0.785 430 024 824 850 447 689 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.785 430 024 824 850 447 689(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
0.785 430 024 824 850 447 689(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: 0.785 430 024 824 850 447 689.

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) 0.785 430 024 824 850 447 689 × 2 = 1 + 0.570 860 049 649 700 895 378;
  • 2) 0.570 860 049 649 700 895 378 × 2 = 1 + 0.141 720 099 299 401 790 756;
  • 3) 0.141 720 099 299 401 790 756 × 2 = 0 + 0.283 440 198 598 803 581 512;
  • 4) 0.283 440 198 598 803 581 512 × 2 = 0 + 0.566 880 397 197 607 163 024;
  • 5) 0.566 880 397 197 607 163 024 × 2 = 1 + 0.133 760 794 395 214 326 048;
  • 6) 0.133 760 794 395 214 326 048 × 2 = 0 + 0.267 521 588 790 428 652 096;
  • 7) 0.267 521 588 790 428 652 096 × 2 = 0 + 0.535 043 177 580 857 304 192;
  • 8) 0.535 043 177 580 857 304 192 × 2 = 1 + 0.070 086 355 161 714 608 384;
  • 9) 0.070 086 355 161 714 608 384 × 2 = 0 + 0.140 172 710 323 429 216 768;
  • 10) 0.140 172 710 323 429 216 768 × 2 = 0 + 0.280 345 420 646 858 433 536;
  • 11) 0.280 345 420 646 858 433 536 × 2 = 0 + 0.560 690 841 293 716 867 072;
  • 12) 0.560 690 841 293 716 867 072 × 2 = 1 + 0.121 381 682 587 433 734 144;
  • 13) 0.121 381 682 587 433 734 144 × 2 = 0 + 0.242 763 365 174 867 468 288;
  • 14) 0.242 763 365 174 867 468 288 × 2 = 0 + 0.485 526 730 349 734 936 576;
  • 15) 0.485 526 730 349 734 936 576 × 2 = 0 + 0.971 053 460 699 469 873 152;
  • 16) 0.971 053 460 699 469 873 152 × 2 = 1 + 0.942 106 921 398 939 746 304;
  • 17) 0.942 106 921 398 939 746 304 × 2 = 1 + 0.884 213 842 797 879 492 608;
  • 18) 0.884 213 842 797 879 492 608 × 2 = 1 + 0.768 427 685 595 758 985 216;
  • 19) 0.768 427 685 595 758 985 216 × 2 = 1 + 0.536 855 371 191 517 970 432;
  • 20) 0.536 855 371 191 517 970 432 × 2 = 1 + 0.073 710 742 383 035 940 864;
  • 21) 0.073 710 742 383 035 940 864 × 2 = 0 + 0.147 421 484 766 071 881 728;
  • 22) 0.147 421 484 766 071 881 728 × 2 = 0 + 0.294 842 969 532 143 763 456;
  • 23) 0.294 842 969 532 143 763 456 × 2 = 0 + 0.589 685 939 064 287 526 912;
  • 24) 0.589 685 939 064 287 526 912 × 2 = 1 + 0.179 371 878 128 575 053 824;
  • 25) 0.179 371 878 128 575 053 824 × 2 = 0 + 0.358 743 756 257 150 107 648;
  • 26) 0.358 743 756 257 150 107 648 × 2 = 0 + 0.717 487 512 514 300 215 296;
  • 27) 0.717 487 512 514 300 215 296 × 2 = 1 + 0.434 975 025 028 600 430 592;
  • 28) 0.434 975 025 028 600 430 592 × 2 = 0 + 0.869 950 050 057 200 861 184;
  • 29) 0.869 950 050 057 200 861 184 × 2 = 1 + 0.739 900 100 114 401 722 368;
  • 30) 0.739 900 100 114 401 722 368 × 2 = 1 + 0.479 800 200 228 803 444 736;
  • 31) 0.479 800 200 228 803 444 736 × 2 = 0 + 0.959 600 400 457 606 889 472;
  • 32) 0.959 600 400 457 606 889 472 × 2 = 1 + 0.919 200 800 915 213 778 944;
  • 33) 0.919 200 800 915 213 778 944 × 2 = 1 + 0.838 401 601 830 427 557 888;
  • 34) 0.838 401 601 830 427 557 888 × 2 = 1 + 0.676 803 203 660 855 115 776;
  • 35) 0.676 803 203 660 855 115 776 × 2 = 1 + 0.353 606 407 321 710 231 552;
  • 36) 0.353 606 407 321 710 231 552 × 2 = 0 + 0.707 212 814 643 420 463 104;
  • 37) 0.707 212 814 643 420 463 104 × 2 = 1 + 0.414 425 629 286 840 926 208;
  • 38) 0.414 425 629 286 840 926 208 × 2 = 0 + 0.828 851 258 573 681 852 416;
  • 39) 0.828 851 258 573 681 852 416 × 2 = 1 + 0.657 702 517 147 363 704 832;
  • 40) 0.657 702 517 147 363 704 832 × 2 = 1 + 0.315 405 034 294 727 409 664;
  • 41) 0.315 405 034 294 727 409 664 × 2 = 0 + 0.630 810 068 589 454 819 328;
  • 42) 0.630 810 068 589 454 819 328 × 2 = 1 + 0.261 620 137 178 909 638 656;
  • 43) 0.261 620 137 178 909 638 656 × 2 = 0 + 0.523 240 274 357 819 277 312;
  • 44) 0.523 240 274 357 819 277 312 × 2 = 1 + 0.046 480 548 715 638 554 624;
  • 45) 0.046 480 548 715 638 554 624 × 2 = 0 + 0.092 961 097 431 277 109 248;
  • 46) 0.092 961 097 431 277 109 248 × 2 = 0 + 0.185 922 194 862 554 218 496;
  • 47) 0.185 922 194 862 554 218 496 × 2 = 0 + 0.371 844 389 725 108 436 992;
  • 48) 0.371 844 389 725 108 436 992 × 2 = 0 + 0.743 688 779 450 216 873 984;
  • 49) 0.743 688 779 450 216 873 984 × 2 = 1 + 0.487 377 558 900 433 747 968;
  • 50) 0.487 377 558 900 433 747 968 × 2 = 0 + 0.974 755 117 800 867 495 936;
  • 51) 0.974 755 117 800 867 495 936 × 2 = 1 + 0.949 510 235 601 734 991 872;
  • 52) 0.949 510 235 601 734 991 872 × 2 = 1 + 0.899 020 471 203 469 983 744;
  • 53) 0.899 020 471 203 469 983 744 × 2 = 1 + 0.798 040 942 406 939 967 488;

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:


0.785 430 024 824 850 447 689(10) =

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

5. Positive number before normalization:

0.785 430 024 824 850 447 689(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:


0.785 430 024 824 850 447 689(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 0.785 430 024 824 850 447 689 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 0.785 430 024 824 850 447 714 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: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 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