0.785 430 024 824 850 447 853 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.785 430 024 824 850 447 853(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
0.785 430 024 824 850 447 853(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: 0.785 430 024 824 850 447 853.

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) 0.785 430 024 824 850 447 853 × 2 = 1 + 0.570 860 049 649 700 895 706;
  • 2) 0.570 860 049 649 700 895 706 × 2 = 1 + 0.141 720 099 299 401 791 412;
  • 3) 0.141 720 099 299 401 791 412 × 2 = 0 + 0.283 440 198 598 803 582 824;
  • 4) 0.283 440 198 598 803 582 824 × 2 = 0 + 0.566 880 397 197 607 165 648;
  • 5) 0.566 880 397 197 607 165 648 × 2 = 1 + 0.133 760 794 395 214 331 296;
  • 6) 0.133 760 794 395 214 331 296 × 2 = 0 + 0.267 521 588 790 428 662 592;
  • 7) 0.267 521 588 790 428 662 592 × 2 = 0 + 0.535 043 177 580 857 325 184;
  • 8) 0.535 043 177 580 857 325 184 × 2 = 1 + 0.070 086 355 161 714 650 368;
  • 9) 0.070 086 355 161 714 650 368 × 2 = 0 + 0.140 172 710 323 429 300 736;
  • 10) 0.140 172 710 323 429 300 736 × 2 = 0 + 0.280 345 420 646 858 601 472;
  • 11) 0.280 345 420 646 858 601 472 × 2 = 0 + 0.560 690 841 293 717 202 944;
  • 12) 0.560 690 841 293 717 202 944 × 2 = 1 + 0.121 381 682 587 434 405 888;
  • 13) 0.121 381 682 587 434 405 888 × 2 = 0 + 0.242 763 365 174 868 811 776;
  • 14) 0.242 763 365 174 868 811 776 × 2 = 0 + 0.485 526 730 349 737 623 552;
  • 15) 0.485 526 730 349 737 623 552 × 2 = 0 + 0.971 053 460 699 475 247 104;
  • 16) 0.971 053 460 699 475 247 104 × 2 = 1 + 0.942 106 921 398 950 494 208;
  • 17) 0.942 106 921 398 950 494 208 × 2 = 1 + 0.884 213 842 797 900 988 416;
  • 18) 0.884 213 842 797 900 988 416 × 2 = 1 + 0.768 427 685 595 801 976 832;
  • 19) 0.768 427 685 595 801 976 832 × 2 = 1 + 0.536 855 371 191 603 953 664;
  • 20) 0.536 855 371 191 603 953 664 × 2 = 1 + 0.073 710 742 383 207 907 328;
  • 21) 0.073 710 742 383 207 907 328 × 2 = 0 + 0.147 421 484 766 415 814 656;
  • 22) 0.147 421 484 766 415 814 656 × 2 = 0 + 0.294 842 969 532 831 629 312;
  • 23) 0.294 842 969 532 831 629 312 × 2 = 0 + 0.589 685 939 065 663 258 624;
  • 24) 0.589 685 939 065 663 258 624 × 2 = 1 + 0.179 371 878 131 326 517 248;
  • 25) 0.179 371 878 131 326 517 248 × 2 = 0 + 0.358 743 756 262 653 034 496;
  • 26) 0.358 743 756 262 653 034 496 × 2 = 0 + 0.717 487 512 525 306 068 992;
  • 27) 0.717 487 512 525 306 068 992 × 2 = 1 + 0.434 975 025 050 612 137 984;
  • 28) 0.434 975 025 050 612 137 984 × 2 = 0 + 0.869 950 050 101 224 275 968;
  • 29) 0.869 950 050 101 224 275 968 × 2 = 1 + 0.739 900 100 202 448 551 936;
  • 30) 0.739 900 100 202 448 551 936 × 2 = 1 + 0.479 800 200 404 897 103 872;
  • 31) 0.479 800 200 404 897 103 872 × 2 = 0 + 0.959 600 400 809 794 207 744;
  • 32) 0.959 600 400 809 794 207 744 × 2 = 1 + 0.919 200 801 619 588 415 488;
  • 33) 0.919 200 801 619 588 415 488 × 2 = 1 + 0.838 401 603 239 176 830 976;
  • 34) 0.838 401 603 239 176 830 976 × 2 = 1 + 0.676 803 206 478 353 661 952;
  • 35) 0.676 803 206 478 353 661 952 × 2 = 1 + 0.353 606 412 956 707 323 904;
  • 36) 0.353 606 412 956 707 323 904 × 2 = 0 + 0.707 212 825 913 414 647 808;
  • 37) 0.707 212 825 913 414 647 808 × 2 = 1 + 0.414 425 651 826 829 295 616;
  • 38) 0.414 425 651 826 829 295 616 × 2 = 0 + 0.828 851 303 653 658 591 232;
  • 39) 0.828 851 303 653 658 591 232 × 2 = 1 + 0.657 702 607 307 317 182 464;
  • 40) 0.657 702 607 307 317 182 464 × 2 = 1 + 0.315 405 214 614 634 364 928;
  • 41) 0.315 405 214 614 634 364 928 × 2 = 0 + 0.630 810 429 229 268 729 856;
  • 42) 0.630 810 429 229 268 729 856 × 2 = 1 + 0.261 620 858 458 537 459 712;
  • 43) 0.261 620 858 458 537 459 712 × 2 = 0 + 0.523 241 716 917 074 919 424;
  • 44) 0.523 241 716 917 074 919 424 × 2 = 1 + 0.046 483 433 834 149 838 848;
  • 45) 0.046 483 433 834 149 838 848 × 2 = 0 + 0.092 966 867 668 299 677 696;
  • 46) 0.092 966 867 668 299 677 696 × 2 = 0 + 0.185 933 735 336 599 355 392;
  • 47) 0.185 933 735 336 599 355 392 × 2 = 0 + 0.371 867 470 673 198 710 784;
  • 48) 0.371 867 470 673 198 710 784 × 2 = 0 + 0.743 734 941 346 397 421 568;
  • 49) 0.743 734 941 346 397 421 568 × 2 = 1 + 0.487 469 882 692 794 843 136;
  • 50) 0.487 469 882 692 794 843 136 × 2 = 0 + 0.974 939 765 385 589 686 272;
  • 51) 0.974 939 765 385 589 686 272 × 2 = 1 + 0.949 879 530 771 179 372 544;
  • 52) 0.949 879 530 771 179 372 544 × 2 = 1 + 0.899 759 061 542 358 745 088;
  • 53) 0.899 759 061 542 358 745 088 × 2 = 1 + 0.799 518 123 084 717 490 176;

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:


0.785 430 024 824 850 447 853(10) =

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

5. Positive number before normalization:

0.785 430 024 824 850 447 853(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:


0.785 430 024 824 850 447 853(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 0.785 430 024 824 850 447 853 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

Share

» Convert decimal 0.785 430 024 824 850 447 931 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: Your Greatest Competitor, Your Most Powerful Ally. Gemini Notebook YouTube Video of 7.54 minutes

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