-0.000 035 666 93 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 93(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 93(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 93| = 0.000 035 666 93


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 93.

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 93 × 2 = 0 + 0.000 071 333 86;
  • 2) 0.000 071 333 86 × 2 = 0 + 0.000 142 667 72;
  • 3) 0.000 142 667 72 × 2 = 0 + 0.000 285 335 44;
  • 4) 0.000 285 335 44 × 2 = 0 + 0.000 570 670 88;
  • 5) 0.000 570 670 88 × 2 = 0 + 0.001 141 341 76;
  • 6) 0.001 141 341 76 × 2 = 0 + 0.002 282 683 52;
  • 7) 0.002 282 683 52 × 2 = 0 + 0.004 565 367 04;
  • 8) 0.004 565 367 04 × 2 = 0 + 0.009 130 734 08;
  • 9) 0.009 130 734 08 × 2 = 0 + 0.018 261 468 16;
  • 10) 0.018 261 468 16 × 2 = 0 + 0.036 522 936 32;
  • 11) 0.036 522 936 32 × 2 = 0 + 0.073 045 872 64;
  • 12) 0.073 045 872 64 × 2 = 0 + 0.146 091 745 28;
  • 13) 0.146 091 745 28 × 2 = 0 + 0.292 183 490 56;
  • 14) 0.292 183 490 56 × 2 = 0 + 0.584 366 981 12;
  • 15) 0.584 366 981 12 × 2 = 1 + 0.168 733 962 24;
  • 16) 0.168 733 962 24 × 2 = 0 + 0.337 467 924 48;
  • 17) 0.337 467 924 48 × 2 = 0 + 0.674 935 848 96;
  • 18) 0.674 935 848 96 × 2 = 1 + 0.349 871 697 92;
  • 19) 0.349 871 697 92 × 2 = 0 + 0.699 743 395 84;
  • 20) 0.699 743 395 84 × 2 = 1 + 0.399 486 791 68;
  • 21) 0.399 486 791 68 × 2 = 0 + 0.798 973 583 36;
  • 22) 0.798 973 583 36 × 2 = 1 + 0.597 947 166 72;
  • 23) 0.597 947 166 72 × 2 = 1 + 0.195 894 333 44;
  • 24) 0.195 894 333 44 × 2 = 0 + 0.391 788 666 88;
  • 25) 0.391 788 666 88 × 2 = 0 + 0.783 577 333 76;
  • 26) 0.783 577 333 76 × 2 = 1 + 0.567 154 667 52;
  • 27) 0.567 154 667 52 × 2 = 1 + 0.134 309 335 04;
  • 28) 0.134 309 335 04 × 2 = 0 + 0.268 618 670 08;
  • 29) 0.268 618 670 08 × 2 = 0 + 0.537 237 340 16;
  • 30) 0.537 237 340 16 × 2 = 1 + 0.074 474 680 32;
  • 31) 0.074 474 680 32 × 2 = 0 + 0.148 949 360 64;
  • 32) 0.148 949 360 64 × 2 = 0 + 0.297 898 721 28;
  • 33) 0.297 898 721 28 × 2 = 0 + 0.595 797 442 56;
  • 34) 0.595 797 442 56 × 2 = 1 + 0.191 594 885 12;
  • 35) 0.191 594 885 12 × 2 = 0 + 0.383 189 770 24;
  • 36) 0.383 189 770 24 × 2 = 0 + 0.766 379 540 48;
  • 37) 0.766 379 540 48 × 2 = 1 + 0.532 759 080 96;
  • 38) 0.532 759 080 96 × 2 = 1 + 0.065 518 161 92;
  • 39) 0.065 518 161 92 × 2 = 0 + 0.131 036 323 84;
  • 40) 0.131 036 323 84 × 2 = 0 + 0.262 072 647 68;
  • 41) 0.262 072 647 68 × 2 = 0 + 0.524 145 295 36;
  • 42) 0.524 145 295 36 × 2 = 1 + 0.048 290 590 72;
  • 43) 0.048 290 590 72 × 2 = 0 + 0.096 581 181 44;
  • 44) 0.096 581 181 44 × 2 = 0 + 0.193 162 362 88;
  • 45) 0.193 162 362 88 × 2 = 0 + 0.386 324 725 76;
  • 46) 0.386 324 725 76 × 2 = 0 + 0.772 649 451 52;
  • 47) 0.772 649 451 52 × 2 = 1 + 0.545 298 903 04;
  • 48) 0.545 298 903 04 × 2 = 1 + 0.090 597 806 08;
  • 49) 0.090 597 806 08 × 2 = 0 + 0.181 195 612 16;
  • 50) 0.181 195 612 16 × 2 = 0 + 0.362 391 224 32;
  • 51) 0.362 391 224 32 × 2 = 0 + 0.724 782 448 64;
  • 52) 0.724 782 448 64 × 2 = 1 + 0.449 564 897 28;
  • 53) 0.449 564 897 28 × 2 = 0 + 0.899 129 794 56;
  • 54) 0.899 129 794 56 × 2 = 1 + 0.798 259 589 12;
  • 55) 0.798 259 589 12 × 2 = 1 + 0.596 519 178 24;
  • 56) 0.596 519 178 24 × 2 = 1 + 0.193 038 356 48;
  • 57) 0.193 038 356 48 × 2 = 0 + 0.386 076 712 96;
  • 58) 0.386 076 712 96 × 2 = 0 + 0.772 153 425 92;
  • 59) 0.772 153 425 92 × 2 = 1 + 0.544 306 851 84;
  • 60) 0.544 306 851 84 × 2 = 1 + 0.088 613 703 68;
  • 61) 0.088 613 703 68 × 2 = 0 + 0.177 227 407 36;
  • 62) 0.177 227 407 36 × 2 = 0 + 0.354 454 814 72;
  • 63) 0.354 454 814 72 × 2 = 0 + 0.708 909 629 44;
  • 64) 0.708 909 629 44 × 2 = 1 + 0.417 819 258 88;
  • 65) 0.417 819 258 88 × 2 = 0 + 0.835 638 517 76;
  • 66) 0.835 638 517 76 × 2 = 1 + 0.671 277 035 52;
  • 67) 0.671 277 035 52 × 2 = 1 + 0.342 554 071 04;

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 93(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0100 1100 0100 0011 0001 0111 0011 0001 011(2)

6. Positive number before normalization:

0.000 035 666 93(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0100 1100 0100 0011 0001 0111 0011 0001 011(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 93(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0100 1100 0100 0011 0001 0111 0011 0001 011(2) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0100 1100 0100 0011 0001 0111 0011 0001 011(2) × 20 =

1.0010 1011 0011 0010 0010 0110 0010 0001 1000 1011 1001 1000 1011(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 0010 0110 0010 0001 1000 1011 1001 1000 1011


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 0010 0110 0010 0001 1000 1011 1001 1000 1011 =

0010 1011 0011 0010 0010 0110 0010 0001 1000 1011 1001 1000 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 0000

Mantissa (52 bits) =
0010 1011 0011 0010 0010 0110 0010 0001 1000 1011 1001 1000 1011


Decimal number -0.000 035 666 93 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0010 0010 0110 0010 0001 1000 1011 1001 1000 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