64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 1 024.968 599 999 999 923 966 242 931 9 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 1 024.968 599 999 999 923 966 242 931 9(10) converted and written in 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: 1 024.
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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 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;

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.


1 024(10) =


100 0000 0000(2)


3. Convert to binary (base 2) the fractional part: 0.968 599 999 999 923 966 242 931 9.

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) 0.968 599 999 999 923 966 242 931 9 × 2 = 1 + 0.937 199 999 999 847 932 485 863 8;
  • 2) 0.937 199 999 999 847 932 485 863 8 × 2 = 1 + 0.874 399 999 999 695 864 971 727 6;
  • 3) 0.874 399 999 999 695 864 971 727 6 × 2 = 1 + 0.748 799 999 999 391 729 943 455 2;
  • 4) 0.748 799 999 999 391 729 943 455 2 × 2 = 1 + 0.497 599 999 998 783 459 886 910 4;
  • 5) 0.497 599 999 998 783 459 886 910 4 × 2 = 0 + 0.995 199 999 997 566 919 773 820 8;
  • 6) 0.995 199 999 997 566 919 773 820 8 × 2 = 1 + 0.990 399 999 995 133 839 547 641 6;
  • 7) 0.990 399 999 995 133 839 547 641 6 × 2 = 1 + 0.980 799 999 990 267 679 095 283 2;
  • 8) 0.980 799 999 990 267 679 095 283 2 × 2 = 1 + 0.961 599 999 980 535 358 190 566 4;
  • 9) 0.961 599 999 980 535 358 190 566 4 × 2 = 1 + 0.923 199 999 961 070 716 381 132 8;
  • 10) 0.923 199 999 961 070 716 381 132 8 × 2 = 1 + 0.846 399 999 922 141 432 762 265 6;
  • 11) 0.846 399 999 922 141 432 762 265 6 × 2 = 1 + 0.692 799 999 844 282 865 524 531 2;
  • 12) 0.692 799 999 844 282 865 524 531 2 × 2 = 1 + 0.385 599 999 688 565 731 049 062 4;
  • 13) 0.385 599 999 688 565 731 049 062 4 × 2 = 0 + 0.771 199 999 377 131 462 098 124 8;
  • 14) 0.771 199 999 377 131 462 098 124 8 × 2 = 1 + 0.542 399 998 754 262 924 196 249 6;
  • 15) 0.542 399 998 754 262 924 196 249 6 × 2 = 1 + 0.084 799 997 508 525 848 392 499 2;
  • 16) 0.084 799 997 508 525 848 392 499 2 × 2 = 0 + 0.169 599 995 017 051 696 784 998 4;
  • 17) 0.169 599 995 017 051 696 784 998 4 × 2 = 0 + 0.339 199 990 034 103 393 569 996 8;
  • 18) 0.339 199 990 034 103 393 569 996 8 × 2 = 0 + 0.678 399 980 068 206 787 139 993 6;
  • 19) 0.678 399 980 068 206 787 139 993 6 × 2 = 1 + 0.356 799 960 136 413 574 279 987 2;
  • 20) 0.356 799 960 136 413 574 279 987 2 × 2 = 0 + 0.713 599 920 272 827 148 559 974 4;
  • 21) 0.713 599 920 272 827 148 559 974 4 × 2 = 1 + 0.427 199 840 545 654 297 119 948 8;
  • 22) 0.427 199 840 545 654 297 119 948 8 × 2 = 0 + 0.854 399 681 091 308 594 239 897 6;
  • 23) 0.854 399 681 091 308 594 239 897 6 × 2 = 1 + 0.708 799 362 182 617 188 479 795 2;
  • 24) 0.708 799 362 182 617 188 479 795 2 × 2 = 1 + 0.417 598 724 365 234 376 959 590 4;
  • 25) 0.417 598 724 365 234 376 959 590 4 × 2 = 0 + 0.835 197 448 730 468 753 919 180 8;
  • 26) 0.835 197 448 730 468 753 919 180 8 × 2 = 1 + 0.670 394 897 460 937 507 838 361 6;
  • 27) 0.670 394 897 460 937 507 838 361 6 × 2 = 1 + 0.340 789 794 921 875 015 676 723 2;
  • 28) 0.340 789 794 921 875 015 676 723 2 × 2 = 0 + 0.681 579 589 843 750 031 353 446 4;
  • 29) 0.681 579 589 843 750 031 353 446 4 × 2 = 1 + 0.363 159 179 687 500 062 706 892 8;
  • 30) 0.363 159 179 687 500 062 706 892 8 × 2 = 0 + 0.726 318 359 375 000 125 413 785 6;
  • 31) 0.726 318 359 375 000 125 413 785 6 × 2 = 1 + 0.452 636 718 750 000 250 827 571 2;
  • 32) 0.452 636 718 750 000 250 827 571 2 × 2 = 0 + 0.905 273 437 500 000 501 655 142 4;
  • 33) 0.905 273 437 500 000 501 655 142 4 × 2 = 1 + 0.810 546 875 000 001 003 310 284 8;
  • 34) 0.810 546 875 000 001 003 310 284 8 × 2 = 1 + 0.621 093 750 000 002 006 620 569 6;
  • 35) 0.621 093 750 000 002 006 620 569 6 × 2 = 1 + 0.242 187 500 000 004 013 241 139 2;
  • 36) 0.242 187 500 000 004 013 241 139 2 × 2 = 0 + 0.484 375 000 000 008 026 482 278 4;
  • 37) 0.484 375 000 000 008 026 482 278 4 × 2 = 0 + 0.968 750 000 000 016 052 964 556 8;
  • 38) 0.968 750 000 000 016 052 964 556 8 × 2 = 1 + 0.937 500 000 000 032 105 929 113 6;
  • 39) 0.937 500 000 000 032 105 929 113 6 × 2 = 1 + 0.875 000 000 000 064 211 858 227 2;
  • 40) 0.875 000 000 000 064 211 858 227 2 × 2 = 1 + 0.750 000 000 000 128 423 716 454 4;
  • 41) 0.750 000 000 000 128 423 716 454 4 × 2 = 1 + 0.500 000 000 000 256 847 432 908 8;
  • 42) 0.500 000 000 000 256 847 432 908 8 × 2 = 1 + 0.000 000 000 000 513 694 865 817 6;
  • 43) 0.000 000 000 000 513 694 865 817 6 × 2 = 0 + 0.000 000 000 001 027 389 731 635 2;
  • 44) 0.000 000 000 001 027 389 731 635 2 × 2 = 0 + 0.000 000 000 002 054 779 463 270 4;
  • 45) 0.000 000 000 002 054 779 463 270 4 × 2 = 0 + 0.000 000 000 004 109 558 926 540 8;
  • 46) 0.000 000 000 004 109 558 926 540 8 × 2 = 0 + 0.000 000 000 008 219 117 853 081 6;
  • 47) 0.000 000 000 008 219 117 853 081 6 × 2 = 0 + 0.000 000 000 016 438 235 706 163 2;
  • 48) 0.000 000 000 016 438 235 706 163 2 × 2 = 0 + 0.000 000 000 032 876 471 412 326 4;
  • 49) 0.000 000 000 032 876 471 412 326 4 × 2 = 0 + 0.000 000 000 065 752 942 824 652 8;
  • 50) 0.000 000 000 065 752 942 824 652 8 × 2 = 0 + 0.000 000 000 131 505 885 649 305 6;
  • 51) 0.000 000 000 131 505 885 649 305 6 × 2 = 0 + 0.000 000 000 263 011 771 298 611 2;
  • 52) 0.000 000 000 263 011 771 298 611 2 × 2 = 0 + 0.000 000 000 526 023 542 597 222 4;
  • 53) 0.000 000 000 526 023 542 597 222 4 × 2 = 0 + 0.000 000 001 052 047 085 194 444 8;

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...)


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:


0.968 599 999 999 923 966 242 931 9(10) =


0.1111 0111 1111 0110 0010 1011 0110 1010 1110 0111 1100 0000 0000 0(2)


5. Positive number before normalization:

1 024.968 599 999 999 923 966 242 931 9(10) =


100 0000 0000.1111 0111 1111 0110 0010 1011 0110 1010 1110 0111 1100 0000 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the left, so that only one non zero digit remains to the left of it:


1 024.968 599 999 999 923 966 242 931 9(10) =


100 0000 0000.1111 0111 1111 0110 0010 1011 0110 1010 1110 0111 1100 0000 0000 0(2) =


100 0000 0000.1111 0111 1111 0110 0010 1011 0110 1010 1110 0111 1100 0000 0000 0(2) × 20 =


1.0000 0000 0011 1101 1111 1101 1000 1010 1101 1010 1011 1001 1111 0000 0000 000(2) × 210


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): 10


Mantissa (not normalized):
1.0000 0000 0011 1101 1111 1101 1000 1010 1101 1010 1011 1001 1111 0000 0000 000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


10 + 2(11-1) - 1 =


(10 + 1 023)(10) =


1 033(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;
  • 1 033 ÷ 2 = 516 + 1;
  • 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) =


1033(10) =


100 0000 1001(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. 0000 0000 0011 1101 1111 1101 1000 1010 1101 1010 1011 1001 1111 000 0000 0000 =


0000 0000 0011 1101 1111 1101 1000 1010 1101 1010 1011 1001 1111


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 1001


Mantissa (52 bits) =
0000 0000 0011 1101 1111 1101 1000 1010 1101 1010 1011 1001 1111


The base ten decimal number 1 024.968 599 999 999 923 966 242 931 9 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 1001 - 0000 0000 0011 1101 1111 1101 1000 1010 1101 1010 1011 1001 1111

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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