-0.000 035 666 835 262 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 835 262(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
-0.000 035 666 835 262(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:

|-0.000 035 666 835 262| = 0.000 035 666 835 262


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: 0.000 035 666 835 262.

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) 0.000 035 666 835 262 × 2 = 0 + 0.000 071 333 670 524;
  • 2) 0.000 071 333 670 524 × 2 = 0 + 0.000 142 667 341 048;
  • 3) 0.000 142 667 341 048 × 2 = 0 + 0.000 285 334 682 096;
  • 4) 0.000 285 334 682 096 × 2 = 0 + 0.000 570 669 364 192;
  • 5) 0.000 570 669 364 192 × 2 = 0 + 0.001 141 338 728 384;
  • 6) 0.001 141 338 728 384 × 2 = 0 + 0.002 282 677 456 768;
  • 7) 0.002 282 677 456 768 × 2 = 0 + 0.004 565 354 913 536;
  • 8) 0.004 565 354 913 536 × 2 = 0 + 0.009 130 709 827 072;
  • 9) 0.009 130 709 827 072 × 2 = 0 + 0.018 261 419 654 144;
  • 10) 0.018 261 419 654 144 × 2 = 0 + 0.036 522 839 308 288;
  • 11) 0.036 522 839 308 288 × 2 = 0 + 0.073 045 678 616 576;
  • 12) 0.073 045 678 616 576 × 2 = 0 + 0.146 091 357 233 152;
  • 13) 0.146 091 357 233 152 × 2 = 0 + 0.292 182 714 466 304;
  • 14) 0.292 182 714 466 304 × 2 = 0 + 0.584 365 428 932 608;
  • 15) 0.584 365 428 932 608 × 2 = 1 + 0.168 730 857 865 216;
  • 16) 0.168 730 857 865 216 × 2 = 0 + 0.337 461 715 730 432;
  • 17) 0.337 461 715 730 432 × 2 = 0 + 0.674 923 431 460 864;
  • 18) 0.674 923 431 460 864 × 2 = 1 + 0.349 846 862 921 728;
  • 19) 0.349 846 862 921 728 × 2 = 0 + 0.699 693 725 843 456;
  • 20) 0.699 693 725 843 456 × 2 = 1 + 0.399 387 451 686 912;
  • 21) 0.399 387 451 686 912 × 2 = 0 + 0.798 774 903 373 824;
  • 22) 0.798 774 903 373 824 × 2 = 1 + 0.597 549 806 747 648;
  • 23) 0.597 549 806 747 648 × 2 = 1 + 0.195 099 613 495 296;
  • 24) 0.195 099 613 495 296 × 2 = 0 + 0.390 199 226 990 592;
  • 25) 0.390 199 226 990 592 × 2 = 0 + 0.780 398 453 981 184;
  • 26) 0.780 398 453 981 184 × 2 = 1 + 0.560 796 907 962 368;
  • 27) 0.560 796 907 962 368 × 2 = 1 + 0.121 593 815 924 736;
  • 28) 0.121 593 815 924 736 × 2 = 0 + 0.243 187 631 849 472;
  • 29) 0.243 187 631 849 472 × 2 = 0 + 0.486 375 263 698 944;
  • 30) 0.486 375 263 698 944 × 2 = 0 + 0.972 750 527 397 888;
  • 31) 0.972 750 527 397 888 × 2 = 1 + 0.945 501 054 795 776;
  • 32) 0.945 501 054 795 776 × 2 = 1 + 0.891 002 109 591 552;
  • 33) 0.891 002 109 591 552 × 2 = 1 + 0.782 004 219 183 104;
  • 34) 0.782 004 219 183 104 × 2 = 1 + 0.564 008 438 366 208;
  • 35) 0.564 008 438 366 208 × 2 = 1 + 0.128 016 876 732 416;
  • 36) 0.128 016 876 732 416 × 2 = 0 + 0.256 033 753 464 832;
  • 37) 0.256 033 753 464 832 × 2 = 0 + 0.512 067 506 929 664;
  • 38) 0.512 067 506 929 664 × 2 = 1 + 0.024 135 013 859 328;
  • 39) 0.024 135 013 859 328 × 2 = 0 + 0.048 270 027 718 656;
  • 40) 0.048 270 027 718 656 × 2 = 0 + 0.096 540 055 437 312;
  • 41) 0.096 540 055 437 312 × 2 = 0 + 0.193 080 110 874 624;
  • 42) 0.193 080 110 874 624 × 2 = 0 + 0.386 160 221 749 248;
  • 43) 0.386 160 221 749 248 × 2 = 0 + 0.772 320 443 498 496;
  • 44) 0.772 320 443 498 496 × 2 = 1 + 0.544 640 886 996 992;
  • 45) 0.544 640 886 996 992 × 2 = 1 + 0.089 281 773 993 984;
  • 46) 0.089 281 773 993 984 × 2 = 0 + 0.178 563 547 987 968;
  • 47) 0.178 563 547 987 968 × 2 = 0 + 0.357 127 095 975 936;
  • 48) 0.357 127 095 975 936 × 2 = 0 + 0.714 254 191 951 872;
  • 49) 0.714 254 191 951 872 × 2 = 1 + 0.428 508 383 903 744;
  • 50) 0.428 508 383 903 744 × 2 = 0 + 0.857 016 767 807 488;
  • 51) 0.857 016 767 807 488 × 2 = 1 + 0.714 033 535 614 976;
  • 52) 0.714 033 535 614 976 × 2 = 1 + 0.428 067 071 229 952;
  • 53) 0.428 067 071 229 952 × 2 = 0 + 0.856 134 142 459 904;
  • 54) 0.856 134 142 459 904 × 2 = 1 + 0.712 268 284 919 808;
  • 55) 0.712 268 284 919 808 × 2 = 1 + 0.424 536 569 839 616;
  • 56) 0.424 536 569 839 616 × 2 = 0 + 0.849 073 139 679 232;
  • 57) 0.849 073 139 679 232 × 2 = 1 + 0.698 146 279 358 464;
  • 58) 0.698 146 279 358 464 × 2 = 1 + 0.396 292 558 716 928;
  • 59) 0.396 292 558 716 928 × 2 = 0 + 0.792 585 117 433 856;
  • 60) 0.792 585 117 433 856 × 2 = 1 + 0.585 170 234 867 712;
  • 61) 0.585 170 234 867 712 × 2 = 1 + 0.170 340 469 735 424;
  • 62) 0.170 340 469 735 424 × 2 = 0 + 0.340 680 939 470 848;
  • 63) 0.340 680 939 470 848 × 2 = 0 + 0.681 361 878 941 696;
  • 64) 0.681 361 878 941 696 × 2 = 1 + 0.362 723 757 883 392;
  • 65) 0.362 723 757 883 392 × 2 = 0 + 0.725 447 515 766 784;
  • 66) 0.725 447 515 766 784 × 2 = 1 + 0.450 895 031 533 568;
  • 67) 0.450 895 031 533 568 × 2 = 0 + 0.901 790 063 067 136;

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:


0.000 035 666 835 262(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0001 1000 1011 0110 1101 1001 010(2)

6. Positive number before normalization:

0.000 035 666 835 262(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0001 1000 1011 0110 1101 1001 010(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:


0.000 035 666 835 262(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0001 1000 1011 0110 1101 1001 010(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0001 1000 1011 0110 1101 1001 010(2) × 20 =

1.0010 1011 0011 0001 1111 0010 0000 1100 0101 1011 0110 1100 1010(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 0010 0000 1100 0101 1011 0110 1100 1010


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 0010 0000 1100 0101 1011 0110 1100 1010 =

0010 1011 0011 0001 1111 0010 0000 1100 0101 1011 0110 1100 1010


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 0010 0000 1100 0101 1011 0110 1100 1010


Decimal number -0.000 035 666 835 262 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 0010 0000 1100 0101 1011 0110 1100 1010

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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