0.052 631 578 947 368 418 130 992 040 460 114 367 05 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.052 631 578 947 368 418 130 992 040 460 114 367 05(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
0.052 631 578 947 368 418 130 992 040 460 114 367 05(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: 0.052 631 578 947 368 418 130 992 040 460 114 367 05.

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) 0.052 631 578 947 368 418 130 992 040 460 114 367 05 × 2 = 0 + 0.105 263 157 894 736 836 261 984 080 920 228 734 1;
  • 2) 0.105 263 157 894 736 836 261 984 080 920 228 734 1 × 2 = 0 + 0.210 526 315 789 473 672 523 968 161 840 457 468 2;
  • 3) 0.210 526 315 789 473 672 523 968 161 840 457 468 2 × 2 = 0 + 0.421 052 631 578 947 345 047 936 323 680 914 936 4;
  • 4) 0.421 052 631 578 947 345 047 936 323 680 914 936 4 × 2 = 0 + 0.842 105 263 157 894 690 095 872 647 361 829 872 8;
  • 5) 0.842 105 263 157 894 690 095 872 647 361 829 872 8 × 2 = 1 + 0.684 210 526 315 789 380 191 745 294 723 659 745 6;
  • 6) 0.684 210 526 315 789 380 191 745 294 723 659 745 6 × 2 = 1 + 0.368 421 052 631 578 760 383 490 589 447 319 491 2;
  • 7) 0.368 421 052 631 578 760 383 490 589 447 319 491 2 × 2 = 0 + 0.736 842 105 263 157 520 766 981 178 894 638 982 4;
  • 8) 0.736 842 105 263 157 520 766 981 178 894 638 982 4 × 2 = 1 + 0.473 684 210 526 315 041 533 962 357 789 277 964 8;
  • 9) 0.473 684 210 526 315 041 533 962 357 789 277 964 8 × 2 = 0 + 0.947 368 421 052 630 083 067 924 715 578 555 929 6;
  • 10) 0.947 368 421 052 630 083 067 924 715 578 555 929 6 × 2 = 1 + 0.894 736 842 105 260 166 135 849 431 157 111 859 2;
  • 11) 0.894 736 842 105 260 166 135 849 431 157 111 859 2 × 2 = 1 + 0.789 473 684 210 520 332 271 698 862 314 223 718 4;
  • 12) 0.789 473 684 210 520 332 271 698 862 314 223 718 4 × 2 = 1 + 0.578 947 368 421 040 664 543 397 724 628 447 436 8;
  • 13) 0.578 947 368 421 040 664 543 397 724 628 447 436 8 × 2 = 1 + 0.157 894 736 842 081 329 086 795 449 256 894 873 6;
  • 14) 0.157 894 736 842 081 329 086 795 449 256 894 873 6 × 2 = 0 + 0.315 789 473 684 162 658 173 590 898 513 789 747 2;
  • 15) 0.315 789 473 684 162 658 173 590 898 513 789 747 2 × 2 = 0 + 0.631 578 947 368 325 316 347 181 797 027 579 494 4;
  • 16) 0.631 578 947 368 325 316 347 181 797 027 579 494 4 × 2 = 1 + 0.263 157 894 736 650 632 694 363 594 055 158 988 8;
  • 17) 0.263 157 894 736 650 632 694 363 594 055 158 988 8 × 2 = 0 + 0.526 315 789 473 301 265 388 727 188 110 317 977 6;
  • 18) 0.526 315 789 473 301 265 388 727 188 110 317 977 6 × 2 = 1 + 0.052 631 578 946 602 530 777 454 376 220 635 955 2;
  • 19) 0.052 631 578 946 602 530 777 454 376 220 635 955 2 × 2 = 0 + 0.105 263 157 893 205 061 554 908 752 441 271 910 4;
  • 20) 0.105 263 157 893 205 061 554 908 752 441 271 910 4 × 2 = 0 + 0.210 526 315 786 410 123 109 817 504 882 543 820 8;
  • 21) 0.210 526 315 786 410 123 109 817 504 882 543 820 8 × 2 = 0 + 0.421 052 631 572 820 246 219 635 009 765 087 641 6;
  • 22) 0.421 052 631 572 820 246 219 635 009 765 087 641 6 × 2 = 0 + 0.842 105 263 145 640 492 439 270 019 530 175 283 2;
  • 23) 0.842 105 263 145 640 492 439 270 019 530 175 283 2 × 2 = 1 + 0.684 210 526 291 280 984 878 540 039 060 350 566 4;
  • 24) 0.684 210 526 291 280 984 878 540 039 060 350 566 4 × 2 = 1 + 0.368 421 052 582 561 969 757 080 078 120 701 132 8;
  • 25) 0.368 421 052 582 561 969 757 080 078 120 701 132 8 × 2 = 0 + 0.736 842 105 165 123 939 514 160 156 241 402 265 6;
  • 26) 0.736 842 105 165 123 939 514 160 156 241 402 265 6 × 2 = 1 + 0.473 684 210 330 247 879 028 320 312 482 804 531 2;
  • 27) 0.473 684 210 330 247 879 028 320 312 482 804 531 2 × 2 = 0 + 0.947 368 420 660 495 758 056 640 624 965 609 062 4;
  • 28) 0.947 368 420 660 495 758 056 640 624 965 609 062 4 × 2 = 1 + 0.894 736 841 320 991 516 113 281 249 931 218 124 8;
  • 29) 0.894 736 841 320 991 516 113 281 249 931 218 124 8 × 2 = 1 + 0.789 473 682 641 983 032 226 562 499 862 436 249 6;
  • 30) 0.789 473 682 641 983 032 226 562 499 862 436 249 6 × 2 = 1 + 0.578 947 365 283 966 064 453 124 999 724 872 499 2;
  • 31) 0.578 947 365 283 966 064 453 124 999 724 872 499 2 × 2 = 1 + 0.157 894 730 567 932 128 906 249 999 449 744 998 4;
  • 32) 0.157 894 730 567 932 128 906 249 999 449 744 998 4 × 2 = 0 + 0.315 789 461 135 864 257 812 499 998 899 489 996 8;
  • 33) 0.315 789 461 135 864 257 812 499 998 899 489 996 8 × 2 = 0 + 0.631 578 922 271 728 515 624 999 997 798 979 993 6;
  • 34) 0.631 578 922 271 728 515 624 999 997 798 979 993 6 × 2 = 1 + 0.263 157 844 543 457 031 249 999 995 597 959 987 2;
  • 35) 0.263 157 844 543 457 031 249 999 995 597 959 987 2 × 2 = 0 + 0.526 315 689 086 914 062 499 999 991 195 919 974 4;
  • 36) 0.526 315 689 086 914 062 499 999 991 195 919 974 4 × 2 = 1 + 0.052 631 378 173 828 124 999 999 982 391 839 948 8;
  • 37) 0.052 631 378 173 828 124 999 999 982 391 839 948 8 × 2 = 0 + 0.105 262 756 347 656 249 999 999 964 783 679 897 6;
  • 38) 0.105 262 756 347 656 249 999 999 964 783 679 897 6 × 2 = 0 + 0.210 525 512 695 312 499 999 999 929 567 359 795 2;
  • 39) 0.210 525 512 695 312 499 999 999 929 567 359 795 2 × 2 = 0 + 0.421 051 025 390 624 999 999 999 859 134 719 590 4;
  • 40) 0.421 051 025 390 624 999 999 999 859 134 719 590 4 × 2 = 0 + 0.842 102 050 781 249 999 999 999 718 269 439 180 8;
  • 41) 0.842 102 050 781 249 999 999 999 718 269 439 180 8 × 2 = 1 + 0.684 204 101 562 499 999 999 999 436 538 878 361 6;
  • 42) 0.684 204 101 562 499 999 999 999 436 538 878 361 6 × 2 = 1 + 0.368 408 203 124 999 999 999 998 873 077 756 723 2;
  • 43) 0.368 408 203 124 999 999 999 998 873 077 756 723 2 × 2 = 0 + 0.736 816 406 249 999 999 999 997 746 155 513 446 4;
  • 44) 0.736 816 406 249 999 999 999 997 746 155 513 446 4 × 2 = 1 + 0.473 632 812 499 999 999 999 995 492 311 026 892 8;
  • 45) 0.473 632 812 499 999 999 999 995 492 311 026 892 8 × 2 = 0 + 0.947 265 624 999 999 999 999 990 984 622 053 785 6;
  • 46) 0.947 265 624 999 999 999 999 990 984 622 053 785 6 × 2 = 1 + 0.894 531 249 999 999 999 999 981 969 244 107 571 2;
  • 47) 0.894 531 249 999 999 999 999 981 969 244 107 571 2 × 2 = 1 + 0.789 062 499 999 999 999 999 963 938 488 215 142 4;
  • 48) 0.789 062 499 999 999 999 999 963 938 488 215 142 4 × 2 = 1 + 0.578 124 999 999 999 999 999 927 876 976 430 284 8;
  • 49) 0.578 124 999 999 999 999 999 927 876 976 430 284 8 × 2 = 1 + 0.156 249 999 999 999 999 999 855 753 952 860 569 6;
  • 50) 0.156 249 999 999 999 999 999 855 753 952 860 569 6 × 2 = 0 + 0.312 499 999 999 999 999 999 711 507 905 721 139 2;
  • 51) 0.312 499 999 999 999 999 999 711 507 905 721 139 2 × 2 = 0 + 0.624 999 999 999 999 999 999 423 015 811 442 278 4;
  • 52) 0.624 999 999 999 999 999 999 423 015 811 442 278 4 × 2 = 1 + 0.249 999 999 999 999 999 998 846 031 622 884 556 8;
  • 53) 0.249 999 999 999 999 999 998 846 031 622 884 556 8 × 2 = 0 + 0.499 999 999 999 999 999 997 692 063 245 769 113 6;
  • 54) 0.499 999 999 999 999 999 997 692 063 245 769 113 6 × 2 = 0 + 0.999 999 999 999 999 999 995 384 126 491 538 227 2;
  • 55) 0.999 999 999 999 999 999 995 384 126 491 538 227 2 × 2 = 1 + 0.999 999 999 999 999 999 990 768 252 983 076 454 4;
  • 56) 0.999 999 999 999 999 999 990 768 252 983 076 454 4 × 2 = 1 + 0.999 999 999 999 999 999 981 536 505 966 152 908 8;
  • 57) 0.999 999 999 999 999 999 981 536 505 966 152 908 8 × 2 = 1 + 0.999 999 999 999 999 999 963 073 011 932 305 817 6;

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:


0.052 631 578 947 368 418 130 992 040 460 114 367 05(10) =

0.0000 1101 0111 1001 0100 0011 0101 1110 0101 0000 1101 0111 1001 0011 1(2)

5. Positive number before normalization:

0.052 631 578 947 368 418 130 992 040 460 114 367 05(10) =

0.0000 1101 0111 1001 0100 0011 0101 1110 0101 0000 1101 0111 1001 0011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.052 631 578 947 368 418 130 992 040 460 114 367 05(10) =

0.0000 1101 0111 1001 0100 0011 0101 1110 0101 0000 1101 0111 1001 0011 1(2) =

0.0000 1101 0111 1001 0100 0011 0101 1110 0101 0000 1101 0111 1001 0011 1(2) × 20 =

1.1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 0111(2) × 2-5


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): -5

Mantissa (not normalized):
1.1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1018(10) =

011 1111 1010(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. 1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 0111 =

1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 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 1111 1010

Mantissa (52 bits) =
1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 0111


Decimal number 0.052 631 578 947 368 418 130 992 040 460 114 367 05 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1010 - 1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 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