1.000 000 021 979 552 668 138 406 918 054 535 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.000 000 021 979 552 668 138 406 918 054 535(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
1.000 000 021 979 552 668 138 406 918 054 535(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 021 979 552 668 138 406 918 054 535.

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) 0.000 000 021 979 552 668 138 406 918 054 535 × 2 = 0 + 0.000 000 043 959 105 336 276 813 836 109 07;
  • 2) 0.000 000 043 959 105 336 276 813 836 109 07 × 2 = 0 + 0.000 000 087 918 210 672 553 627 672 218 14;
  • 3) 0.000 000 087 918 210 672 553 627 672 218 14 × 2 = 0 + 0.000 000 175 836 421 345 107 255 344 436 28;
  • 4) 0.000 000 175 836 421 345 107 255 344 436 28 × 2 = 0 + 0.000 000 351 672 842 690 214 510 688 872 56;
  • 5) 0.000 000 351 672 842 690 214 510 688 872 56 × 2 = 0 + 0.000 000 703 345 685 380 429 021 377 745 12;
  • 6) 0.000 000 703 345 685 380 429 021 377 745 12 × 2 = 0 + 0.000 001 406 691 370 760 858 042 755 490 24;
  • 7) 0.000 001 406 691 370 760 858 042 755 490 24 × 2 = 0 + 0.000 002 813 382 741 521 716 085 510 980 48;
  • 8) 0.000 002 813 382 741 521 716 085 510 980 48 × 2 = 0 + 0.000 005 626 765 483 043 432 171 021 960 96;
  • 9) 0.000 005 626 765 483 043 432 171 021 960 96 × 2 = 0 + 0.000 011 253 530 966 086 864 342 043 921 92;
  • 10) 0.000 011 253 530 966 086 864 342 043 921 92 × 2 = 0 + 0.000 022 507 061 932 173 728 684 087 843 84;
  • 11) 0.000 022 507 061 932 173 728 684 087 843 84 × 2 = 0 + 0.000 045 014 123 864 347 457 368 175 687 68;
  • 12) 0.000 045 014 123 864 347 457 368 175 687 68 × 2 = 0 + 0.000 090 028 247 728 694 914 736 351 375 36;
  • 13) 0.000 090 028 247 728 694 914 736 351 375 36 × 2 = 0 + 0.000 180 056 495 457 389 829 472 702 750 72;
  • 14) 0.000 180 056 495 457 389 829 472 702 750 72 × 2 = 0 + 0.000 360 112 990 914 779 658 945 405 501 44;
  • 15) 0.000 360 112 990 914 779 658 945 405 501 44 × 2 = 0 + 0.000 720 225 981 829 559 317 890 811 002 88;
  • 16) 0.000 720 225 981 829 559 317 890 811 002 88 × 2 = 0 + 0.001 440 451 963 659 118 635 781 622 005 76;
  • 17) 0.001 440 451 963 659 118 635 781 622 005 76 × 2 = 0 + 0.002 880 903 927 318 237 271 563 244 011 52;
  • 18) 0.002 880 903 927 318 237 271 563 244 011 52 × 2 = 0 + 0.005 761 807 854 636 474 543 126 488 023 04;
  • 19) 0.005 761 807 854 636 474 543 126 488 023 04 × 2 = 0 + 0.011 523 615 709 272 949 086 252 976 046 08;
  • 20) 0.011 523 615 709 272 949 086 252 976 046 08 × 2 = 0 + 0.023 047 231 418 545 898 172 505 952 092 16;
  • 21) 0.023 047 231 418 545 898 172 505 952 092 16 × 2 = 0 + 0.046 094 462 837 091 796 345 011 904 184 32;
  • 22) 0.046 094 462 837 091 796 345 011 904 184 32 × 2 = 0 + 0.092 188 925 674 183 592 690 023 808 368 64;
  • 23) 0.092 188 925 674 183 592 690 023 808 368 64 × 2 = 0 + 0.184 377 851 348 367 185 380 047 616 737 28;
  • 24) 0.184 377 851 348 367 185 380 047 616 737 28 × 2 = 0 + 0.368 755 702 696 734 370 760 095 233 474 56;
  • 25) 0.368 755 702 696 734 370 760 095 233 474 56 × 2 = 0 + 0.737 511 405 393 468 741 520 190 466 949 12;
  • 26) 0.737 511 405 393 468 741 520 190 466 949 12 × 2 = 1 + 0.475 022 810 786 937 483 040 380 933 898 24;
  • 27) 0.475 022 810 786 937 483 040 380 933 898 24 × 2 = 0 + 0.950 045 621 573 874 966 080 761 867 796 48;
  • 28) 0.950 045 621 573 874 966 080 761 867 796 48 × 2 = 1 + 0.900 091 243 147 749 932 161 523 735 592 96;
  • 29) 0.900 091 243 147 749 932 161 523 735 592 96 × 2 = 1 + 0.800 182 486 295 499 864 323 047 471 185 92;
  • 30) 0.800 182 486 295 499 864 323 047 471 185 92 × 2 = 1 + 0.600 364 972 590 999 728 646 094 942 371 84;
  • 31) 0.600 364 972 590 999 728 646 094 942 371 84 × 2 = 1 + 0.200 729 945 181 999 457 292 189 884 743 68;
  • 32) 0.200 729 945 181 999 457 292 189 884 743 68 × 2 = 0 + 0.401 459 890 363 998 914 584 379 769 487 36;
  • 33) 0.401 459 890 363 998 914 584 379 769 487 36 × 2 = 0 + 0.802 919 780 727 997 829 168 759 538 974 72;
  • 34) 0.802 919 780 727 997 829 168 759 538 974 72 × 2 = 1 + 0.605 839 561 455 995 658 337 519 077 949 44;
  • 35) 0.605 839 561 455 995 658 337 519 077 949 44 × 2 = 1 + 0.211 679 122 911 991 316 675 038 155 898 88;
  • 36) 0.211 679 122 911 991 316 675 038 155 898 88 × 2 = 0 + 0.423 358 245 823 982 633 350 076 311 797 76;
  • 37) 0.423 358 245 823 982 633 350 076 311 797 76 × 2 = 0 + 0.846 716 491 647 965 266 700 152 623 595 52;
  • 38) 0.846 716 491 647 965 266 700 152 623 595 52 × 2 = 1 + 0.693 432 983 295 930 533 400 305 247 191 04;
  • 39) 0.693 432 983 295 930 533 400 305 247 191 04 × 2 = 1 + 0.386 865 966 591 861 066 800 610 494 382 08;
  • 40) 0.386 865 966 591 861 066 800 610 494 382 08 × 2 = 0 + 0.773 731 933 183 722 133 601 220 988 764 16;
  • 41) 0.773 731 933 183 722 133 601 220 988 764 16 × 2 = 1 + 0.547 463 866 367 444 267 202 441 977 528 32;
  • 42) 0.547 463 866 367 444 267 202 441 977 528 32 × 2 = 1 + 0.094 927 732 734 888 534 404 883 955 056 64;
  • 43) 0.094 927 732 734 888 534 404 883 955 056 64 × 2 = 0 + 0.189 855 465 469 777 068 809 767 910 113 28;
  • 44) 0.189 855 465 469 777 068 809 767 910 113 28 × 2 = 0 + 0.379 710 930 939 554 137 619 535 820 226 56;
  • 45) 0.379 710 930 939 554 137 619 535 820 226 56 × 2 = 0 + 0.759 421 861 879 108 275 239 071 640 453 12;
  • 46) 0.759 421 861 879 108 275 239 071 640 453 12 × 2 = 1 + 0.518 843 723 758 216 550 478 143 280 906 24;
  • 47) 0.518 843 723 758 216 550 478 143 280 906 24 × 2 = 1 + 0.037 687 447 516 433 100 956 286 561 812 48;
  • 48) 0.037 687 447 516 433 100 956 286 561 812 48 × 2 = 0 + 0.075 374 895 032 866 201 912 573 123 624 96;
  • 49) 0.075 374 895 032 866 201 912 573 123 624 96 × 2 = 0 + 0.150 749 790 065 732 403 825 146 247 249 92;
  • 50) 0.150 749 790 065 732 403 825 146 247 249 92 × 2 = 0 + 0.301 499 580 131 464 807 650 292 494 499 84;
  • 51) 0.301 499 580 131 464 807 650 292 494 499 84 × 2 = 0 + 0.602 999 160 262 929 615 300 584 988 999 68;
  • 52) 0.602 999 160 262 929 615 300 584 988 999 68 × 2 = 1 + 0.205 998 320 525 859 230 601 169 977 999 36;
  • 53) 0.205 998 320 525 859 230 601 169 977 999 36 × 2 = 0 + 0.411 996 641 051 718 461 202 339 955 998 72;

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:


0.000 000 021 979 552 668 138 406 918 054 535(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0(2)

5. Positive number before normalization:

1.000 000 021 979 552 668 138 406 918 054 535(10) =

1.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.000 000 021 979 552 668 138 406 918 054 535(10) =

1.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0(2) =

1.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0(2) × 20


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): 0

Mantissa (not normalized):
1.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

1023(10) =

011 1111 1111(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. 0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0 =

0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001


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 1111 1111

Mantissa (52 bits) =
0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001


Decimal number 1.000 000 021 979 552 668 138 406 918 054 535 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001

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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