0.000 000 000 000 123 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 123(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
0.000 000 000 000 123(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: 0.000 000 000 000 123.

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) 0.000 000 000 000 123 × 2 = 0 + 0.000 000 000 000 246;
  • 2) 0.000 000 000 000 246 × 2 = 0 + 0.000 000 000 000 492;
  • 3) 0.000 000 000 000 492 × 2 = 0 + 0.000 000 000 000 984;
  • 4) 0.000 000 000 000 984 × 2 = 0 + 0.000 000 000 001 968;
  • 5) 0.000 000 000 001 968 × 2 = 0 + 0.000 000 000 003 936;
  • 6) 0.000 000 000 003 936 × 2 = 0 + 0.000 000 000 007 872;
  • 7) 0.000 000 000 007 872 × 2 = 0 + 0.000 000 000 015 744;
  • 8) 0.000 000 000 015 744 × 2 = 0 + 0.000 000 000 031 488;
  • 9) 0.000 000 000 031 488 × 2 = 0 + 0.000 000 000 062 976;
  • 10) 0.000 000 000 062 976 × 2 = 0 + 0.000 000 000 125 952;
  • 11) 0.000 000 000 125 952 × 2 = 0 + 0.000 000 000 251 904;
  • 12) 0.000 000 000 251 904 × 2 = 0 + 0.000 000 000 503 808;
  • 13) 0.000 000 000 503 808 × 2 = 0 + 0.000 000 001 007 616;
  • 14) 0.000 000 001 007 616 × 2 = 0 + 0.000 000 002 015 232;
  • 15) 0.000 000 002 015 232 × 2 = 0 + 0.000 000 004 030 464;
  • 16) 0.000 000 004 030 464 × 2 = 0 + 0.000 000 008 060 928;
  • 17) 0.000 000 008 060 928 × 2 = 0 + 0.000 000 016 121 856;
  • 18) 0.000 000 016 121 856 × 2 = 0 + 0.000 000 032 243 712;
  • 19) 0.000 000 032 243 712 × 2 = 0 + 0.000 000 064 487 424;
  • 20) 0.000 000 064 487 424 × 2 = 0 + 0.000 000 128 974 848;
  • 21) 0.000 000 128 974 848 × 2 = 0 + 0.000 000 257 949 696;
  • 22) 0.000 000 257 949 696 × 2 = 0 + 0.000 000 515 899 392;
  • 23) 0.000 000 515 899 392 × 2 = 0 + 0.000 001 031 798 784;
  • 24) 0.000 001 031 798 784 × 2 = 0 + 0.000 002 063 597 568;
  • 25) 0.000 002 063 597 568 × 2 = 0 + 0.000 004 127 195 136;
  • 26) 0.000 004 127 195 136 × 2 = 0 + 0.000 008 254 390 272;
  • 27) 0.000 008 254 390 272 × 2 = 0 + 0.000 016 508 780 544;
  • 28) 0.000 016 508 780 544 × 2 = 0 + 0.000 033 017 561 088;
  • 29) 0.000 033 017 561 088 × 2 = 0 + 0.000 066 035 122 176;
  • 30) 0.000 066 035 122 176 × 2 = 0 + 0.000 132 070 244 352;
  • 31) 0.000 132 070 244 352 × 2 = 0 + 0.000 264 140 488 704;
  • 32) 0.000 264 140 488 704 × 2 = 0 + 0.000 528 280 977 408;
  • 33) 0.000 528 280 977 408 × 2 = 0 + 0.001 056 561 954 816;
  • 34) 0.001 056 561 954 816 × 2 = 0 + 0.002 113 123 909 632;
  • 35) 0.002 113 123 909 632 × 2 = 0 + 0.004 226 247 819 264;
  • 36) 0.004 226 247 819 264 × 2 = 0 + 0.008 452 495 638 528;
  • 37) 0.008 452 495 638 528 × 2 = 0 + 0.016 904 991 277 056;
  • 38) 0.016 904 991 277 056 × 2 = 0 + 0.033 809 982 554 112;
  • 39) 0.033 809 982 554 112 × 2 = 0 + 0.067 619 965 108 224;
  • 40) 0.067 619 965 108 224 × 2 = 0 + 0.135 239 930 216 448;
  • 41) 0.135 239 930 216 448 × 2 = 0 + 0.270 479 860 432 896;
  • 42) 0.270 479 860 432 896 × 2 = 0 + 0.540 959 720 865 792;
  • 43) 0.540 959 720 865 792 × 2 = 1 + 0.081 919 441 731 584;
  • 44) 0.081 919 441 731 584 × 2 = 0 + 0.163 838 883 463 168;
  • 45) 0.163 838 883 463 168 × 2 = 0 + 0.327 677 766 926 336;
  • 46) 0.327 677 766 926 336 × 2 = 0 + 0.655 355 533 852 672;
  • 47) 0.655 355 533 852 672 × 2 = 1 + 0.310 711 067 705 344;
  • 48) 0.310 711 067 705 344 × 2 = 0 + 0.621 422 135 410 688;
  • 49) 0.621 422 135 410 688 × 2 = 1 + 0.242 844 270 821 376;
  • 50) 0.242 844 270 821 376 × 2 = 0 + 0.485 688 541 642 752;
  • 51) 0.485 688 541 642 752 × 2 = 0 + 0.971 377 083 285 504;
  • 52) 0.971 377 083 285 504 × 2 = 1 + 0.942 754 166 571 008;
  • 53) 0.942 754 166 571 008 × 2 = 1 + 0.885 508 333 142 016;
  • 54) 0.885 508 333 142 016 × 2 = 1 + 0.771 016 666 284 032;
  • 55) 0.771 016 666 284 032 × 2 = 1 + 0.542 033 332 568 064;
  • 56) 0.542 033 332 568 064 × 2 = 1 + 0.084 066 665 136 128;
  • 57) 0.084 066 665 136 128 × 2 = 0 + 0.168 133 330 272 256;
  • 58) 0.168 133 330 272 256 × 2 = 0 + 0.336 266 660 544 512;
  • 59) 0.336 266 660 544 512 × 2 = 0 + 0.672 533 321 089 024;
  • 60) 0.672 533 321 089 024 × 2 = 1 + 0.345 066 642 178 048;
  • 61) 0.345 066 642 178 048 × 2 = 0 + 0.690 133 284 356 096;
  • 62) 0.690 133 284 356 096 × 2 = 1 + 0.380 266 568 712 192;
  • 63) 0.380 266 568 712 192 × 2 = 0 + 0.760 533 137 424 384;
  • 64) 0.760 533 137 424 384 × 2 = 1 + 0.521 066 274 848 768;
  • 65) 0.521 066 274 848 768 × 2 = 1 + 0.042 132 549 697 536;
  • 66) 0.042 132 549 697 536 × 2 = 0 + 0.084 265 099 395 072;
  • 67) 0.084 265 099 395 072 × 2 = 0 + 0.168 530 198 790 144;
  • 68) 0.168 530 198 790 144 × 2 = 0 + 0.337 060 397 580 288;
  • 69) 0.337 060 397 580 288 × 2 = 0 + 0.674 120 795 160 576;
  • 70) 0.674 120 795 160 576 × 2 = 1 + 0.348 241 590 321 152;
  • 71) 0.348 241 590 321 152 × 2 = 0 + 0.696 483 180 642 304;
  • 72) 0.696 483 180 642 304 × 2 = 1 + 0.392 966 361 284 608;
  • 73) 0.392 966 361 284 608 × 2 = 0 + 0.785 932 722 569 216;
  • 74) 0.785 932 722 569 216 × 2 = 1 + 0.571 865 445 138 432;
  • 75) 0.571 865 445 138 432 × 2 = 1 + 0.143 730 890 276 864;
  • 76) 0.143 730 890 276 864 × 2 = 0 + 0.287 461 780 553 728;
  • 77) 0.287 461 780 553 728 × 2 = 0 + 0.574 923 561 107 456;
  • 78) 0.574 923 561 107 456 × 2 = 1 + 0.149 847 122 214 912;
  • 79) 0.149 847 122 214 912 × 2 = 0 + 0.299 694 244 429 824;
  • 80) 0.299 694 244 429 824 × 2 = 0 + 0.599 388 488 859 648;
  • 81) 0.599 388 488 859 648 × 2 = 1 + 0.198 776 977 719 296;
  • 82) 0.198 776 977 719 296 × 2 = 0 + 0.397 553 955 438 592;
  • 83) 0.397 553 955 438 592 × 2 = 0 + 0.795 107 910 877 184;
  • 84) 0.795 107 910 877 184 × 2 = 1 + 0.590 215 821 754 368;
  • 85) 0.590 215 821 754 368 × 2 = 1 + 0.180 431 643 508 736;
  • 86) 0.180 431 643 508 736 × 2 = 0 + 0.360 863 287 017 472;
  • 87) 0.360 863 287 017 472 × 2 = 0 + 0.721 726 574 034 944;
  • 88) 0.721 726 574 034 944 × 2 = 1 + 0.443 453 148 069 888;
  • 89) 0.443 453 148 069 888 × 2 = 0 + 0.886 906 296 139 776;
  • 90) 0.886 906 296 139 776 × 2 = 1 + 0.773 812 592 279 552;
  • 91) 0.773 812 592 279 552 × 2 = 1 + 0.547 625 184 559 104;
  • 92) 0.547 625 184 559 104 × 2 = 1 + 0.095 250 369 118 208;
  • 93) 0.095 250 369 118 208 × 2 = 0 + 0.190 500 738 236 416;
  • 94) 0.190 500 738 236 416 × 2 = 0 + 0.381 001 476 472 832;
  • 95) 0.381 001 476 472 832 × 2 = 0 + 0.762 002 952 945 664;

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:


0.000 000 000 000 123(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0010 1001 1111 0001 0101 1000 0101 0110 0100 1001 1001 0111 000(2)

5. Positive number before normalization:

0.000 000 000 000 123(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0010 1001 1111 0001 0101 1000 0101 0110 0100 1001 1001 0111 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:


0.000 000 000 000 123(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0010 1001 1111 0001 0101 1000 0101 0110 0100 1001 1001 0111 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0010 1001 1111 0001 0101 1000 0101 0110 0100 1001 1001 0111 000(2) × 20 =

1.0001 0100 1111 1000 1010 1100 0010 1011 0010 0100 1100 1011 1000(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):
1.0001 0100 1111 1000 1010 1100 0010 1011 0010 0100 1100 1011 1000


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. 0001 0100 1111 1000 1010 1100 0010 1011 0010 0100 1100 1011 1000 =

0001 0100 1111 1000 1010 1100 0010 1011 0010 0100 1100 1011 1000


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) =
0001 0100 1111 1000 1010 1100 0010 1011 0010 0100 1100 1011 1000


Decimal number 0.000 000 000 000 123 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0100 - 0001 0100 1111 1000 1010 1100 0010 1011 0010 0100 1100 1011 1000

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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