763.859 999 999 999 899 955 582 804 977 893 791 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 763.859 999 999 999 899 955 582 804 977 893 791(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
763.859 999 999 999 899 955 582 804 977 893 791(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: 763.
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;
  • 763 ÷ 2 = 381 + 1;
  • 381 ÷ 2 = 190 + 1;
  • 190 ÷ 2 = 95 + 0;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

763(10) =

10 1111 1011(2)


3. Convert to binary (base 2) the fractional part: 0.859 999 999 999 899 955 582 804 977 893 791.

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) 0.859 999 999 999 899 955 582 804 977 893 791 × 2 = 1 + 0.719 999 999 999 799 911 165 609 955 787 582;
  • 2) 0.719 999 999 999 799 911 165 609 955 787 582 × 2 = 1 + 0.439 999 999 999 599 822 331 219 911 575 164;
  • 3) 0.439 999 999 999 599 822 331 219 911 575 164 × 2 = 0 + 0.879 999 999 999 199 644 662 439 823 150 328;
  • 4) 0.879 999 999 999 199 644 662 439 823 150 328 × 2 = 1 + 0.759 999 999 998 399 289 324 879 646 300 656;
  • 5) 0.759 999 999 998 399 289 324 879 646 300 656 × 2 = 1 + 0.519 999 999 996 798 578 649 759 292 601 312;
  • 6) 0.519 999 999 996 798 578 649 759 292 601 312 × 2 = 1 + 0.039 999 999 993 597 157 299 518 585 202 624;
  • 7) 0.039 999 999 993 597 157 299 518 585 202 624 × 2 = 0 + 0.079 999 999 987 194 314 599 037 170 405 248;
  • 8) 0.079 999 999 987 194 314 599 037 170 405 248 × 2 = 0 + 0.159 999 999 974 388 629 198 074 340 810 496;
  • 9) 0.159 999 999 974 388 629 198 074 340 810 496 × 2 = 0 + 0.319 999 999 948 777 258 396 148 681 620 992;
  • 10) 0.319 999 999 948 777 258 396 148 681 620 992 × 2 = 0 + 0.639 999 999 897 554 516 792 297 363 241 984;
  • 11) 0.639 999 999 897 554 516 792 297 363 241 984 × 2 = 1 + 0.279 999 999 795 109 033 584 594 726 483 968;
  • 12) 0.279 999 999 795 109 033 584 594 726 483 968 × 2 = 0 + 0.559 999 999 590 218 067 169 189 452 967 936;
  • 13) 0.559 999 999 590 218 067 169 189 452 967 936 × 2 = 1 + 0.119 999 999 180 436 134 338 378 905 935 872;
  • 14) 0.119 999 999 180 436 134 338 378 905 935 872 × 2 = 0 + 0.239 999 998 360 872 268 676 757 811 871 744;
  • 15) 0.239 999 998 360 872 268 676 757 811 871 744 × 2 = 0 + 0.479 999 996 721 744 537 353 515 623 743 488;
  • 16) 0.479 999 996 721 744 537 353 515 623 743 488 × 2 = 0 + 0.959 999 993 443 489 074 707 031 247 486 976;
  • 17) 0.959 999 993 443 489 074 707 031 247 486 976 × 2 = 1 + 0.919 999 986 886 978 149 414 062 494 973 952;
  • 18) 0.919 999 986 886 978 149 414 062 494 973 952 × 2 = 1 + 0.839 999 973 773 956 298 828 124 989 947 904;
  • 19) 0.839 999 973 773 956 298 828 124 989 947 904 × 2 = 1 + 0.679 999 947 547 912 597 656 249 979 895 808;
  • 20) 0.679 999 947 547 912 597 656 249 979 895 808 × 2 = 1 + 0.359 999 895 095 825 195 312 499 959 791 616;
  • 21) 0.359 999 895 095 825 195 312 499 959 791 616 × 2 = 0 + 0.719 999 790 191 650 390 624 999 919 583 232;
  • 22) 0.719 999 790 191 650 390 624 999 919 583 232 × 2 = 1 + 0.439 999 580 383 300 781 249 999 839 166 464;
  • 23) 0.439 999 580 383 300 781 249 999 839 166 464 × 2 = 0 + 0.879 999 160 766 601 562 499 999 678 332 928;
  • 24) 0.879 999 160 766 601 562 499 999 678 332 928 × 2 = 1 + 0.759 998 321 533 203 124 999 999 356 665 856;
  • 25) 0.759 998 321 533 203 124 999 999 356 665 856 × 2 = 1 + 0.519 996 643 066 406 249 999 998 713 331 712;
  • 26) 0.519 996 643 066 406 249 999 998 713 331 712 × 2 = 1 + 0.039 993 286 132 812 499 999 997 426 663 424;
  • 27) 0.039 993 286 132 812 499 999 997 426 663 424 × 2 = 0 + 0.079 986 572 265 624 999 999 994 853 326 848;
  • 28) 0.079 986 572 265 624 999 999 994 853 326 848 × 2 = 0 + 0.159 973 144 531 249 999 999 989 706 653 696;
  • 29) 0.159 973 144 531 249 999 999 989 706 653 696 × 2 = 0 + 0.319 946 289 062 499 999 999 979 413 307 392;
  • 30) 0.319 946 289 062 499 999 999 979 413 307 392 × 2 = 0 + 0.639 892 578 124 999 999 999 958 826 614 784;
  • 31) 0.639 892 578 124 999 999 999 958 826 614 784 × 2 = 1 + 0.279 785 156 249 999 999 999 917 653 229 568;
  • 32) 0.279 785 156 249 999 999 999 917 653 229 568 × 2 = 0 + 0.559 570 312 499 999 999 999 835 306 459 136;
  • 33) 0.559 570 312 499 999 999 999 835 306 459 136 × 2 = 1 + 0.119 140 624 999 999 999 999 670 612 918 272;
  • 34) 0.119 140 624 999 999 999 999 670 612 918 272 × 2 = 0 + 0.238 281 249 999 999 999 999 341 225 836 544;
  • 35) 0.238 281 249 999 999 999 999 341 225 836 544 × 2 = 0 + 0.476 562 499 999 999 999 998 682 451 673 088;
  • 36) 0.476 562 499 999 999 999 998 682 451 673 088 × 2 = 0 + 0.953 124 999 999 999 999 997 364 903 346 176;
  • 37) 0.953 124 999 999 999 999 997 364 903 346 176 × 2 = 1 + 0.906 249 999 999 999 999 994 729 806 692 352;
  • 38) 0.906 249 999 999 999 999 994 729 806 692 352 × 2 = 1 + 0.812 499 999 999 999 999 989 459 613 384 704;
  • 39) 0.812 499 999 999 999 999 989 459 613 384 704 × 2 = 1 + 0.624 999 999 999 999 999 978 919 226 769 408;
  • 40) 0.624 999 999 999 999 999 978 919 226 769 408 × 2 = 1 + 0.249 999 999 999 999 999 957 838 453 538 816;
  • 41) 0.249 999 999 999 999 999 957 838 453 538 816 × 2 = 0 + 0.499 999 999 999 999 999 915 676 907 077 632;
  • 42) 0.499 999 999 999 999 999 915 676 907 077 632 × 2 = 0 + 0.999 999 999 999 999 999 831 353 814 155 264;
  • 43) 0.999 999 999 999 999 999 831 353 814 155 264 × 2 = 1 + 0.999 999 999 999 999 999 662 707 628 310 528;
  • 44) 0.999 999 999 999 999 999 662 707 628 310 528 × 2 = 1 + 0.999 999 999 999 999 999 325 415 256 621 056;
  • 45) 0.999 999 999 999 999 999 325 415 256 621 056 × 2 = 1 + 0.999 999 999 999 999 998 650 830 513 242 112;
  • 46) 0.999 999 999 999 999 998 650 830 513 242 112 × 2 = 1 + 0.999 999 999 999 999 997 301 661 026 484 224;
  • 47) 0.999 999 999 999 999 997 301 661 026 484 224 × 2 = 1 + 0.999 999 999 999 999 994 603 322 052 968 448;
  • 48) 0.999 999 999 999 999 994 603 322 052 968 448 × 2 = 1 + 0.999 999 999 999 999 989 206 644 105 936 896;
  • 49) 0.999 999 999 999 999 989 206 644 105 936 896 × 2 = 1 + 0.999 999 999 999 999 978 413 288 211 873 792;
  • 50) 0.999 999 999 999 999 978 413 288 211 873 792 × 2 = 1 + 0.999 999 999 999 999 956 826 576 423 747 584;
  • 51) 0.999 999 999 999 999 956 826 576 423 747 584 × 2 = 1 + 0.999 999 999 999 999 913 653 152 847 495 168;
  • 52) 0.999 999 999 999 999 913 653 152 847 495 168 × 2 = 1 + 0.999 999 999 999 999 827 306 305 694 990 336;
  • 53) 0.999 999 999 999 999 827 306 305 694 990 336 × 2 = 1 + 0.999 999 999 999 999 654 612 611 389 980 672;

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:


0.859 999 999 999 899 955 582 804 977 893 791(10) =

0.1101 1100 0010 1000 1111 0101 1100 0010 1000 1111 0011 1111 1111 1(2)

5. Positive number before normalization:

763.859 999 999 999 899 955 582 804 977 893 791(10) =

10 1111 1011.1101 1100 0010 1000 1111 0101 1100 0010 1000 1111 0011 1111 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the left, so that only one non zero digit remains to the left of it:


763.859 999 999 999 899 955 582 804 977 893 791(10) =

10 1111 1011.1101 1100 0010 1000 1111 0101 1100 0010 1000 1111 0011 1111 1111 1(2) =

10 1111 1011.1101 1100 0010 1000 1111 0101 1100 0010 1000 1111 0011 1111 1111 1(2) × 20 =

1.0111 1101 1110 1110 0001 0100 0111 1010 1110 0001 0100 0111 1001 1111 1111 11(2) × 29


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): 9

Mantissa (not normalized):
1.0111 1101 1110 1110 0001 0100 0111 1010 1110 0001 0100 0111 1001 1111 1111 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

9 + 2(11-1) - 1 =

(9 + 1 023)(10) =

1 032(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;
  • 1 032 ÷ 2 = 516 + 0;
  • 516 ÷ 2 = 258 + 0;
  • 258 ÷ 2 = 129 + 0;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1032(10) =

100 0000 1000(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0111 1101 1110 1110 0001 0100 0111 1010 1110 0001 0100 0111 1001 11 1111 1111 =

0111 1101 1110 1110 0001 0100 0111 1010 1110 0001 0100 0111 1001


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) =
100 0000 1000

Mantissa (52 bits) =
0111 1101 1110 1110 0001 0100 0111 1010 1110 0001 0100 0111 1001


Decimal number 763.859 999 999 999 899 955 582 804 977 893 791 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 1000 - 0111 1101 1110 1110 0001 0100 0111 1010 1110 0001 0100 0111 1001

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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