0.785 430 024 824 850 447 628 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.785 430 024 824 850 447 628(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
0.785 430 024 824 850 447 628(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: 0.785 430 024 824 850 447 628.

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) 0.785 430 024 824 850 447 628 × 2 = 1 + 0.570 860 049 649 700 895 256;
  • 2) 0.570 860 049 649 700 895 256 × 2 = 1 + 0.141 720 099 299 401 790 512;
  • 3) 0.141 720 099 299 401 790 512 × 2 = 0 + 0.283 440 198 598 803 581 024;
  • 4) 0.283 440 198 598 803 581 024 × 2 = 0 + 0.566 880 397 197 607 162 048;
  • 5) 0.566 880 397 197 607 162 048 × 2 = 1 + 0.133 760 794 395 214 324 096;
  • 6) 0.133 760 794 395 214 324 096 × 2 = 0 + 0.267 521 588 790 428 648 192;
  • 7) 0.267 521 588 790 428 648 192 × 2 = 0 + 0.535 043 177 580 857 296 384;
  • 8) 0.535 043 177 580 857 296 384 × 2 = 1 + 0.070 086 355 161 714 592 768;
  • 9) 0.070 086 355 161 714 592 768 × 2 = 0 + 0.140 172 710 323 429 185 536;
  • 10) 0.140 172 710 323 429 185 536 × 2 = 0 + 0.280 345 420 646 858 371 072;
  • 11) 0.280 345 420 646 858 371 072 × 2 = 0 + 0.560 690 841 293 716 742 144;
  • 12) 0.560 690 841 293 716 742 144 × 2 = 1 + 0.121 381 682 587 433 484 288;
  • 13) 0.121 381 682 587 433 484 288 × 2 = 0 + 0.242 763 365 174 866 968 576;
  • 14) 0.242 763 365 174 866 968 576 × 2 = 0 + 0.485 526 730 349 733 937 152;
  • 15) 0.485 526 730 349 733 937 152 × 2 = 0 + 0.971 053 460 699 467 874 304;
  • 16) 0.971 053 460 699 467 874 304 × 2 = 1 + 0.942 106 921 398 935 748 608;
  • 17) 0.942 106 921 398 935 748 608 × 2 = 1 + 0.884 213 842 797 871 497 216;
  • 18) 0.884 213 842 797 871 497 216 × 2 = 1 + 0.768 427 685 595 742 994 432;
  • 19) 0.768 427 685 595 742 994 432 × 2 = 1 + 0.536 855 371 191 485 988 864;
  • 20) 0.536 855 371 191 485 988 864 × 2 = 1 + 0.073 710 742 382 971 977 728;
  • 21) 0.073 710 742 382 971 977 728 × 2 = 0 + 0.147 421 484 765 943 955 456;
  • 22) 0.147 421 484 765 943 955 456 × 2 = 0 + 0.294 842 969 531 887 910 912;
  • 23) 0.294 842 969 531 887 910 912 × 2 = 0 + 0.589 685 939 063 775 821 824;
  • 24) 0.589 685 939 063 775 821 824 × 2 = 1 + 0.179 371 878 127 551 643 648;
  • 25) 0.179 371 878 127 551 643 648 × 2 = 0 + 0.358 743 756 255 103 287 296;
  • 26) 0.358 743 756 255 103 287 296 × 2 = 0 + 0.717 487 512 510 206 574 592;
  • 27) 0.717 487 512 510 206 574 592 × 2 = 1 + 0.434 975 025 020 413 149 184;
  • 28) 0.434 975 025 020 413 149 184 × 2 = 0 + 0.869 950 050 040 826 298 368;
  • 29) 0.869 950 050 040 826 298 368 × 2 = 1 + 0.739 900 100 081 652 596 736;
  • 30) 0.739 900 100 081 652 596 736 × 2 = 1 + 0.479 800 200 163 305 193 472;
  • 31) 0.479 800 200 163 305 193 472 × 2 = 0 + 0.959 600 400 326 610 386 944;
  • 32) 0.959 600 400 326 610 386 944 × 2 = 1 + 0.919 200 800 653 220 773 888;
  • 33) 0.919 200 800 653 220 773 888 × 2 = 1 + 0.838 401 601 306 441 547 776;
  • 34) 0.838 401 601 306 441 547 776 × 2 = 1 + 0.676 803 202 612 883 095 552;
  • 35) 0.676 803 202 612 883 095 552 × 2 = 1 + 0.353 606 405 225 766 191 104;
  • 36) 0.353 606 405 225 766 191 104 × 2 = 0 + 0.707 212 810 451 532 382 208;
  • 37) 0.707 212 810 451 532 382 208 × 2 = 1 + 0.414 425 620 903 064 764 416;
  • 38) 0.414 425 620 903 064 764 416 × 2 = 0 + 0.828 851 241 806 129 528 832;
  • 39) 0.828 851 241 806 129 528 832 × 2 = 1 + 0.657 702 483 612 259 057 664;
  • 40) 0.657 702 483 612 259 057 664 × 2 = 1 + 0.315 404 967 224 518 115 328;
  • 41) 0.315 404 967 224 518 115 328 × 2 = 0 + 0.630 809 934 449 036 230 656;
  • 42) 0.630 809 934 449 036 230 656 × 2 = 1 + 0.261 619 868 898 072 461 312;
  • 43) 0.261 619 868 898 072 461 312 × 2 = 0 + 0.523 239 737 796 144 922 624;
  • 44) 0.523 239 737 796 144 922 624 × 2 = 1 + 0.046 479 475 592 289 845 248;
  • 45) 0.046 479 475 592 289 845 248 × 2 = 0 + 0.092 958 951 184 579 690 496;
  • 46) 0.092 958 951 184 579 690 496 × 2 = 0 + 0.185 917 902 369 159 380 992;
  • 47) 0.185 917 902 369 159 380 992 × 2 = 0 + 0.371 835 804 738 318 761 984;
  • 48) 0.371 835 804 738 318 761 984 × 2 = 0 + 0.743 671 609 476 637 523 968;
  • 49) 0.743 671 609 476 637 523 968 × 2 = 1 + 0.487 343 218 953 275 047 936;
  • 50) 0.487 343 218 953 275 047 936 × 2 = 0 + 0.974 686 437 906 550 095 872;
  • 51) 0.974 686 437 906 550 095 872 × 2 = 1 + 0.949 372 875 813 100 191 744;
  • 52) 0.949 372 875 813 100 191 744 × 2 = 1 + 0.898 745 751 626 200 383 488;
  • 53) 0.898 745 751 626 200 383 488 × 2 = 1 + 0.797 491 503 252 400 766 976;

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:


0.785 430 024 824 850 447 628(10) =

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

5. Positive number before normalization:

0.785 430 024 824 850 447 628(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:


0.785 430 024 824 850 447 628(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 0.785 430 024 824 850 447 628 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