5.919 999 999 999 999 928 945 726 423 985 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 5.919 999 999 999 999 928 945 726 423 985 3(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
5.919 999 999 999 999 928 945 726 423 985 3(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: 5.
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;
  • 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.

5(10) =

101(2)


3. Convert to binary (base 2) the fractional part: 0.919 999 999 999 999 928 945 726 423 985 3.

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) 0.919 999 999 999 999 928 945 726 423 985 3 × 2 = 1 + 0.839 999 999 999 999 857 891 452 847 970 6;
  • 2) 0.839 999 999 999 999 857 891 452 847 970 6 × 2 = 1 + 0.679 999 999 999 999 715 782 905 695 941 2;
  • 3) 0.679 999 999 999 999 715 782 905 695 941 2 × 2 = 1 + 0.359 999 999 999 999 431 565 811 391 882 4;
  • 4) 0.359 999 999 999 999 431 565 811 391 882 4 × 2 = 0 + 0.719 999 999 999 998 863 131 622 783 764 8;
  • 5) 0.719 999 999 999 998 863 131 622 783 764 8 × 2 = 1 + 0.439 999 999 999 997 726 263 245 567 529 6;
  • 6) 0.439 999 999 999 997 726 263 245 567 529 6 × 2 = 0 + 0.879 999 999 999 995 452 526 491 135 059 2;
  • 7) 0.879 999 999 999 995 452 526 491 135 059 2 × 2 = 1 + 0.759 999 999 999 990 905 052 982 270 118 4;
  • 8) 0.759 999 999 999 990 905 052 982 270 118 4 × 2 = 1 + 0.519 999 999 999 981 810 105 964 540 236 8;
  • 9) 0.519 999 999 999 981 810 105 964 540 236 8 × 2 = 1 + 0.039 999 999 999 963 620 211 929 080 473 6;
  • 10) 0.039 999 999 999 963 620 211 929 080 473 6 × 2 = 0 + 0.079 999 999 999 927 240 423 858 160 947 2;
  • 11) 0.079 999 999 999 927 240 423 858 160 947 2 × 2 = 0 + 0.159 999 999 999 854 480 847 716 321 894 4;
  • 12) 0.159 999 999 999 854 480 847 716 321 894 4 × 2 = 0 + 0.319 999 999 999 708 961 695 432 643 788 8;
  • 13) 0.319 999 999 999 708 961 695 432 643 788 8 × 2 = 0 + 0.639 999 999 999 417 923 390 865 287 577 6;
  • 14) 0.639 999 999 999 417 923 390 865 287 577 6 × 2 = 1 + 0.279 999 999 998 835 846 781 730 575 155 2;
  • 15) 0.279 999 999 998 835 846 781 730 575 155 2 × 2 = 0 + 0.559 999 999 997 671 693 563 461 150 310 4;
  • 16) 0.559 999 999 997 671 693 563 461 150 310 4 × 2 = 1 + 0.119 999 999 995 343 387 126 922 300 620 8;
  • 17) 0.119 999 999 995 343 387 126 922 300 620 8 × 2 = 0 + 0.239 999 999 990 686 774 253 844 601 241 6;
  • 18) 0.239 999 999 990 686 774 253 844 601 241 6 × 2 = 0 + 0.479 999 999 981 373 548 507 689 202 483 2;
  • 19) 0.479 999 999 981 373 548 507 689 202 483 2 × 2 = 0 + 0.959 999 999 962 747 097 015 378 404 966 4;
  • 20) 0.959 999 999 962 747 097 015 378 404 966 4 × 2 = 1 + 0.919 999 999 925 494 194 030 756 809 932 8;
  • 21) 0.919 999 999 925 494 194 030 756 809 932 8 × 2 = 1 + 0.839 999 999 850 988 388 061 513 619 865 6;
  • 22) 0.839 999 999 850 988 388 061 513 619 865 6 × 2 = 1 + 0.679 999 999 701 976 776 123 027 239 731 2;
  • 23) 0.679 999 999 701 976 776 123 027 239 731 2 × 2 = 1 + 0.359 999 999 403 953 552 246 054 479 462 4;
  • 24) 0.359 999 999 403 953 552 246 054 479 462 4 × 2 = 0 + 0.719 999 998 807 907 104 492 108 958 924 8;
  • 25) 0.719 999 998 807 907 104 492 108 958 924 8 × 2 = 1 + 0.439 999 997 615 814 208 984 217 917 849 6;
  • 26) 0.439 999 997 615 814 208 984 217 917 849 6 × 2 = 0 + 0.879 999 995 231 628 417 968 435 835 699 2;
  • 27) 0.879 999 995 231 628 417 968 435 835 699 2 × 2 = 1 + 0.759 999 990 463 256 835 936 871 671 398 4;
  • 28) 0.759 999 990 463 256 835 936 871 671 398 4 × 2 = 1 + 0.519 999 980 926 513 671 873 743 342 796 8;
  • 29) 0.519 999 980 926 513 671 873 743 342 796 8 × 2 = 1 + 0.039 999 961 853 027 343 747 486 685 593 6;
  • 30) 0.039 999 961 853 027 343 747 486 685 593 6 × 2 = 0 + 0.079 999 923 706 054 687 494 973 371 187 2;
  • 31) 0.079 999 923 706 054 687 494 973 371 187 2 × 2 = 0 + 0.159 999 847 412 109 374 989 946 742 374 4;
  • 32) 0.159 999 847 412 109 374 989 946 742 374 4 × 2 = 0 + 0.319 999 694 824 218 749 979 893 484 748 8;
  • 33) 0.319 999 694 824 218 749 979 893 484 748 8 × 2 = 0 + 0.639 999 389 648 437 499 959 786 969 497 6;
  • 34) 0.639 999 389 648 437 499 959 786 969 497 6 × 2 = 1 + 0.279 998 779 296 874 999 919 573 938 995 2;
  • 35) 0.279 998 779 296 874 999 919 573 938 995 2 × 2 = 0 + 0.559 997 558 593 749 999 839 147 877 990 4;
  • 36) 0.559 997 558 593 749 999 839 147 877 990 4 × 2 = 1 + 0.119 995 117 187 499 999 678 295 755 980 8;
  • 37) 0.119 995 117 187 499 999 678 295 755 980 8 × 2 = 0 + 0.239 990 234 374 999 999 356 591 511 961 6;
  • 38) 0.239 990 234 374 999 999 356 591 511 961 6 × 2 = 0 + 0.479 980 468 749 999 998 713 183 023 923 2;
  • 39) 0.479 980 468 749 999 998 713 183 023 923 2 × 2 = 0 + 0.959 960 937 499 999 997 426 366 047 846 4;
  • 40) 0.959 960 937 499 999 997 426 366 047 846 4 × 2 = 1 + 0.919 921 874 999 999 994 852 732 095 692 8;
  • 41) 0.919 921 874 999 999 994 852 732 095 692 8 × 2 = 1 + 0.839 843 749 999 999 989 705 464 191 385 6;
  • 42) 0.839 843 749 999 999 989 705 464 191 385 6 × 2 = 1 + 0.679 687 499 999 999 979 410 928 382 771 2;
  • 43) 0.679 687 499 999 999 979 410 928 382 771 2 × 2 = 1 + 0.359 374 999 999 999 958 821 856 765 542 4;
  • 44) 0.359 374 999 999 999 958 821 856 765 542 4 × 2 = 0 + 0.718 749 999 999 999 917 643 713 531 084 8;
  • 45) 0.718 749 999 999 999 917 643 713 531 084 8 × 2 = 1 + 0.437 499 999 999 999 835 287 427 062 169 6;
  • 46) 0.437 499 999 999 999 835 287 427 062 169 6 × 2 = 0 + 0.874 999 999 999 999 670 574 854 124 339 2;
  • 47) 0.874 999 999 999 999 670 574 854 124 339 2 × 2 = 1 + 0.749 999 999 999 999 341 149 708 248 678 4;
  • 48) 0.749 999 999 999 999 341 149 708 248 678 4 × 2 = 1 + 0.499 999 999 999 998 682 299 416 497 356 8;
  • 49) 0.499 999 999 999 998 682 299 416 497 356 8 × 2 = 0 + 0.999 999 999 999 997 364 598 832 994 713 6;
  • 50) 0.999 999 999 999 997 364 598 832 994 713 6 × 2 = 1 + 0.999 999 999 999 994 729 197 665 989 427 2;
  • 51) 0.999 999 999 999 994 729 197 665 989 427 2 × 2 = 1 + 0.999 999 999 999 989 458 395 331 978 854 4;
  • 52) 0.999 999 999 999 989 458 395 331 978 854 4 × 2 = 1 + 0.999 999 999 999 978 916 790 663 957 708 8;
  • 53) 0.999 999 999 999 978 916 790 663 957 708 8 × 2 = 1 + 0.999 999 999 999 957 833 581 327 915 417 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:


0.919 999 999 999 999 928 945 726 423 985 3(10) =

0.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2)

5. Positive number before normalization:

5.919 999 999 999 999 928 945 726 423 985 3(10) =

101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


5.919 999 999 999 999 928 945 726 423 985 3(10) =

101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2) =

101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2) × 20 =

1.0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 111(2) × 22


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): 2

Mantissa (not normalized):
1.0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 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) =

1025(10) =

100 0000 0001(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. 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 111 =

0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101


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 0001

Mantissa (52 bits) =
0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101


Decimal number 5.919 999 999 999 999 928 945 726 423 985 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101

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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