0.000 000 000 000 249 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 249(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
0.000 000 000 000 249(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: 0.000 000 000 000 249.

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) 0.000 000 000 000 249 × 2 = 0 + 0.000 000 000 000 498;
  • 2) 0.000 000 000 000 498 × 2 = 0 + 0.000 000 000 000 996;
  • 3) 0.000 000 000 000 996 × 2 = 0 + 0.000 000 000 001 992;
  • 4) 0.000 000 000 001 992 × 2 = 0 + 0.000 000 000 003 984;
  • 5) 0.000 000 000 003 984 × 2 = 0 + 0.000 000 000 007 968;
  • 6) 0.000 000 000 007 968 × 2 = 0 + 0.000 000 000 015 936;
  • 7) 0.000 000 000 015 936 × 2 = 0 + 0.000 000 000 031 872;
  • 8) 0.000 000 000 031 872 × 2 = 0 + 0.000 000 000 063 744;
  • 9) 0.000 000 000 063 744 × 2 = 0 + 0.000 000 000 127 488;
  • 10) 0.000 000 000 127 488 × 2 = 0 + 0.000 000 000 254 976;
  • 11) 0.000 000 000 254 976 × 2 = 0 + 0.000 000 000 509 952;
  • 12) 0.000 000 000 509 952 × 2 = 0 + 0.000 000 001 019 904;
  • 13) 0.000 000 001 019 904 × 2 = 0 + 0.000 000 002 039 808;
  • 14) 0.000 000 002 039 808 × 2 = 0 + 0.000 000 004 079 616;
  • 15) 0.000 000 004 079 616 × 2 = 0 + 0.000 000 008 159 232;
  • 16) 0.000 000 008 159 232 × 2 = 0 + 0.000 000 016 318 464;
  • 17) 0.000 000 016 318 464 × 2 = 0 + 0.000 000 032 636 928;
  • 18) 0.000 000 032 636 928 × 2 = 0 + 0.000 000 065 273 856;
  • 19) 0.000 000 065 273 856 × 2 = 0 + 0.000 000 130 547 712;
  • 20) 0.000 000 130 547 712 × 2 = 0 + 0.000 000 261 095 424;
  • 21) 0.000 000 261 095 424 × 2 = 0 + 0.000 000 522 190 848;
  • 22) 0.000 000 522 190 848 × 2 = 0 + 0.000 001 044 381 696;
  • 23) 0.000 001 044 381 696 × 2 = 0 + 0.000 002 088 763 392;
  • 24) 0.000 002 088 763 392 × 2 = 0 + 0.000 004 177 526 784;
  • 25) 0.000 004 177 526 784 × 2 = 0 + 0.000 008 355 053 568;
  • 26) 0.000 008 355 053 568 × 2 = 0 + 0.000 016 710 107 136;
  • 27) 0.000 016 710 107 136 × 2 = 0 + 0.000 033 420 214 272;
  • 28) 0.000 033 420 214 272 × 2 = 0 + 0.000 066 840 428 544;
  • 29) 0.000 066 840 428 544 × 2 = 0 + 0.000 133 680 857 088;
  • 30) 0.000 133 680 857 088 × 2 = 0 + 0.000 267 361 714 176;
  • 31) 0.000 267 361 714 176 × 2 = 0 + 0.000 534 723 428 352;
  • 32) 0.000 534 723 428 352 × 2 = 0 + 0.001 069 446 856 704;
  • 33) 0.001 069 446 856 704 × 2 = 0 + 0.002 138 893 713 408;
  • 34) 0.002 138 893 713 408 × 2 = 0 + 0.004 277 787 426 816;
  • 35) 0.004 277 787 426 816 × 2 = 0 + 0.008 555 574 853 632;
  • 36) 0.008 555 574 853 632 × 2 = 0 + 0.017 111 149 707 264;
  • 37) 0.017 111 149 707 264 × 2 = 0 + 0.034 222 299 414 528;
  • 38) 0.034 222 299 414 528 × 2 = 0 + 0.068 444 598 829 056;
  • 39) 0.068 444 598 829 056 × 2 = 0 + 0.136 889 197 658 112;
  • 40) 0.136 889 197 658 112 × 2 = 0 + 0.273 778 395 316 224;
  • 41) 0.273 778 395 316 224 × 2 = 0 + 0.547 556 790 632 448;
  • 42) 0.547 556 790 632 448 × 2 = 1 + 0.095 113 581 264 896;
  • 43) 0.095 113 581 264 896 × 2 = 0 + 0.190 227 162 529 792;
  • 44) 0.190 227 162 529 792 × 2 = 0 + 0.380 454 325 059 584;
  • 45) 0.380 454 325 059 584 × 2 = 0 + 0.760 908 650 119 168;
  • 46) 0.760 908 650 119 168 × 2 = 1 + 0.521 817 300 238 336;
  • 47) 0.521 817 300 238 336 × 2 = 1 + 0.043 634 600 476 672;
  • 48) 0.043 634 600 476 672 × 2 = 0 + 0.087 269 200 953 344;
  • 49) 0.087 269 200 953 344 × 2 = 0 + 0.174 538 401 906 688;
  • 50) 0.174 538 401 906 688 × 2 = 0 + 0.349 076 803 813 376;
  • 51) 0.349 076 803 813 376 × 2 = 0 + 0.698 153 607 626 752;
  • 52) 0.698 153 607 626 752 × 2 = 1 + 0.396 307 215 253 504;
  • 53) 0.396 307 215 253 504 × 2 = 0 + 0.792 614 430 507 008;
  • 54) 0.792 614 430 507 008 × 2 = 1 + 0.585 228 861 014 016;
  • 55) 0.585 228 861 014 016 × 2 = 1 + 0.170 457 722 028 032;
  • 56) 0.170 457 722 028 032 × 2 = 0 + 0.340 915 444 056 064;
  • 57) 0.340 915 444 056 064 × 2 = 0 + 0.681 830 888 112 128;
  • 58) 0.681 830 888 112 128 × 2 = 1 + 0.363 661 776 224 256;
  • 59) 0.363 661 776 224 256 × 2 = 0 + 0.727 323 552 448 512;
  • 60) 0.727 323 552 448 512 × 2 = 1 + 0.454 647 104 897 024;
  • 61) 0.454 647 104 897 024 × 2 = 0 + 0.909 294 209 794 048;
  • 62) 0.909 294 209 794 048 × 2 = 1 + 0.818 588 419 588 096;
  • 63) 0.818 588 419 588 096 × 2 = 1 + 0.637 176 839 176 192;
  • 64) 0.637 176 839 176 192 × 2 = 1 + 0.274 353 678 352 384;
  • 65) 0.274 353 678 352 384 × 2 = 0 + 0.548 707 356 704 768;
  • 66) 0.548 707 356 704 768 × 2 = 1 + 0.097 414 713 409 536;
  • 67) 0.097 414 713 409 536 × 2 = 0 + 0.194 829 426 819 072;
  • 68) 0.194 829 426 819 072 × 2 = 0 + 0.389 658 853 638 144;
  • 69) 0.389 658 853 638 144 × 2 = 0 + 0.779 317 707 276 288;
  • 70) 0.779 317 707 276 288 × 2 = 1 + 0.558 635 414 552 576;
  • 71) 0.558 635 414 552 576 × 2 = 1 + 0.117 270 829 105 152;
  • 72) 0.117 270 829 105 152 × 2 = 0 + 0.234 541 658 210 304;
  • 73) 0.234 541 658 210 304 × 2 = 0 + 0.469 083 316 420 608;
  • 74) 0.469 083 316 420 608 × 2 = 0 + 0.938 166 632 841 216;
  • 75) 0.938 166 632 841 216 × 2 = 1 + 0.876 333 265 682 432;
  • 76) 0.876 333 265 682 432 × 2 = 1 + 0.752 666 531 364 864;
  • 77) 0.752 666 531 364 864 × 2 = 1 + 0.505 333 062 729 728;
  • 78) 0.505 333 062 729 728 × 2 = 1 + 0.010 666 125 459 456;
  • 79) 0.010 666 125 459 456 × 2 = 0 + 0.021 332 250 918 912;
  • 80) 0.021 332 250 918 912 × 2 = 0 + 0.042 664 501 837 824;
  • 81) 0.042 664 501 837 824 × 2 = 0 + 0.085 329 003 675 648;
  • 82) 0.085 329 003 675 648 × 2 = 0 + 0.170 658 007 351 296;
  • 83) 0.170 658 007 351 296 × 2 = 0 + 0.341 316 014 702 592;
  • 84) 0.341 316 014 702 592 × 2 = 0 + 0.682 632 029 405 184;
  • 85) 0.682 632 029 405 184 × 2 = 1 + 0.365 264 058 810 368;
  • 86) 0.365 264 058 810 368 × 2 = 0 + 0.730 528 117 620 736;
  • 87) 0.730 528 117 620 736 × 2 = 1 + 0.461 056 235 241 472;
  • 88) 0.461 056 235 241 472 × 2 = 0 + 0.922 112 470 482 944;
  • 89) 0.922 112 470 482 944 × 2 = 1 + 0.844 224 940 965 888;
  • 90) 0.844 224 940 965 888 × 2 = 1 + 0.688 449 881 931 776;
  • 91) 0.688 449 881 931 776 × 2 = 1 + 0.376 899 763 863 552;
  • 92) 0.376 899 763 863 552 × 2 = 0 + 0.753 799 527 727 104;
  • 93) 0.753 799 527 727 104 × 2 = 1 + 0.507 599 055 454 208;
  • 94) 0.507 599 055 454 208 × 2 = 1 + 0.015 198 110 908 416;

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:


0.000 000 000 000 249(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0110 0001 0110 0101 0111 0100 0110 0011 1100 0000 1010 1110 11(2)

5. Positive number before normalization:

0.000 000 000 000 249(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0110 0001 0110 0101 0111 0100 0110 0011 1100 0000 1010 1110 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:


0.000 000 000 000 249(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0110 0001 0110 0101 0111 0100 0110 0011 1100 0000 1010 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0110 0001 0110 0101 0111 0100 0110 0011 1100 0000 1010 1110 11(2) × 20 =

1.0001 1000 0101 1001 0101 1101 0001 1000 1111 0000 0010 1011 1011(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):
1.0001 1000 0101 1001 0101 1101 0001 1000 1111 0000 0010 1011 1011


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. 0001 1000 0101 1001 0101 1101 0001 1000 1111 0000 0010 1011 1011 =

0001 1000 0101 1001 0101 1101 0001 1000 1111 0000 0010 1011 1011


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) =
0001 1000 0101 1001 0101 1101 0001 1000 1111 0000 0010 1011 1011


Decimal number 0.000 000 000 000 249 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0101 - 0001 1000 0101 1001 0101 1101 0001 1000 1111 0000 0010 1011 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