0.785 430 024 824 850 447 561 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.785 430 024 824 850 447 561(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
0.785 430 024 824 850 447 561(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: 0.785 430 024 824 850 447 561.

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) 0.785 430 024 824 850 447 561 × 2 = 1 + 0.570 860 049 649 700 895 122;
  • 2) 0.570 860 049 649 700 895 122 × 2 = 1 + 0.141 720 099 299 401 790 244;
  • 3) 0.141 720 099 299 401 790 244 × 2 = 0 + 0.283 440 198 598 803 580 488;
  • 4) 0.283 440 198 598 803 580 488 × 2 = 0 + 0.566 880 397 197 607 160 976;
  • 5) 0.566 880 397 197 607 160 976 × 2 = 1 + 0.133 760 794 395 214 321 952;
  • 6) 0.133 760 794 395 214 321 952 × 2 = 0 + 0.267 521 588 790 428 643 904;
  • 7) 0.267 521 588 790 428 643 904 × 2 = 0 + 0.535 043 177 580 857 287 808;
  • 8) 0.535 043 177 580 857 287 808 × 2 = 1 + 0.070 086 355 161 714 575 616;
  • 9) 0.070 086 355 161 714 575 616 × 2 = 0 + 0.140 172 710 323 429 151 232;
  • 10) 0.140 172 710 323 429 151 232 × 2 = 0 + 0.280 345 420 646 858 302 464;
  • 11) 0.280 345 420 646 858 302 464 × 2 = 0 + 0.560 690 841 293 716 604 928;
  • 12) 0.560 690 841 293 716 604 928 × 2 = 1 + 0.121 381 682 587 433 209 856;
  • 13) 0.121 381 682 587 433 209 856 × 2 = 0 + 0.242 763 365 174 866 419 712;
  • 14) 0.242 763 365 174 866 419 712 × 2 = 0 + 0.485 526 730 349 732 839 424;
  • 15) 0.485 526 730 349 732 839 424 × 2 = 0 + 0.971 053 460 699 465 678 848;
  • 16) 0.971 053 460 699 465 678 848 × 2 = 1 + 0.942 106 921 398 931 357 696;
  • 17) 0.942 106 921 398 931 357 696 × 2 = 1 + 0.884 213 842 797 862 715 392;
  • 18) 0.884 213 842 797 862 715 392 × 2 = 1 + 0.768 427 685 595 725 430 784;
  • 19) 0.768 427 685 595 725 430 784 × 2 = 1 + 0.536 855 371 191 450 861 568;
  • 20) 0.536 855 371 191 450 861 568 × 2 = 1 + 0.073 710 742 382 901 723 136;
  • 21) 0.073 710 742 382 901 723 136 × 2 = 0 + 0.147 421 484 765 803 446 272;
  • 22) 0.147 421 484 765 803 446 272 × 2 = 0 + 0.294 842 969 531 606 892 544;
  • 23) 0.294 842 969 531 606 892 544 × 2 = 0 + 0.589 685 939 063 213 785 088;
  • 24) 0.589 685 939 063 213 785 088 × 2 = 1 + 0.179 371 878 126 427 570 176;
  • 25) 0.179 371 878 126 427 570 176 × 2 = 0 + 0.358 743 756 252 855 140 352;
  • 26) 0.358 743 756 252 855 140 352 × 2 = 0 + 0.717 487 512 505 710 280 704;
  • 27) 0.717 487 512 505 710 280 704 × 2 = 1 + 0.434 975 025 011 420 561 408;
  • 28) 0.434 975 025 011 420 561 408 × 2 = 0 + 0.869 950 050 022 841 122 816;
  • 29) 0.869 950 050 022 841 122 816 × 2 = 1 + 0.739 900 100 045 682 245 632;
  • 30) 0.739 900 100 045 682 245 632 × 2 = 1 + 0.479 800 200 091 364 491 264;
  • 31) 0.479 800 200 091 364 491 264 × 2 = 0 + 0.959 600 400 182 728 982 528;
  • 32) 0.959 600 400 182 728 982 528 × 2 = 1 + 0.919 200 800 365 457 965 056;
  • 33) 0.919 200 800 365 457 965 056 × 2 = 1 + 0.838 401 600 730 915 930 112;
  • 34) 0.838 401 600 730 915 930 112 × 2 = 1 + 0.676 803 201 461 831 860 224;
  • 35) 0.676 803 201 461 831 860 224 × 2 = 1 + 0.353 606 402 923 663 720 448;
  • 36) 0.353 606 402 923 663 720 448 × 2 = 0 + 0.707 212 805 847 327 440 896;
  • 37) 0.707 212 805 847 327 440 896 × 2 = 1 + 0.414 425 611 694 654 881 792;
  • 38) 0.414 425 611 694 654 881 792 × 2 = 0 + 0.828 851 223 389 309 763 584;
  • 39) 0.828 851 223 389 309 763 584 × 2 = 1 + 0.657 702 446 778 619 527 168;
  • 40) 0.657 702 446 778 619 527 168 × 2 = 1 + 0.315 404 893 557 239 054 336;
  • 41) 0.315 404 893 557 239 054 336 × 2 = 0 + 0.630 809 787 114 478 108 672;
  • 42) 0.630 809 787 114 478 108 672 × 2 = 1 + 0.261 619 574 228 956 217 344;
  • 43) 0.261 619 574 228 956 217 344 × 2 = 0 + 0.523 239 148 457 912 434 688;
  • 44) 0.523 239 148 457 912 434 688 × 2 = 1 + 0.046 478 296 915 824 869 376;
  • 45) 0.046 478 296 915 824 869 376 × 2 = 0 + 0.092 956 593 831 649 738 752;
  • 46) 0.092 956 593 831 649 738 752 × 2 = 0 + 0.185 913 187 663 299 477 504;
  • 47) 0.185 913 187 663 299 477 504 × 2 = 0 + 0.371 826 375 326 598 955 008;
  • 48) 0.371 826 375 326 598 955 008 × 2 = 0 + 0.743 652 750 653 197 910 016;
  • 49) 0.743 652 750 653 197 910 016 × 2 = 1 + 0.487 305 501 306 395 820 032;
  • 50) 0.487 305 501 306 395 820 032 × 2 = 0 + 0.974 611 002 612 791 640 064;
  • 51) 0.974 611 002 612 791 640 064 × 2 = 1 + 0.949 222 005 225 583 280 128;
  • 52) 0.949 222 005 225 583 280 128 × 2 = 1 + 0.898 444 010 451 166 560 256;
  • 53) 0.898 444 010 451 166 560 256 × 2 = 1 + 0.796 888 020 902 333 120 512;

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:


0.785 430 024 824 850 447 561(10) =

0.1100 1001 0001 0001 1111 0001 0010 1101 1110 1011 0101 0000 1011 1(2)

5. Positive number before normalization:

0.785 430 024 824 850 447 561(10) =

0.1100 1001 0001 0001 1111 0001 0010 1101 1110 1011 0101 0000 1011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.785 430 024 824 850 447 561(10) =

0.1100 1001 0001 0001 1111 0001 0010 1101 1110 1011 0101 0000 1011 1(2) =

0.1100 1001 0001 0001 1111 0001 0010 1101 1110 1011 0101 0000 1011 1(2) × 20 =

1.1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 0111(2) × 2-1


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): -1

Mantissa (not normalized):
1.1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 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) =

1022(10) =

011 1111 1110(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. 1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 0111 =

1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 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 1110

Mantissa (52 bits) =
1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 0111


Decimal number 0.785 430 024 824 850 447 561 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1001 0010 0010 0011 1110 0010 0101 1011 1101 0110 1010 0001 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