0.785 398 163 397 448 163 538 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.785 398 163 397 448 163 538(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
0.785 398 163 397 448 163 538(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: 0.785 398 163 397 448 163 538.

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) 0.785 398 163 397 448 163 538 × 2 = 1 + 0.570 796 326 794 896 327 076;
  • 2) 0.570 796 326 794 896 327 076 × 2 = 1 + 0.141 592 653 589 792 654 152;
  • 3) 0.141 592 653 589 792 654 152 × 2 = 0 + 0.283 185 307 179 585 308 304;
  • 4) 0.283 185 307 179 585 308 304 × 2 = 0 + 0.566 370 614 359 170 616 608;
  • 5) 0.566 370 614 359 170 616 608 × 2 = 1 + 0.132 741 228 718 341 233 216;
  • 6) 0.132 741 228 718 341 233 216 × 2 = 0 + 0.265 482 457 436 682 466 432;
  • 7) 0.265 482 457 436 682 466 432 × 2 = 0 + 0.530 964 914 873 364 932 864;
  • 8) 0.530 964 914 873 364 932 864 × 2 = 1 + 0.061 929 829 746 729 865 728;
  • 9) 0.061 929 829 746 729 865 728 × 2 = 0 + 0.123 859 659 493 459 731 456;
  • 10) 0.123 859 659 493 459 731 456 × 2 = 0 + 0.247 719 318 986 919 462 912;
  • 11) 0.247 719 318 986 919 462 912 × 2 = 0 + 0.495 438 637 973 838 925 824;
  • 12) 0.495 438 637 973 838 925 824 × 2 = 0 + 0.990 877 275 947 677 851 648;
  • 13) 0.990 877 275 947 677 851 648 × 2 = 1 + 0.981 754 551 895 355 703 296;
  • 14) 0.981 754 551 895 355 703 296 × 2 = 1 + 0.963 509 103 790 711 406 592;
  • 15) 0.963 509 103 790 711 406 592 × 2 = 1 + 0.927 018 207 581 422 813 184;
  • 16) 0.927 018 207 581 422 813 184 × 2 = 1 + 0.854 036 415 162 845 626 368;
  • 17) 0.854 036 415 162 845 626 368 × 2 = 1 + 0.708 072 830 325 691 252 736;
  • 18) 0.708 072 830 325 691 252 736 × 2 = 1 + 0.416 145 660 651 382 505 472;
  • 19) 0.416 145 660 651 382 505 472 × 2 = 0 + 0.832 291 321 302 765 010 944;
  • 20) 0.832 291 321 302 765 010 944 × 2 = 1 + 0.664 582 642 605 530 021 888;
  • 21) 0.664 582 642 605 530 021 888 × 2 = 1 + 0.329 165 285 211 060 043 776;
  • 22) 0.329 165 285 211 060 043 776 × 2 = 0 + 0.658 330 570 422 120 087 552;
  • 23) 0.658 330 570 422 120 087 552 × 2 = 1 + 0.316 661 140 844 240 175 104;
  • 24) 0.316 661 140 844 240 175 104 × 2 = 0 + 0.633 322 281 688 480 350 208;
  • 25) 0.633 322 281 688 480 350 208 × 2 = 1 + 0.266 644 563 376 960 700 416;
  • 26) 0.266 644 563 376 960 700 416 × 2 = 0 + 0.533 289 126 753 921 400 832;
  • 27) 0.533 289 126 753 921 400 832 × 2 = 1 + 0.066 578 253 507 842 801 664;
  • 28) 0.066 578 253 507 842 801 664 × 2 = 0 + 0.133 156 507 015 685 603 328;
  • 29) 0.133 156 507 015 685 603 328 × 2 = 0 + 0.266 313 014 031 371 206 656;
  • 30) 0.266 313 014 031 371 206 656 × 2 = 0 + 0.532 626 028 062 742 413 312;
  • 31) 0.532 626 028 062 742 413 312 × 2 = 1 + 0.065 252 056 125 484 826 624;
  • 32) 0.065 252 056 125 484 826 624 × 2 = 0 + 0.130 504 112 250 969 653 248;
  • 33) 0.130 504 112 250 969 653 248 × 2 = 0 + 0.261 008 224 501 939 306 496;
  • 34) 0.261 008 224 501 939 306 496 × 2 = 0 + 0.522 016 449 003 878 612 992;
  • 35) 0.522 016 449 003 878 612 992 × 2 = 1 + 0.044 032 898 007 757 225 984;
  • 36) 0.044 032 898 007 757 225 984 × 2 = 0 + 0.088 065 796 015 514 451 968;
  • 37) 0.088 065 796 015 514 451 968 × 2 = 0 + 0.176 131 592 031 028 903 936;
  • 38) 0.176 131 592 031 028 903 936 × 2 = 0 + 0.352 263 184 062 057 807 872;
  • 39) 0.352 263 184 062 057 807 872 × 2 = 0 + 0.704 526 368 124 115 615 744;
  • 40) 0.704 526 368 124 115 615 744 × 2 = 1 + 0.409 052 736 248 231 231 488;
  • 41) 0.409 052 736 248 231 231 488 × 2 = 0 + 0.818 105 472 496 462 462 976;
  • 42) 0.818 105 472 496 462 462 976 × 2 = 1 + 0.636 210 944 992 924 925 952;
  • 43) 0.636 210 944 992 924 925 952 × 2 = 1 + 0.272 421 889 985 849 851 904;
  • 44) 0.272 421 889 985 849 851 904 × 2 = 0 + 0.544 843 779 971 699 703 808;
  • 45) 0.544 843 779 971 699 703 808 × 2 = 1 + 0.089 687 559 943 399 407 616;
  • 46) 0.089 687 559 943 399 407 616 × 2 = 0 + 0.179 375 119 886 798 815 232;
  • 47) 0.179 375 119 886 798 815 232 × 2 = 0 + 0.358 750 239 773 597 630 464;
  • 48) 0.358 750 239 773 597 630 464 × 2 = 0 + 0.717 500 479 547 195 260 928;
  • 49) 0.717 500 479 547 195 260 928 × 2 = 1 + 0.435 000 959 094 390 521 856;
  • 50) 0.435 000 959 094 390 521 856 × 2 = 0 + 0.870 001 918 188 781 043 712;
  • 51) 0.870 001 918 188 781 043 712 × 2 = 1 + 0.740 003 836 377 562 087 424;
  • 52) 0.740 003 836 377 562 087 424 × 2 = 1 + 0.480 007 672 755 124 174 848;
  • 53) 0.480 007 672 755 124 174 848 × 2 = 0 + 0.960 015 345 510 248 349 696;

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:


0.785 398 163 397 448 163 538(10) =

0.1100 1001 0000 1111 1101 1010 1010 0010 0010 0001 0110 1000 1011 0(2)

5. Positive number before normalization:

0.785 398 163 397 448 163 538(10) =

0.1100 1001 0000 1111 1101 1010 1010 0010 0010 0001 0110 1000 1011 0(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:


0.785 398 163 397 448 163 538(10) =

0.1100 1001 0000 1111 1101 1010 1010 0010 0010 0001 0110 1000 1011 0(2) =

0.1100 1001 0000 1111 1101 1010 1010 0010 0010 0001 0110 1000 1011 0(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110(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 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110


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 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110 =

1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110


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 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110


Decimal number 0.785 398 163 397 448 163 538 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110

» Convert decimal 0.785 398 163 397 448 163 442 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 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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