-0.000 035 666 835 254 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 835 254(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
-0.000 035 666 835 254(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:

|-0.000 035 666 835 254| = 0.000 035 666 835 254


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: 0.000 035 666 835 254.

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) 0.000 035 666 835 254 × 2 = 0 + 0.000 071 333 670 508;
  • 2) 0.000 071 333 670 508 × 2 = 0 + 0.000 142 667 341 016;
  • 3) 0.000 142 667 341 016 × 2 = 0 + 0.000 285 334 682 032;
  • 4) 0.000 285 334 682 032 × 2 = 0 + 0.000 570 669 364 064;
  • 5) 0.000 570 669 364 064 × 2 = 0 + 0.001 141 338 728 128;
  • 6) 0.001 141 338 728 128 × 2 = 0 + 0.002 282 677 456 256;
  • 7) 0.002 282 677 456 256 × 2 = 0 + 0.004 565 354 912 512;
  • 8) 0.004 565 354 912 512 × 2 = 0 + 0.009 130 709 825 024;
  • 9) 0.009 130 709 825 024 × 2 = 0 + 0.018 261 419 650 048;
  • 10) 0.018 261 419 650 048 × 2 = 0 + 0.036 522 839 300 096;
  • 11) 0.036 522 839 300 096 × 2 = 0 + 0.073 045 678 600 192;
  • 12) 0.073 045 678 600 192 × 2 = 0 + 0.146 091 357 200 384;
  • 13) 0.146 091 357 200 384 × 2 = 0 + 0.292 182 714 400 768;
  • 14) 0.292 182 714 400 768 × 2 = 0 + 0.584 365 428 801 536;
  • 15) 0.584 365 428 801 536 × 2 = 1 + 0.168 730 857 603 072;
  • 16) 0.168 730 857 603 072 × 2 = 0 + 0.337 461 715 206 144;
  • 17) 0.337 461 715 206 144 × 2 = 0 + 0.674 923 430 412 288;
  • 18) 0.674 923 430 412 288 × 2 = 1 + 0.349 846 860 824 576;
  • 19) 0.349 846 860 824 576 × 2 = 0 + 0.699 693 721 649 152;
  • 20) 0.699 693 721 649 152 × 2 = 1 + 0.399 387 443 298 304;
  • 21) 0.399 387 443 298 304 × 2 = 0 + 0.798 774 886 596 608;
  • 22) 0.798 774 886 596 608 × 2 = 1 + 0.597 549 773 193 216;
  • 23) 0.597 549 773 193 216 × 2 = 1 + 0.195 099 546 386 432;
  • 24) 0.195 099 546 386 432 × 2 = 0 + 0.390 199 092 772 864;
  • 25) 0.390 199 092 772 864 × 2 = 0 + 0.780 398 185 545 728;
  • 26) 0.780 398 185 545 728 × 2 = 1 + 0.560 796 371 091 456;
  • 27) 0.560 796 371 091 456 × 2 = 1 + 0.121 592 742 182 912;
  • 28) 0.121 592 742 182 912 × 2 = 0 + 0.243 185 484 365 824;
  • 29) 0.243 185 484 365 824 × 2 = 0 + 0.486 370 968 731 648;
  • 30) 0.486 370 968 731 648 × 2 = 0 + 0.972 741 937 463 296;
  • 31) 0.972 741 937 463 296 × 2 = 1 + 0.945 483 874 926 592;
  • 32) 0.945 483 874 926 592 × 2 = 1 + 0.890 967 749 853 184;
  • 33) 0.890 967 749 853 184 × 2 = 1 + 0.781 935 499 706 368;
  • 34) 0.781 935 499 706 368 × 2 = 1 + 0.563 870 999 412 736;
  • 35) 0.563 870 999 412 736 × 2 = 1 + 0.127 741 998 825 472;
  • 36) 0.127 741 998 825 472 × 2 = 0 + 0.255 483 997 650 944;
  • 37) 0.255 483 997 650 944 × 2 = 0 + 0.510 967 995 301 888;
  • 38) 0.510 967 995 301 888 × 2 = 1 + 0.021 935 990 603 776;
  • 39) 0.021 935 990 603 776 × 2 = 0 + 0.043 871 981 207 552;
  • 40) 0.043 871 981 207 552 × 2 = 0 + 0.087 743 962 415 104;
  • 41) 0.087 743 962 415 104 × 2 = 0 + 0.175 487 924 830 208;
  • 42) 0.175 487 924 830 208 × 2 = 0 + 0.350 975 849 660 416;
  • 43) 0.350 975 849 660 416 × 2 = 0 + 0.701 951 699 320 832;
  • 44) 0.701 951 699 320 832 × 2 = 1 + 0.403 903 398 641 664;
  • 45) 0.403 903 398 641 664 × 2 = 0 + 0.807 806 797 283 328;
  • 46) 0.807 806 797 283 328 × 2 = 1 + 0.615 613 594 566 656;
  • 47) 0.615 613 594 566 656 × 2 = 1 + 0.231 227 189 133 312;
  • 48) 0.231 227 189 133 312 × 2 = 0 + 0.462 454 378 266 624;
  • 49) 0.462 454 378 266 624 × 2 = 0 + 0.924 908 756 533 248;
  • 50) 0.924 908 756 533 248 × 2 = 1 + 0.849 817 513 066 496;
  • 51) 0.849 817 513 066 496 × 2 = 1 + 0.699 635 026 132 992;
  • 52) 0.699 635 026 132 992 × 2 = 1 + 0.399 270 052 265 984;
  • 53) 0.399 270 052 265 984 × 2 = 0 + 0.798 540 104 531 968;
  • 54) 0.798 540 104 531 968 × 2 = 1 + 0.597 080 209 063 936;
  • 55) 0.597 080 209 063 936 × 2 = 1 + 0.194 160 418 127 872;
  • 56) 0.194 160 418 127 872 × 2 = 0 + 0.388 320 836 255 744;
  • 57) 0.388 320 836 255 744 × 2 = 0 + 0.776 641 672 511 488;
  • 58) 0.776 641 672 511 488 × 2 = 1 + 0.553 283 345 022 976;
  • 59) 0.553 283 345 022 976 × 2 = 1 + 0.106 566 690 045 952;
  • 60) 0.106 566 690 045 952 × 2 = 0 + 0.213 133 380 091 904;
  • 61) 0.213 133 380 091 904 × 2 = 0 + 0.426 266 760 183 808;
  • 62) 0.426 266 760 183 808 × 2 = 0 + 0.852 533 520 367 616;
  • 63) 0.852 533 520 367 616 × 2 = 1 + 0.705 067 040 735 232;
  • 64) 0.705 067 040 735 232 × 2 = 1 + 0.410 134 081 470 464;
  • 65) 0.410 134 081 470 464 × 2 = 0 + 0.820 268 162 940 928;
  • 66) 0.820 268 162 940 928 × 2 = 1 + 0.640 536 325 881 856;
  • 67) 0.640 536 325 881 856 × 2 = 1 + 0.281 072 651 763 712;

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:


0.000 035 666 835 254(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0001 0110 0111 0110 0110 0011 011(2)

6. Positive number before normalization:

0.000 035 666 835 254(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0001 0110 0111 0110 0110 0011 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:


0.000 035 666 835 254(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0001 0110 0111 0110 0110 0011 011(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0001 0110 0111 0110 0110 0011 011(2) × 20 =

1.0010 1011 0011 0001 1111 0010 0000 1011 0011 1011 0011 0001 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 0001 1111 0010 0000 1011 0011 1011 0011 0001 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 0001 1111 0010 0000 1011 0011 1011 0011 0001 1011 =

0010 1011 0011 0001 1111 0010 0000 1011 0011 1011 0011 0001 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 0001 1111 0010 0000 1011 0011 1011 0011 0001 1011


Decimal number -0.000 035 666 835 254 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 0010 0000 1011 0011 1011 0011 0001 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