0.000 000 000 000 322 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 322(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
0.000 000 000 000 322(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: 0.000 000 000 000 322.

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) 0.000 000 000 000 322 × 2 = 0 + 0.000 000 000 000 644;
  • 2) 0.000 000 000 000 644 × 2 = 0 + 0.000 000 000 001 288;
  • 3) 0.000 000 000 001 288 × 2 = 0 + 0.000 000 000 002 576;
  • 4) 0.000 000 000 002 576 × 2 = 0 + 0.000 000 000 005 152;
  • 5) 0.000 000 000 005 152 × 2 = 0 + 0.000 000 000 010 304;
  • 6) 0.000 000 000 010 304 × 2 = 0 + 0.000 000 000 020 608;
  • 7) 0.000 000 000 020 608 × 2 = 0 + 0.000 000 000 041 216;
  • 8) 0.000 000 000 041 216 × 2 = 0 + 0.000 000 000 082 432;
  • 9) 0.000 000 000 082 432 × 2 = 0 + 0.000 000 000 164 864;
  • 10) 0.000 000 000 164 864 × 2 = 0 + 0.000 000 000 329 728;
  • 11) 0.000 000 000 329 728 × 2 = 0 + 0.000 000 000 659 456;
  • 12) 0.000 000 000 659 456 × 2 = 0 + 0.000 000 001 318 912;
  • 13) 0.000 000 001 318 912 × 2 = 0 + 0.000 000 002 637 824;
  • 14) 0.000 000 002 637 824 × 2 = 0 + 0.000 000 005 275 648;
  • 15) 0.000 000 005 275 648 × 2 = 0 + 0.000 000 010 551 296;
  • 16) 0.000 000 010 551 296 × 2 = 0 + 0.000 000 021 102 592;
  • 17) 0.000 000 021 102 592 × 2 = 0 + 0.000 000 042 205 184;
  • 18) 0.000 000 042 205 184 × 2 = 0 + 0.000 000 084 410 368;
  • 19) 0.000 000 084 410 368 × 2 = 0 + 0.000 000 168 820 736;
  • 20) 0.000 000 168 820 736 × 2 = 0 + 0.000 000 337 641 472;
  • 21) 0.000 000 337 641 472 × 2 = 0 + 0.000 000 675 282 944;
  • 22) 0.000 000 675 282 944 × 2 = 0 + 0.000 001 350 565 888;
  • 23) 0.000 001 350 565 888 × 2 = 0 + 0.000 002 701 131 776;
  • 24) 0.000 002 701 131 776 × 2 = 0 + 0.000 005 402 263 552;
  • 25) 0.000 005 402 263 552 × 2 = 0 + 0.000 010 804 527 104;
  • 26) 0.000 010 804 527 104 × 2 = 0 + 0.000 021 609 054 208;
  • 27) 0.000 021 609 054 208 × 2 = 0 + 0.000 043 218 108 416;
  • 28) 0.000 043 218 108 416 × 2 = 0 + 0.000 086 436 216 832;
  • 29) 0.000 086 436 216 832 × 2 = 0 + 0.000 172 872 433 664;
  • 30) 0.000 172 872 433 664 × 2 = 0 + 0.000 345 744 867 328;
  • 31) 0.000 345 744 867 328 × 2 = 0 + 0.000 691 489 734 656;
  • 32) 0.000 691 489 734 656 × 2 = 0 + 0.001 382 979 469 312;
  • 33) 0.001 382 979 469 312 × 2 = 0 + 0.002 765 958 938 624;
  • 34) 0.002 765 958 938 624 × 2 = 0 + 0.005 531 917 877 248;
  • 35) 0.005 531 917 877 248 × 2 = 0 + 0.011 063 835 754 496;
  • 36) 0.011 063 835 754 496 × 2 = 0 + 0.022 127 671 508 992;
  • 37) 0.022 127 671 508 992 × 2 = 0 + 0.044 255 343 017 984;
  • 38) 0.044 255 343 017 984 × 2 = 0 + 0.088 510 686 035 968;
  • 39) 0.088 510 686 035 968 × 2 = 0 + 0.177 021 372 071 936;
  • 40) 0.177 021 372 071 936 × 2 = 0 + 0.354 042 744 143 872;
  • 41) 0.354 042 744 143 872 × 2 = 0 + 0.708 085 488 287 744;
  • 42) 0.708 085 488 287 744 × 2 = 1 + 0.416 170 976 575 488;
  • 43) 0.416 170 976 575 488 × 2 = 0 + 0.832 341 953 150 976;
  • 44) 0.832 341 953 150 976 × 2 = 1 + 0.664 683 906 301 952;
  • 45) 0.664 683 906 301 952 × 2 = 1 + 0.329 367 812 603 904;
  • 46) 0.329 367 812 603 904 × 2 = 0 + 0.658 735 625 207 808;
  • 47) 0.658 735 625 207 808 × 2 = 1 + 0.317 471 250 415 616;
  • 48) 0.317 471 250 415 616 × 2 = 0 + 0.634 942 500 831 232;
  • 49) 0.634 942 500 831 232 × 2 = 1 + 0.269 885 001 662 464;
  • 50) 0.269 885 001 662 464 × 2 = 0 + 0.539 770 003 324 928;
  • 51) 0.539 770 003 324 928 × 2 = 1 + 0.079 540 006 649 856;
  • 52) 0.079 540 006 649 856 × 2 = 0 + 0.159 080 013 299 712;
  • 53) 0.159 080 013 299 712 × 2 = 0 + 0.318 160 026 599 424;
  • 54) 0.318 160 026 599 424 × 2 = 0 + 0.636 320 053 198 848;
  • 55) 0.636 320 053 198 848 × 2 = 1 + 0.272 640 106 397 696;
  • 56) 0.272 640 106 397 696 × 2 = 0 + 0.545 280 212 795 392;
  • 57) 0.545 280 212 795 392 × 2 = 1 + 0.090 560 425 590 784;
  • 58) 0.090 560 425 590 784 × 2 = 0 + 0.181 120 851 181 568;
  • 59) 0.181 120 851 181 568 × 2 = 0 + 0.362 241 702 363 136;
  • 60) 0.362 241 702 363 136 × 2 = 0 + 0.724 483 404 726 272;
  • 61) 0.724 483 404 726 272 × 2 = 1 + 0.448 966 809 452 544;
  • 62) 0.448 966 809 452 544 × 2 = 0 + 0.897 933 618 905 088;
  • 63) 0.897 933 618 905 088 × 2 = 1 + 0.795 867 237 810 176;
  • 64) 0.795 867 237 810 176 × 2 = 1 + 0.591 734 475 620 352;
  • 65) 0.591 734 475 620 352 × 2 = 1 + 0.183 468 951 240 704;
  • 66) 0.183 468 951 240 704 × 2 = 0 + 0.366 937 902 481 408;
  • 67) 0.366 937 902 481 408 × 2 = 0 + 0.733 875 804 962 816;
  • 68) 0.733 875 804 962 816 × 2 = 1 + 0.467 751 609 925 632;
  • 69) 0.467 751 609 925 632 × 2 = 0 + 0.935 503 219 851 264;
  • 70) 0.935 503 219 851 264 × 2 = 1 + 0.871 006 439 702 528;
  • 71) 0.871 006 439 702 528 × 2 = 1 + 0.742 012 879 405 056;
  • 72) 0.742 012 879 405 056 × 2 = 1 + 0.484 025 758 810 112;
  • 73) 0.484 025 758 810 112 × 2 = 0 + 0.968 051 517 620 224;
  • 74) 0.968 051 517 620 224 × 2 = 1 + 0.936 103 035 240 448;
  • 75) 0.936 103 035 240 448 × 2 = 1 + 0.872 206 070 480 896;
  • 76) 0.872 206 070 480 896 × 2 = 1 + 0.744 412 140 961 792;
  • 77) 0.744 412 140 961 792 × 2 = 1 + 0.488 824 281 923 584;
  • 78) 0.488 824 281 923 584 × 2 = 0 + 0.977 648 563 847 168;
  • 79) 0.977 648 563 847 168 × 2 = 1 + 0.955 297 127 694 336;
  • 80) 0.955 297 127 694 336 × 2 = 1 + 0.910 594 255 388 672;
  • 81) 0.910 594 255 388 672 × 2 = 1 + 0.821 188 510 777 344;
  • 82) 0.821 188 510 777 344 × 2 = 1 + 0.642 377 021 554 688;
  • 83) 0.642 377 021 554 688 × 2 = 1 + 0.284 754 043 109 376;
  • 84) 0.284 754 043 109 376 × 2 = 0 + 0.569 508 086 218 752;
  • 85) 0.569 508 086 218 752 × 2 = 1 + 0.139 016 172 437 504;
  • 86) 0.139 016 172 437 504 × 2 = 0 + 0.278 032 344 875 008;
  • 87) 0.278 032 344 875 008 × 2 = 0 + 0.556 064 689 750 016;
  • 88) 0.556 064 689 750 016 × 2 = 1 + 0.112 129 379 500 032;
  • 89) 0.112 129 379 500 032 × 2 = 0 + 0.224 258 759 000 064;
  • 90) 0.224 258 759 000 064 × 2 = 0 + 0.448 517 518 000 128;
  • 91) 0.448 517 518 000 128 × 2 = 0 + 0.897 035 036 000 256;
  • 92) 0.897 035 036 000 256 × 2 = 1 + 0.794 070 072 000 512;
  • 93) 0.794 070 072 000 512 × 2 = 1 + 0.588 140 144 001 024;
  • 94) 0.588 140 144 001 024 × 2 = 1 + 0.176 280 288 002 048;

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:


0.000 000 000 000 322(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1010 1010 0010 1000 1011 1001 0111 0111 1011 1110 1001 0001 11(2)

5. Positive number before normalization:

0.000 000 000 000 322(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1010 1010 0010 1000 1011 1001 0111 0111 1011 1110 1001 0001 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 42 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 322(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1010 1010 0010 1000 1011 1001 0111 0111 1011 1110 1001 0001 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1010 1010 0010 1000 1011 1001 0111 0111 1011 1110 1001 0001 11(2) × 20 =

1.0110 1010 1000 1010 0010 1110 0101 1101 1110 1111 1010 0100 0111(2) × 2-42


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): -42

Mantissa (not normalized):
1.0110 1010 1000 1010 0010 1110 0101 1101 1110 1111 1010 0100 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-42 + 2(11-1) - 1 =

(-42 + 1 023)(10) =

981(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;
  • 981 ÷ 2 = 490 + 1;
  • 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) =

981(10) =

011 1101 0101(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. 0110 1010 1000 1010 0010 1110 0101 1101 1110 1111 1010 0100 0111 =

0110 1010 1000 1010 0010 1110 0101 1101 1110 1111 1010 0100 0111


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 0101

Mantissa (52 bits) =
0110 1010 1000 1010 0010 1110 0101 1101 1110 1111 1010 0100 0111


Decimal number 0.000 000 000 000 322 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0101 - 0110 1010 1000 1010 0010 1110 0101 1101 1110 1111 1010 0100 0111

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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