0.785 430 024 824 850 447 695 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.785 430 024 824 850 447 695(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
0.785 430 024 824 850 447 695(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: 0.785 430 024 824 850 447 695.

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) 0.785 430 024 824 850 447 695 × 2 = 1 + 0.570 860 049 649 700 895 39;
  • 2) 0.570 860 049 649 700 895 39 × 2 = 1 + 0.141 720 099 299 401 790 78;
  • 3) 0.141 720 099 299 401 790 78 × 2 = 0 + 0.283 440 198 598 803 581 56;
  • 4) 0.283 440 198 598 803 581 56 × 2 = 0 + 0.566 880 397 197 607 163 12;
  • 5) 0.566 880 397 197 607 163 12 × 2 = 1 + 0.133 760 794 395 214 326 24;
  • 6) 0.133 760 794 395 214 326 24 × 2 = 0 + 0.267 521 588 790 428 652 48;
  • 7) 0.267 521 588 790 428 652 48 × 2 = 0 + 0.535 043 177 580 857 304 96;
  • 8) 0.535 043 177 580 857 304 96 × 2 = 1 + 0.070 086 355 161 714 609 92;
  • 9) 0.070 086 355 161 714 609 92 × 2 = 0 + 0.140 172 710 323 429 219 84;
  • 10) 0.140 172 710 323 429 219 84 × 2 = 0 + 0.280 345 420 646 858 439 68;
  • 11) 0.280 345 420 646 858 439 68 × 2 = 0 + 0.560 690 841 293 716 879 36;
  • 12) 0.560 690 841 293 716 879 36 × 2 = 1 + 0.121 381 682 587 433 758 72;
  • 13) 0.121 381 682 587 433 758 72 × 2 = 0 + 0.242 763 365 174 867 517 44;
  • 14) 0.242 763 365 174 867 517 44 × 2 = 0 + 0.485 526 730 349 735 034 88;
  • 15) 0.485 526 730 349 735 034 88 × 2 = 0 + 0.971 053 460 699 470 069 76;
  • 16) 0.971 053 460 699 470 069 76 × 2 = 1 + 0.942 106 921 398 940 139 52;
  • 17) 0.942 106 921 398 940 139 52 × 2 = 1 + 0.884 213 842 797 880 279 04;
  • 18) 0.884 213 842 797 880 279 04 × 2 = 1 + 0.768 427 685 595 760 558 08;
  • 19) 0.768 427 685 595 760 558 08 × 2 = 1 + 0.536 855 371 191 521 116 16;
  • 20) 0.536 855 371 191 521 116 16 × 2 = 1 + 0.073 710 742 383 042 232 32;
  • 21) 0.073 710 742 383 042 232 32 × 2 = 0 + 0.147 421 484 766 084 464 64;
  • 22) 0.147 421 484 766 084 464 64 × 2 = 0 + 0.294 842 969 532 168 929 28;
  • 23) 0.294 842 969 532 168 929 28 × 2 = 0 + 0.589 685 939 064 337 858 56;
  • 24) 0.589 685 939 064 337 858 56 × 2 = 1 + 0.179 371 878 128 675 717 12;
  • 25) 0.179 371 878 128 675 717 12 × 2 = 0 + 0.358 743 756 257 351 434 24;
  • 26) 0.358 743 756 257 351 434 24 × 2 = 0 + 0.717 487 512 514 702 868 48;
  • 27) 0.717 487 512 514 702 868 48 × 2 = 1 + 0.434 975 025 029 405 736 96;
  • 28) 0.434 975 025 029 405 736 96 × 2 = 0 + 0.869 950 050 058 811 473 92;
  • 29) 0.869 950 050 058 811 473 92 × 2 = 1 + 0.739 900 100 117 622 947 84;
  • 30) 0.739 900 100 117 622 947 84 × 2 = 1 + 0.479 800 200 235 245 895 68;
  • 31) 0.479 800 200 235 245 895 68 × 2 = 0 + 0.959 600 400 470 491 791 36;
  • 32) 0.959 600 400 470 491 791 36 × 2 = 1 + 0.919 200 800 940 983 582 72;
  • 33) 0.919 200 800 940 983 582 72 × 2 = 1 + 0.838 401 601 881 967 165 44;
  • 34) 0.838 401 601 881 967 165 44 × 2 = 1 + 0.676 803 203 763 934 330 88;
  • 35) 0.676 803 203 763 934 330 88 × 2 = 1 + 0.353 606 407 527 868 661 76;
  • 36) 0.353 606 407 527 868 661 76 × 2 = 0 + 0.707 212 815 055 737 323 52;
  • 37) 0.707 212 815 055 737 323 52 × 2 = 1 + 0.414 425 630 111 474 647 04;
  • 38) 0.414 425 630 111 474 647 04 × 2 = 0 + 0.828 851 260 222 949 294 08;
  • 39) 0.828 851 260 222 949 294 08 × 2 = 1 + 0.657 702 520 445 898 588 16;
  • 40) 0.657 702 520 445 898 588 16 × 2 = 1 + 0.315 405 040 891 797 176 32;
  • 41) 0.315 405 040 891 797 176 32 × 2 = 0 + 0.630 810 081 783 594 352 64;
  • 42) 0.630 810 081 783 594 352 64 × 2 = 1 + 0.261 620 163 567 188 705 28;
  • 43) 0.261 620 163 567 188 705 28 × 2 = 0 + 0.523 240 327 134 377 410 56;
  • 44) 0.523 240 327 134 377 410 56 × 2 = 1 + 0.046 480 654 268 754 821 12;
  • 45) 0.046 480 654 268 754 821 12 × 2 = 0 + 0.092 961 308 537 509 642 24;
  • 46) 0.092 961 308 537 509 642 24 × 2 = 0 + 0.185 922 617 075 019 284 48;
  • 47) 0.185 922 617 075 019 284 48 × 2 = 0 + 0.371 845 234 150 038 568 96;
  • 48) 0.371 845 234 150 038 568 96 × 2 = 0 + 0.743 690 468 300 077 137 92;
  • 49) 0.743 690 468 300 077 137 92 × 2 = 1 + 0.487 380 936 600 154 275 84;
  • 50) 0.487 380 936 600 154 275 84 × 2 = 0 + 0.974 761 873 200 308 551 68;
  • 51) 0.974 761 873 200 308 551 68 × 2 = 1 + 0.949 523 746 400 617 103 36;
  • 52) 0.949 523 746 400 617 103 36 × 2 = 1 + 0.899 047 492 801 234 206 72;
  • 53) 0.899 047 492 801 234 206 72 × 2 = 1 + 0.798 094 985 602 468 413 44;

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:


0.785 430 024 824 850 447 695(10) =

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

5. Positive number before normalization:

0.785 430 024 824 850 447 695(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:


0.785 430 024 824 850 447 695(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 0.785 430 024 824 850 447 695 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