0.000 000 000 000 217 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 217(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
0.000 000 000 000 217(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 217.

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) 0.000 000 000 000 217 × 2 = 0 + 0.000 000 000 000 434;
  • 2) 0.000 000 000 000 434 × 2 = 0 + 0.000 000 000 000 868;
  • 3) 0.000 000 000 000 868 × 2 = 0 + 0.000 000 000 001 736;
  • 4) 0.000 000 000 001 736 × 2 = 0 + 0.000 000 000 003 472;
  • 5) 0.000 000 000 003 472 × 2 = 0 + 0.000 000 000 006 944;
  • 6) 0.000 000 000 006 944 × 2 = 0 + 0.000 000 000 013 888;
  • 7) 0.000 000 000 013 888 × 2 = 0 + 0.000 000 000 027 776;
  • 8) 0.000 000 000 027 776 × 2 = 0 + 0.000 000 000 055 552;
  • 9) 0.000 000 000 055 552 × 2 = 0 + 0.000 000 000 111 104;
  • 10) 0.000 000 000 111 104 × 2 = 0 + 0.000 000 000 222 208;
  • 11) 0.000 000 000 222 208 × 2 = 0 + 0.000 000 000 444 416;
  • 12) 0.000 000 000 444 416 × 2 = 0 + 0.000 000 000 888 832;
  • 13) 0.000 000 000 888 832 × 2 = 0 + 0.000 000 001 777 664;
  • 14) 0.000 000 001 777 664 × 2 = 0 + 0.000 000 003 555 328;
  • 15) 0.000 000 003 555 328 × 2 = 0 + 0.000 000 007 110 656;
  • 16) 0.000 000 007 110 656 × 2 = 0 + 0.000 000 014 221 312;
  • 17) 0.000 000 014 221 312 × 2 = 0 + 0.000 000 028 442 624;
  • 18) 0.000 000 028 442 624 × 2 = 0 + 0.000 000 056 885 248;
  • 19) 0.000 000 056 885 248 × 2 = 0 + 0.000 000 113 770 496;
  • 20) 0.000 000 113 770 496 × 2 = 0 + 0.000 000 227 540 992;
  • 21) 0.000 000 227 540 992 × 2 = 0 + 0.000 000 455 081 984;
  • 22) 0.000 000 455 081 984 × 2 = 0 + 0.000 000 910 163 968;
  • 23) 0.000 000 910 163 968 × 2 = 0 + 0.000 001 820 327 936;
  • 24) 0.000 001 820 327 936 × 2 = 0 + 0.000 003 640 655 872;
  • 25) 0.000 003 640 655 872 × 2 = 0 + 0.000 007 281 311 744;
  • 26) 0.000 007 281 311 744 × 2 = 0 + 0.000 014 562 623 488;
  • 27) 0.000 014 562 623 488 × 2 = 0 + 0.000 029 125 246 976;
  • 28) 0.000 029 125 246 976 × 2 = 0 + 0.000 058 250 493 952;
  • 29) 0.000 058 250 493 952 × 2 = 0 + 0.000 116 500 987 904;
  • 30) 0.000 116 500 987 904 × 2 = 0 + 0.000 233 001 975 808;
  • 31) 0.000 233 001 975 808 × 2 = 0 + 0.000 466 003 951 616;
  • 32) 0.000 466 003 951 616 × 2 = 0 + 0.000 932 007 903 232;
  • 33) 0.000 932 007 903 232 × 2 = 0 + 0.001 864 015 806 464;
  • 34) 0.001 864 015 806 464 × 2 = 0 + 0.003 728 031 612 928;
  • 35) 0.003 728 031 612 928 × 2 = 0 + 0.007 456 063 225 856;
  • 36) 0.007 456 063 225 856 × 2 = 0 + 0.014 912 126 451 712;
  • 37) 0.014 912 126 451 712 × 2 = 0 + 0.029 824 252 903 424;
  • 38) 0.029 824 252 903 424 × 2 = 0 + 0.059 648 505 806 848;
  • 39) 0.059 648 505 806 848 × 2 = 0 + 0.119 297 011 613 696;
  • 40) 0.119 297 011 613 696 × 2 = 0 + 0.238 594 023 227 392;
  • 41) 0.238 594 023 227 392 × 2 = 0 + 0.477 188 046 454 784;
  • 42) 0.477 188 046 454 784 × 2 = 0 + 0.954 376 092 909 568;
  • 43) 0.954 376 092 909 568 × 2 = 1 + 0.908 752 185 819 136;
  • 44) 0.908 752 185 819 136 × 2 = 1 + 0.817 504 371 638 272;
  • 45) 0.817 504 371 638 272 × 2 = 1 + 0.635 008 743 276 544;
  • 46) 0.635 008 743 276 544 × 2 = 1 + 0.270 017 486 553 088;
  • 47) 0.270 017 486 553 088 × 2 = 0 + 0.540 034 973 106 176;
  • 48) 0.540 034 973 106 176 × 2 = 1 + 0.080 069 946 212 352;
  • 49) 0.080 069 946 212 352 × 2 = 0 + 0.160 139 892 424 704;
  • 50) 0.160 139 892 424 704 × 2 = 0 + 0.320 279 784 849 408;
  • 51) 0.320 279 784 849 408 × 2 = 0 + 0.640 559 569 698 816;
  • 52) 0.640 559 569 698 816 × 2 = 1 + 0.281 119 139 397 632;
  • 53) 0.281 119 139 397 632 × 2 = 0 + 0.562 238 278 795 264;
  • 54) 0.562 238 278 795 264 × 2 = 1 + 0.124 476 557 590 528;
  • 55) 0.124 476 557 590 528 × 2 = 0 + 0.248 953 115 181 056;
  • 56) 0.248 953 115 181 056 × 2 = 0 + 0.497 906 230 362 112;
  • 57) 0.497 906 230 362 112 × 2 = 0 + 0.995 812 460 724 224;
  • 58) 0.995 812 460 724 224 × 2 = 1 + 0.991 624 921 448 448;
  • 59) 0.991 624 921 448 448 × 2 = 1 + 0.983 249 842 896 896;
  • 60) 0.983 249 842 896 896 × 2 = 1 + 0.966 499 685 793 792;
  • 61) 0.966 499 685 793 792 × 2 = 1 + 0.932 999 371 587 584;
  • 62) 0.932 999 371 587 584 × 2 = 1 + 0.865 998 743 175 168;
  • 63) 0.865 998 743 175 168 × 2 = 1 + 0.731 997 486 350 336;
  • 64) 0.731 997 486 350 336 × 2 = 1 + 0.463 994 972 700 672;
  • 65) 0.463 994 972 700 672 × 2 = 0 + 0.927 989 945 401 344;
  • 66) 0.927 989 945 401 344 × 2 = 1 + 0.855 979 890 802 688;
  • 67) 0.855 979 890 802 688 × 2 = 1 + 0.711 959 781 605 376;
  • 68) 0.711 959 781 605 376 × 2 = 1 + 0.423 919 563 210 752;
  • 69) 0.423 919 563 210 752 × 2 = 0 + 0.847 839 126 421 504;
  • 70) 0.847 839 126 421 504 × 2 = 1 + 0.695 678 252 843 008;
  • 71) 0.695 678 252 843 008 × 2 = 1 + 0.391 356 505 686 016;
  • 72) 0.391 356 505 686 016 × 2 = 0 + 0.782 713 011 372 032;
  • 73) 0.782 713 011 372 032 × 2 = 1 + 0.565 426 022 744 064;
  • 74) 0.565 426 022 744 064 × 2 = 1 + 0.130 852 045 488 128;
  • 75) 0.130 852 045 488 128 × 2 = 0 + 0.261 704 090 976 256;
  • 76) 0.261 704 090 976 256 × 2 = 0 + 0.523 408 181 952 512;
  • 77) 0.523 408 181 952 512 × 2 = 1 + 0.046 816 363 905 024;
  • 78) 0.046 816 363 905 024 × 2 = 0 + 0.093 632 727 810 048;
  • 79) 0.093 632 727 810 048 × 2 = 0 + 0.187 265 455 620 096;
  • 80) 0.187 265 455 620 096 × 2 = 0 + 0.374 530 911 240 192;
  • 81) 0.374 530 911 240 192 × 2 = 0 + 0.749 061 822 480 384;
  • 82) 0.749 061 822 480 384 × 2 = 1 + 0.498 123 644 960 768;
  • 83) 0.498 123 644 960 768 × 2 = 0 + 0.996 247 289 921 536;
  • 84) 0.996 247 289 921 536 × 2 = 1 + 0.992 494 579 843 072;
  • 85) 0.992 494 579 843 072 × 2 = 1 + 0.984 989 159 686 144;
  • 86) 0.984 989 159 686 144 × 2 = 1 + 0.969 978 319 372 288;
  • 87) 0.969 978 319 372 288 × 2 = 1 + 0.939 956 638 744 576;
  • 88) 0.939 956 638 744 576 × 2 = 1 + 0.879 913 277 489 152;
  • 89) 0.879 913 277 489 152 × 2 = 1 + 0.759 826 554 978 304;
  • 90) 0.759 826 554 978 304 × 2 = 1 + 0.519 653 109 956 608;
  • 91) 0.519 653 109 956 608 × 2 = 1 + 0.039 306 219 913 216;
  • 92) 0.039 306 219 913 216 × 2 = 0 + 0.078 612 439 826 432;
  • 93) 0.078 612 439 826 432 × 2 = 0 + 0.157 224 879 652 864;
  • 94) 0.157 224 879 652 864 × 2 = 0 + 0.314 449 759 305 728;
  • 95) 0.314 449 759 305 728 × 2 = 0 + 0.628 899 518 611 456;

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).


4. 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:


0.000 000 000 000 217(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101 0001 0100 0111 1111 0111 0110 1100 1000 0101 1111 1110 000(2)

5. Positive number before normalization:

0.000 000 000 000 217(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101 0001 0100 0111 1111 0111 0110 1100 1000 0101 1111 1110 000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 217(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101 0001 0100 0111 1111 0111 0110 1100 1000 0101 1111 1110 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101 0001 0100 0111 1111 0111 0110 1100 1000 0101 1111 1110 000(2) × 20 =

1.1110 1000 1010 0011 1111 1011 1011 0110 0100 0010 1111 1111 0000(2) × 2-43


7. 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 0 (a positive number)

Exponent (unadjusted): -43

Mantissa (not normalized):
1.1110 1000 1010 0011 1111 1011 1011 0110 0100 0010 1111 1111 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(10)


9. 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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

980(10) =

011 1101 0100(2)


11. 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. 1110 1000 1010 0011 1111 1011 1011 0110 0100 0010 1111 1111 0000 =

1110 1000 1010 0011 1111 1011 1011 0110 0100 0010 1111 1111 0000


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1101 0100

Mantissa (52 bits) =
1110 1000 1010 0011 1111 1011 1011 0110 0100 0010 1111 1111 0000


Decimal number 0.000 000 000 000 217 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0100 - 1110 1000 1010 0011 1111 1011 1011 0110 0100 0010 1111 1111 0000

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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