-0.000 035 666 927 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 927(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
-0.000 035 666 927(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:

|-0.000 035 666 927| = 0.000 035 666 927


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: 0.000 035 666 927.

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) 0.000 035 666 927 × 2 = 0 + 0.000 071 333 854;
  • 2) 0.000 071 333 854 × 2 = 0 + 0.000 142 667 708;
  • 3) 0.000 142 667 708 × 2 = 0 + 0.000 285 335 416;
  • 4) 0.000 285 335 416 × 2 = 0 + 0.000 570 670 832;
  • 5) 0.000 570 670 832 × 2 = 0 + 0.001 141 341 664;
  • 6) 0.001 141 341 664 × 2 = 0 + 0.002 282 683 328;
  • 7) 0.002 282 683 328 × 2 = 0 + 0.004 565 366 656;
  • 8) 0.004 565 366 656 × 2 = 0 + 0.009 130 733 312;
  • 9) 0.009 130 733 312 × 2 = 0 + 0.018 261 466 624;
  • 10) 0.018 261 466 624 × 2 = 0 + 0.036 522 933 248;
  • 11) 0.036 522 933 248 × 2 = 0 + 0.073 045 866 496;
  • 12) 0.073 045 866 496 × 2 = 0 + 0.146 091 732 992;
  • 13) 0.146 091 732 992 × 2 = 0 + 0.292 183 465 984;
  • 14) 0.292 183 465 984 × 2 = 0 + 0.584 366 931 968;
  • 15) 0.584 366 931 968 × 2 = 1 + 0.168 733 863 936;
  • 16) 0.168 733 863 936 × 2 = 0 + 0.337 467 727 872;
  • 17) 0.337 467 727 872 × 2 = 0 + 0.674 935 455 744;
  • 18) 0.674 935 455 744 × 2 = 1 + 0.349 870 911 488;
  • 19) 0.349 870 911 488 × 2 = 0 + 0.699 741 822 976;
  • 20) 0.699 741 822 976 × 2 = 1 + 0.399 483 645 952;
  • 21) 0.399 483 645 952 × 2 = 0 + 0.798 967 291 904;
  • 22) 0.798 967 291 904 × 2 = 1 + 0.597 934 583 808;
  • 23) 0.597 934 583 808 × 2 = 1 + 0.195 869 167 616;
  • 24) 0.195 869 167 616 × 2 = 0 + 0.391 738 335 232;
  • 25) 0.391 738 335 232 × 2 = 0 + 0.783 476 670 464;
  • 26) 0.783 476 670 464 × 2 = 1 + 0.566 953 340 928;
  • 27) 0.566 953 340 928 × 2 = 1 + 0.133 906 681 856;
  • 28) 0.133 906 681 856 × 2 = 0 + 0.267 813 363 712;
  • 29) 0.267 813 363 712 × 2 = 0 + 0.535 626 727 424;
  • 30) 0.535 626 727 424 × 2 = 1 + 0.071 253 454 848;
  • 31) 0.071 253 454 848 × 2 = 0 + 0.142 506 909 696;
  • 32) 0.142 506 909 696 × 2 = 0 + 0.285 013 819 392;
  • 33) 0.285 013 819 392 × 2 = 0 + 0.570 027 638 784;
  • 34) 0.570 027 638 784 × 2 = 1 + 0.140 055 277 568;
  • 35) 0.140 055 277 568 × 2 = 0 + 0.280 110 555 136;
  • 36) 0.280 110 555 136 × 2 = 0 + 0.560 221 110 272;
  • 37) 0.560 221 110 272 × 2 = 1 + 0.120 442 220 544;
  • 38) 0.120 442 220 544 × 2 = 0 + 0.240 884 441 088;
  • 39) 0.240 884 441 088 × 2 = 0 + 0.481 768 882 176;
  • 40) 0.481 768 882 176 × 2 = 0 + 0.963 537 764 352;
  • 41) 0.963 537 764 352 × 2 = 1 + 0.927 075 528 704;
  • 42) 0.927 075 528 704 × 2 = 1 + 0.854 151 057 408;
  • 43) 0.854 151 057 408 × 2 = 1 + 0.708 302 114 816;
  • 44) 0.708 302 114 816 × 2 = 1 + 0.416 604 229 632;
  • 45) 0.416 604 229 632 × 2 = 0 + 0.833 208 459 264;
  • 46) 0.833 208 459 264 × 2 = 1 + 0.666 416 918 528;
  • 47) 0.666 416 918 528 × 2 = 1 + 0.332 833 837 056;
  • 48) 0.332 833 837 056 × 2 = 0 + 0.665 667 674 112;
  • 49) 0.665 667 674 112 × 2 = 1 + 0.331 335 348 224;
  • 50) 0.331 335 348 224 × 2 = 0 + 0.662 670 696 448;
  • 51) 0.662 670 696 448 × 2 = 1 + 0.325 341 392 896;
  • 52) 0.325 341 392 896 × 2 = 0 + 0.650 682 785 792;
  • 53) 0.650 682 785 792 × 2 = 1 + 0.301 365 571 584;
  • 54) 0.301 365 571 584 × 2 = 0 + 0.602 731 143 168;
  • 55) 0.602 731 143 168 × 2 = 1 + 0.205 462 286 336;
  • 56) 0.205 462 286 336 × 2 = 0 + 0.410 924 572 672;
  • 57) 0.410 924 572 672 × 2 = 0 + 0.821 849 145 344;
  • 58) 0.821 849 145 344 × 2 = 1 + 0.643 698 290 688;
  • 59) 0.643 698 290 688 × 2 = 1 + 0.287 396 581 376;
  • 60) 0.287 396 581 376 × 2 = 0 + 0.574 793 162 752;
  • 61) 0.574 793 162 752 × 2 = 1 + 0.149 586 325 504;
  • 62) 0.149 586 325 504 × 2 = 0 + 0.299 172 651 008;
  • 63) 0.299 172 651 008 × 2 = 0 + 0.598 345 302 016;
  • 64) 0.598 345 302 016 × 2 = 1 + 0.196 690 604 032;
  • 65) 0.196 690 604 032 × 2 = 0 + 0.393 381 208 064;
  • 66) 0.393 381 208 064 × 2 = 0 + 0.786 762 416 128;
  • 67) 0.786 762 416 128 × 2 = 1 + 0.573 524 832 256;

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:


0.000 035 666 927(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0100 1000 1111 0110 1010 1010 0110 1001 001(2)

6. Positive number before normalization:

0.000 035 666 927(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0100 1000 1111 0110 1010 1010 0110 1001 001(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:


0.000 035 666 927(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0100 1000 1111 0110 1010 1010 0110 1001 001(2) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0100 1000 1111 0110 1010 1010 0110 1001 001(2) × 20 =

1.0010 1011 0011 0010 0010 0100 0111 1011 0101 0101 0011 0100 1001(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 0100 0111 1011 0101 0101 0011 0100 1001


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 0100 0111 1011 0101 0101 0011 0100 1001 =

0010 1011 0011 0010 0010 0100 0111 1011 0101 0101 0011 0100 1001


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 0100 0111 1011 0101 0101 0011 0100 1001


Decimal number -0.000 035 666 927 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0010 0010 0100 0111 1011 0101 0101 0011 0100 1001

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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