 1001 0100 101(2) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0000 0100 1100 1011 0011 0111 1001 0100 101(2) × 20 =

1.0010 1011 0011 0010 0000 0010 0110 0101 1001 1011 1100 1010 0101(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 0010 0000 0010 0110 0101 1001 1011 1100 1010 0101


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 0010 0000 0010 0110 0101 1001 1011 1100 1010 0101 =

0010 1011 0011 0010 0000 0010 0110 0101 1001 1011 1100 1010 0101


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 0010 0000 0010 0110 0101 1001 1011 1100 1010 0101


Decimal number -0.000 035 666 865 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0010 0000 0010 0110 0101 1001 1011 1100 1010 0101

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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