0.785 430 024 824 850 447 806 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.785 430 024 824 850 447 806(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
0.785 430 024 824 850 447 806(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: 0.785 430 024 824 850 447 806.

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) 0.785 430 024 824 850 447 806 × 2 = 1 + 0.570 860 049 649 700 895 612;
  • 2) 0.570 860 049 649 700 895 612 × 2 = 1 + 0.141 720 099 299 401 791 224;
  • 3) 0.141 720 099 299 401 791 224 × 2 = 0 + 0.283 440 198 598 803 582 448;
  • 4) 0.283 440 198 598 803 582 448 × 2 = 0 + 0.566 880 397 197 607 164 896;
  • 5) 0.566 880 397 197 607 164 896 × 2 = 1 + 0.133 760 794 395 214 329 792;
  • 6) 0.133 760 794 395 214 329 792 × 2 = 0 + 0.267 521 588 790 428 659 584;
  • 7) 0.267 521 588 790 428 659 584 × 2 = 0 + 0.535 043 177 580 857 319 168;
  • 8) 0.535 043 177 580 857 319 168 × 2 = 1 + 0.070 086 355 161 714 638 336;
  • 9) 0.070 086 355 161 714 638 336 × 2 = 0 + 0.140 172 710 323 429 276 672;
  • 10) 0.140 172 710 323 429 276 672 × 2 = 0 + 0.280 345 420 646 858 553 344;
  • 11) 0.280 345 420 646 858 553 344 × 2 = 0 + 0.560 690 841 293 717 106 688;
  • 12) 0.560 690 841 293 717 106 688 × 2 = 1 + 0.121 381 682 587 434 213 376;
  • 13) 0.121 381 682 587 434 213 376 × 2 = 0 + 0.242 763 365 174 868 426 752;
  • 14) 0.242 763 365 174 868 426 752 × 2 = 0 + 0.485 526 730 349 736 853 504;
  • 15) 0.485 526 730 349 736 853 504 × 2 = 0 + 0.971 053 460 699 473 707 008;
  • 16) 0.971 053 460 699 473 707 008 × 2 = 1 + 0.942 106 921 398 947 414 016;
  • 17) 0.942 106 921 398 947 414 016 × 2 = 1 + 0.884 213 842 797 894 828 032;
  • 18) 0.884 213 842 797 894 828 032 × 2 = 1 + 0.768 427 685 595 789 656 064;
  • 19) 0.768 427 685 595 789 656 064 × 2 = 1 + 0.536 855 371 191 579 312 128;
  • 20) 0.536 855 371 191 579 312 128 × 2 = 1 + 0.073 710 742 383 158 624 256;
  • 21) 0.073 710 742 383 158 624 256 × 2 = 0 + 0.147 421 484 766 317 248 512;
  • 22) 0.147 421 484 766 317 248 512 × 2 = 0 + 0.294 842 969 532 634 497 024;
  • 23) 0.294 842 969 532 634 497 024 × 2 = 0 + 0.589 685 939 065 268 994 048;
  • 24) 0.589 685 939 065 268 994 048 × 2 = 1 + 0.179 371 878 130 537 988 096;
  • 25) 0.179 371 878 130 537 988 096 × 2 = 0 + 0.358 743 756 261 075 976 192;
  • 26) 0.358 743 756 261 075 976 192 × 2 = 0 + 0.717 487 512 522 151 952 384;
  • 27) 0.717 487 512 522 151 952 384 × 2 = 1 + 0.434 975 025 044 303 904 768;
  • 28) 0.434 975 025 044 303 904 768 × 2 = 0 + 0.869 950 050 088 607 809 536;
  • 29) 0.869 950 050 088 607 809 536 × 2 = 1 + 0.739 900 100 177 215 619 072;
  • 30) 0.739 900 100 177 215 619 072 × 2 = 1 + 0.479 800 200 354 431 238 144;
  • 31) 0.479 800 200 354 431 238 144 × 2 = 0 + 0.959 600 400 708 862 476 288;
  • 32) 0.959 600 400 708 862 476 288 × 2 = 1 + 0.919 200 801 417 724 952 576;
  • 33) 0.919 200 801 417 724 952 576 × 2 = 1 + 0.838 401 602 835 449 905 152;
  • 34) 0.838 401 602 835 449 905 152 × 2 = 1 + 0.676 803 205 670 899 810 304;
  • 35) 0.676 803 205 670 899 810 304 × 2 = 1 + 0.353 606 411 341 799 620 608;
  • 36) 0.353 606 411 341 799 620 608 × 2 = 0 + 0.707 212 822 683 599 241 216;
  • 37) 0.707 212 822 683 599 241 216 × 2 = 1 + 0.414 425 645 367 198 482 432;
  • 38) 0.414 425 645 367 198 482 432 × 2 = 0 + 0.828 851 290 734 396 964 864;
  • 39) 0.828 851 290 734 396 964 864 × 2 = 1 + 0.657 702 581 468 793 929 728;
  • 40) 0.657 702 581 468 793 929 728 × 2 = 1 + 0.315 405 162 937 587 859 456;
  • 41) 0.315 405 162 937 587 859 456 × 2 = 0 + 0.630 810 325 875 175 718 912;
  • 42) 0.630 810 325 875 175 718 912 × 2 = 1 + 0.261 620 651 750 351 437 824;
  • 43) 0.261 620 651 750 351 437 824 × 2 = 0 + 0.523 241 303 500 702 875 648;
  • 44) 0.523 241 303 500 702 875 648 × 2 = 1 + 0.046 482 607 001 405 751 296;
  • 45) 0.046 482 607 001 405 751 296 × 2 = 0 + 0.092 965 214 002 811 502 592;
  • 46) 0.092 965 214 002 811 502 592 × 2 = 0 + 0.185 930 428 005 623 005 184;
  • 47) 0.185 930 428 005 623 005 184 × 2 = 0 + 0.371 860 856 011 246 010 368;
  • 48) 0.371 860 856 011 246 010 368 × 2 = 0 + 0.743 721 712 022 492 020 736;
  • 49) 0.743 721 712 022 492 020 736 × 2 = 1 + 0.487 443 424 044 984 041 472;
  • 50) 0.487 443 424 044 984 041 472 × 2 = 0 + 0.974 886 848 089 968 082 944;
  • 51) 0.974 886 848 089 968 082 944 × 2 = 1 + 0.949 773 696 179 936 165 888;
  • 52) 0.949 773 696 179 936 165 888 × 2 = 1 + 0.899 547 392 359 872 331 776;
  • 53) 0.899 547 392 359 872 331 776 × 2 = 1 + 0.799 094 784 719 744 663 552;

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:


0.785 430 024 824 850 447 806(10) =

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

5. Positive number before normalization:

0.785 430 024 824 850 447 806(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:


0.785 430 024 824 850 447 806(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 0.785 430 024 824 850 447 806 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