22 222.094 819 999 900 209 950 283 169 746 398 924 974 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 22 222.094 819 999 900 209 950 283 169 746 398 924 974(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
22 222.094 819 999 900 209 950 283 169 746 398 924 974(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: 22 222.
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;
  • 22 222 ÷ 2 = 11 111 + 0;
  • 11 111 ÷ 2 = 5 555 + 1;
  • 5 555 ÷ 2 = 2 777 + 1;
  • 2 777 ÷ 2 = 1 388 + 1;
  • 1 388 ÷ 2 = 694 + 0;
  • 694 ÷ 2 = 347 + 0;
  • 347 ÷ 2 = 173 + 1;
  • 173 ÷ 2 = 86 + 1;
  • 86 ÷ 2 = 43 + 0;
  • 43 ÷ 2 = 21 + 1;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

22 222(10) =

101 0110 1100 1110(2)


3. Convert to binary (base 2) the fractional part: 0.094 819 999 900 209 950 283 169 746 398 924 974.

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) 0.094 819 999 900 209 950 283 169 746 398 924 974 × 2 = 0 + 0.189 639 999 800 419 900 566 339 492 797 849 948;
  • 2) 0.189 639 999 800 419 900 566 339 492 797 849 948 × 2 = 0 + 0.379 279 999 600 839 801 132 678 985 595 699 896;
  • 3) 0.379 279 999 600 839 801 132 678 985 595 699 896 × 2 = 0 + 0.758 559 999 201 679 602 265 357 971 191 399 792;
  • 4) 0.758 559 999 201 679 602 265 357 971 191 399 792 × 2 = 1 + 0.517 119 998 403 359 204 530 715 942 382 799 584;
  • 5) 0.517 119 998 403 359 204 530 715 942 382 799 584 × 2 = 1 + 0.034 239 996 806 718 409 061 431 884 765 599 168;
  • 6) 0.034 239 996 806 718 409 061 431 884 765 599 168 × 2 = 0 + 0.068 479 993 613 436 818 122 863 769 531 198 336;
  • 7) 0.068 479 993 613 436 818 122 863 769 531 198 336 × 2 = 0 + 0.136 959 987 226 873 636 245 727 539 062 396 672;
  • 8) 0.136 959 987 226 873 636 245 727 539 062 396 672 × 2 = 0 + 0.273 919 974 453 747 272 491 455 078 124 793 344;
  • 9) 0.273 919 974 453 747 272 491 455 078 124 793 344 × 2 = 0 + 0.547 839 948 907 494 544 982 910 156 249 586 688;
  • 10) 0.547 839 948 907 494 544 982 910 156 249 586 688 × 2 = 1 + 0.095 679 897 814 989 089 965 820 312 499 173 376;
  • 11) 0.095 679 897 814 989 089 965 820 312 499 173 376 × 2 = 0 + 0.191 359 795 629 978 179 931 640 624 998 346 752;
  • 12) 0.191 359 795 629 978 179 931 640 624 998 346 752 × 2 = 0 + 0.382 719 591 259 956 359 863 281 249 996 693 504;
  • 13) 0.382 719 591 259 956 359 863 281 249 996 693 504 × 2 = 0 + 0.765 439 182 519 912 719 726 562 499 993 387 008;
  • 14) 0.765 439 182 519 912 719 726 562 499 993 387 008 × 2 = 1 + 0.530 878 365 039 825 439 453 124 999 986 774 016;
  • 15) 0.530 878 365 039 825 439 453 124 999 986 774 016 × 2 = 1 + 0.061 756 730 079 650 878 906 249 999 973 548 032;
  • 16) 0.061 756 730 079 650 878 906 249 999 973 548 032 × 2 = 0 + 0.123 513 460 159 301 757 812 499 999 947 096 064;
  • 17) 0.123 513 460 159 301 757 812 499 999 947 096 064 × 2 = 0 + 0.247 026 920 318 603 515 624 999 999 894 192 128;
  • 18) 0.247 026 920 318 603 515 624 999 999 894 192 128 × 2 = 0 + 0.494 053 840 637 207 031 249 999 999 788 384 256;
  • 19) 0.494 053 840 637 207 031 249 999 999 788 384 256 × 2 = 0 + 0.988 107 681 274 414 062 499 999 999 576 768 512;
  • 20) 0.988 107 681 274 414 062 499 999 999 576 768 512 × 2 = 1 + 0.976 215 362 548 828 124 999 999 999 153 537 024;
  • 21) 0.976 215 362 548 828 124 999 999 999 153 537 024 × 2 = 1 + 0.952 430 725 097 656 249 999 999 998 307 074 048;
  • 22) 0.952 430 725 097 656 249 999 999 998 307 074 048 × 2 = 1 + 0.904 861 450 195 312 499 999 999 996 614 148 096;
  • 23) 0.904 861 450 195 312 499 999 999 996 614 148 096 × 2 = 1 + 0.809 722 900 390 624 999 999 999 993 228 296 192;
  • 24) 0.809 722 900 390 624 999 999 999 993 228 296 192 × 2 = 1 + 0.619 445 800 781 249 999 999 999 986 456 592 384;
  • 25) 0.619 445 800 781 249 999 999 999 986 456 592 384 × 2 = 1 + 0.238 891 601 562 499 999 999 999 972 913 184 768;
  • 26) 0.238 891 601 562 499 999 999 999 972 913 184 768 × 2 = 0 + 0.477 783 203 124 999 999 999 999 945 826 369 536;
  • 27) 0.477 783 203 124 999 999 999 999 945 826 369 536 × 2 = 0 + 0.955 566 406 249 999 999 999 999 891 652 739 072;
  • 28) 0.955 566 406 249 999 999 999 999 891 652 739 072 × 2 = 1 + 0.911 132 812 499 999 999 999 999 783 305 478 144;
  • 29) 0.911 132 812 499 999 999 999 999 783 305 478 144 × 2 = 1 + 0.822 265 624 999 999 999 999 999 566 610 956 288;
  • 30) 0.822 265 624 999 999 999 999 999 566 610 956 288 × 2 = 1 + 0.644 531 249 999 999 999 999 999 133 221 912 576;
  • 31) 0.644 531 249 999 999 999 999 999 133 221 912 576 × 2 = 1 + 0.289 062 499 999 999 999 999 998 266 443 825 152;
  • 32) 0.289 062 499 999 999 999 999 998 266 443 825 152 × 2 = 0 + 0.578 124 999 999 999 999 999 996 532 887 650 304;
  • 33) 0.578 124 999 999 999 999 999 996 532 887 650 304 × 2 = 1 + 0.156 249 999 999 999 999 999 993 065 775 300 608;
  • 34) 0.156 249 999 999 999 999 999 993 065 775 300 608 × 2 = 0 + 0.312 499 999 999 999 999 999 986 131 550 601 216;
  • 35) 0.312 499 999 999 999 999 999 986 131 550 601 216 × 2 = 0 + 0.624 999 999 999 999 999 999 972 263 101 202 432;
  • 36) 0.624 999 999 999 999 999 999 972 263 101 202 432 × 2 = 1 + 0.249 999 999 999 999 999 999 944 526 202 404 864;
  • 37) 0.249 999 999 999 999 999 999 944 526 202 404 864 × 2 = 0 + 0.499 999 999 999 999 999 999 889 052 404 809 728;
  • 38) 0.499 999 999 999 999 999 999 889 052 404 809 728 × 2 = 0 + 0.999 999 999 999 999 999 999 778 104 809 619 456;
  • 39) 0.999 999 999 999 999 999 999 778 104 809 619 456 × 2 = 1 + 0.999 999 999 999 999 999 999 556 209 619 238 912;
  • 40) 0.999 999 999 999 999 999 999 556 209 619 238 912 × 2 = 1 + 0.999 999 999 999 999 999 999 112 419 238 477 824;
  • 41) 0.999 999 999 999 999 999 999 112 419 238 477 824 × 2 = 1 + 0.999 999 999 999 999 999 998 224 838 476 955 648;
  • 42) 0.999 999 999 999 999 999 998 224 838 476 955 648 × 2 = 1 + 0.999 999 999 999 999 999 996 449 676 953 911 296;
  • 43) 0.999 999 999 999 999 999 996 449 676 953 911 296 × 2 = 1 + 0.999 999 999 999 999 999 992 899 353 907 822 592;
  • 44) 0.999 999 999 999 999 999 992 899 353 907 822 592 × 2 = 1 + 0.999 999 999 999 999 999 985 798 707 815 645 184;
  • 45) 0.999 999 999 999 999 999 985 798 707 815 645 184 × 2 = 1 + 0.999 999 999 999 999 999 971 597 415 631 290 368;
  • 46) 0.999 999 999 999 999 999 971 597 415 631 290 368 × 2 = 1 + 0.999 999 999 999 999 999 943 194 831 262 580 736;
  • 47) 0.999 999 999 999 999 999 943 194 831 262 580 736 × 2 = 1 + 0.999 999 999 999 999 999 886 389 662 525 161 472;
  • 48) 0.999 999 999 999 999 999 886 389 662 525 161 472 × 2 = 1 + 0.999 999 999 999 999 999 772 779 325 050 322 944;
  • 49) 0.999 999 999 999 999 999 772 779 325 050 322 944 × 2 = 1 + 0.999 999 999 999 999 999 545 558 650 100 645 888;
  • 50) 0.999 999 999 999 999 999 545 558 650 100 645 888 × 2 = 1 + 0.999 999 999 999 999 999 091 117 300 201 291 776;
  • 51) 0.999 999 999 999 999 999 091 117 300 201 291 776 × 2 = 1 + 0.999 999 999 999 999 998 182 234 600 402 583 552;
  • 52) 0.999 999 999 999 999 998 182 234 600 402 583 552 × 2 = 1 + 0.999 999 999 999 999 996 364 469 200 805 167 104;
  • 53) 0.999 999 999 999 999 996 364 469 200 805 167 104 × 2 = 1 + 0.999 999 999 999 999 992 728 938 401 610 334 208;

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:


0.094 819 999 900 209 950 283 169 746 398 924 974(10) =

0.0001 1000 0100 0110 0001 1111 1001 1110 1001 0011 1111 1111 1111 1(2)

5. Positive number before normalization:

22 222.094 819 999 900 209 950 283 169 746 398 924 974(10) =

101 0110 1100 1110.0001 1000 0100 0110 0001 1111 1001 1110 1001 0011 1111 1111 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the left, so that only one non zero digit remains to the left of it:


22 222.094 819 999 900 209 950 283 169 746 398 924 974(10) =

101 0110 1100 1110.0001 1000 0100 0110 0001 1111 1001 1110 1001 0011 1111 1111 1111 1(2) =

101 0110 1100 1110.0001 1000 0100 0110 0001 1111 1001 1110 1001 0011 1111 1111 1111 1(2) × 20 =

1.0101 1011 0011 1000 0110 0001 0001 1000 0111 1110 0111 1010 0100 1111 1111 1111 111(2) × 214


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): 14

Mantissa (not normalized):
1.0101 1011 0011 1000 0110 0001 0001 1000 0111 1110 0111 1010 0100 1111 1111 1111 111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

14 + 2(11-1) - 1 =

(14 + 1 023)(10) =

1 037(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;
  • 1 037 ÷ 2 = 518 + 1;
  • 518 ÷ 2 = 259 + 0;
  • 259 ÷ 2 = 129 + 1;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1037(10) =

100 0000 1101(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0101 1011 0011 1000 0110 0001 0001 1000 0111 1110 0111 1010 0100 111 1111 1111 1111 =

0101 1011 0011 1000 0110 0001 0001 1000 0111 1110 0111 1010 0100


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) =
100 0000 1101

Mantissa (52 bits) =
0101 1011 0011 1000 0110 0001 0001 1000 0111 1110 0111 1010 0100


Decimal number 22 222.094 819 999 900 209 950 283 169 746 398 924 974 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 1101 - 0101 1011 0011 1000 0110 0001 0001 1000 0111 1110 0111 1010 0100

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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