Decimal to 64 Bit IEEE 754 Binary: Convert Number 1 024.968 599 999 999 923 966 242 931 783 194 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 1 024.968 599 999 999 923 966 242 931 783 194(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (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: 0.968 599 999 999 923 966 242 931 783 194.

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) 0.968 599 999 999 923 966 242 931 783 194 × 2 = 1 + 0.937 199 999 999 847 932 485 863 566 388;
  • 2) 0.937 199 999 999 847 932 485 863 566 388 × 2 = 1 + 0.874 399 999 999 695 864 971 727 132 776;
  • 3) 0.874 399 999 999 695 864 971 727 132 776 × 2 = 1 + 0.748 799 999 999 391 729 943 454 265 552;
  • 4) 0.748 799 999 999 391 729 943 454 265 552 × 2 = 1 + 0.497 599 999 998 783 459 886 908 531 104;
  • 5) 0.497 599 999 998 783 459 886 908 531 104 × 2 = 0 + 0.995 199 999 997 566 919 773 817 062 208;
  • 6) 0.995 199 999 997 566 919 773 817 062 208 × 2 = 1 + 0.990 399 999 995 133 839 547 634 124 416;
  • 7) 0.990 399 999 995 133 839 547 634 124 416 × 2 = 1 + 0.980 799 999 990 267 679 095 268 248 832;
  • 8) 0.980 799 999 990 267 679 095 268 248 832 × 2 = 1 + 0.961 599 999 980 535 358 190 536 497 664;
  • 9) 0.961 599 999 980 535 358 190 536 497 664 × 2 = 1 + 0.923 199 999 961 070 716 381 072 995 328;
  • 10) 0.923 199 999 961 070 716 381 072 995 328 × 2 = 1 + 0.846 399 999 922 141 432 762 145 990 656;
  • 11) 0.846 399 999 922 141 432 762 145 990 656 × 2 = 1 + 0.692 799 999 844 282 865 524 291 981 312;
  • 12) 0.692 799 999 844 282 865 524 291 981 312 × 2 = 1 + 0.385 599 999 688 565 731 048 583 962 624;
  • 13) 0.385 599 999 688 565 731 048 583 962 624 × 2 = 0 + 0.771 199 999 377 131 462 097 167 925 248;
  • 14) 0.771 199 999 377 131 462 097 167 925 248 × 2 = 1 + 0.542 399 998 754 262 924 194 335 850 496;
  • 15) 0.542 399 998 754 262 924 194 335 850 496 × 2 = 1 + 0.084 799 997 508 525 848 388 671 700 992;
  • 16) 0.084 799 997 508 525 848 388 671 700 992 × 2 = 0 + 0.169 599 995 017 051 696 777 343 401 984;
  • 17) 0.169 599 995 017 051 696 777 343 401 984 × 2 = 0 + 0.339 199 990 034 103 393 554 686 803 968;
  • 18) 0.339 199 990 034 103 393 554 686 803 968 × 2 = 0 + 0.678 399 980 068 206 787 109 373 607 936;
  • 19) 0.678 399 980 068 206 787 109 373 607 936 × 2 = 1 + 0.356 799 960 136 413 574 218 747 215 872;
  • 20) 0.356 799 960 136 413 574 218 747 215 872 × 2 = 0 + 0.713 599 920 272 827 148 437 494 431 744;
  • 21) 0.713 599 920 272 827 148 437 494 431 744 × 2 = 1 + 0.427 199 840 545 654 296 874 988 863 488;
  • 22) 0.427 199 840 545 654 296 874 988 863 488 × 2 = 0 + 0.854 399 681 091 308 593 749 977 726 976;
  • 23) 0.854 399 681 091 308 593 749 977 726 976 × 2 = 1 + 0.708 799 362 182 617 187 499 955 453 952;
  • 24) 0.708 799 362 182 617 187 499 955 453 952 × 2 = 1 + 0.417 598 724 365 234 374 999 910 907 904;
  • 25) 0.417 598 724 365 234 374 999 910 907 904 × 2 = 0 + 0.835 197 448 730 468 749 999 821 815 808;
  • 26) 0.835 197 448 730 468 749 999 821 815 808 × 2 = 1 + 0.670 394 897 460 937 499 999 643 631 616;
  • 27) 0.670 394 897 460 937 499 999 643 631 616 × 2 = 1 + 0.340 789 794 921 874 999 999 287 263 232;
  • 28) 0.340 789 794 921 874 999 999 287 263 232 × 2 = 0 + 0.681 579 589 843 749 999 998 574 526 464;
  • 29) 0.681 579 589 843 749 999 998 574 526 464 × 2 = 1 + 0.363 159 179 687 499 999 997 149 052 928;
  • 30) 0.363 159 179 687 499 999 997 149 052 928 × 2 = 0 + 0.726 318 359 374 999 999 994 298 105 856;
  • 31) 0.726 318 359 374 999 999 994 298 105 856 × 2 = 1 + 0.452 636 718 749 999 999 988 596 211 712;
  • 32) 0.452 636 718 749 999 999 988 596 211 712 × 2 = 0 + 0.905 273 437 499 999 999 977 192 423 424;
  • 33) 0.905 273 437 499 999 999 977 192 423 424 × 2 = 1 + 0.810 546 874 999 999 999 954 384 846 848;
  • 34) 0.810 546 874 999 999 999 954 384 846 848 × 2 = 1 + 0.621 093 749 999 999 999 908 769 693 696;
  • 35) 0.621 093 749 999 999 999 908 769 693 696 × 2 = 1 + 0.242 187 499 999 999 999 817 539 387 392;
  • 36) 0.242 187 499 999 999 999 817 539 387 392 × 2 = 0 + 0.484 374 999 999 999 999 635 078 774 784;
  • 37) 0.484 374 999 999 999 999 635 078 774 784 × 2 = 0 + 0.968 749 999 999 999 999 270 157 549 568;
  • 38) 0.968 749 999 999 999 999 270 157 549 568 × 2 = 1 + 0.937 499 999 999 999 998 540 315 099 136;
  • 39) 0.937 499 999 999 999 998 540 315 099 136 × 2 = 1 + 0.874 999 999 999 999 997 080 630 198 272;
  • 40) 0.874 999 999 999 999 997 080 630 198 272 × 2 = 1 + 0.749 999 999 999 999 994 161 260 396 544;
  • 41) 0.749 999 999 999 999 994 161 260 396 544 × 2 = 1 + 0.499 999 999 999 999 988 322 520 793 088;
  • 42) 0.499 999 999 999 999 988 322 520 793 088 × 2 = 0 + 0.999 999 999 999 999 976 645 041 586 176;
  • 43) 0.999 999 999 999 999 976 645 041 586 176 × 2 = 1 + 0.999 999 999 999 999 953 290 083 172 352;
  • 44) 0.999 999 999 999 999 953 290 083 172 352 × 2 = 1 + 0.999 999 999 999 999 906 580 166 344 704;
  • 45) 0.999 999 999 999 999 906 580 166 344 704 × 2 = 1 + 0.999 999 999 999 999 813 160 332 689 408;
  • 46) 0.999 999 999 999 999 813 160 332 689 408 × 2 = 1 + 0.999 999 999 999 999 626 320 665 378 816;
  • 47) 0.999 999 999 999 999 626 320 665 378 816 × 2 = 1 + 0.999 999 999 999 999 252 641 330 757 632;
  • 48) 0.999 999 999 999 999 252 641 330 757 632 × 2 = 1 + 0.999 999 999 999 998 505 282 661 515 264;
  • 49) 0.999 999 999 999 998 505 282 661 515 264 × 2 = 1 + 0.999 999 999 999 997 010 565 323 030 528;
  • 50) 0.999 999 999 999 997 010 565 323 030 528 × 2 = 1 + 0.999 999 999 999 994 021 130 646 061 056;
  • 51) 0.999 999 999 999 994 021 130 646 061 056 × 2 = 1 + 0.999 999 999 999 988 042 261 292 122 112;
  • 52) 0.999 999 999 999 988 042 261 292 122 112 × 2 = 1 + 0.999 999 999 999 976 084 522 584 244 224;
  • 53) 0.999 999 999 999 976 084 522 584 244 224 × 2 = 1 + 0.999 999 999 999 952 169 045 168 488 448;

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:


0.968 599 999 999 923 966 242 931 783 194(10) =


0.1111 0111 1111 0110 0010 1011 0110 1010 1110 0111 1011 1111 1111 1(2)


5. Positive number before normalization:

1 024.968 599 999 999 923 966 242 931 783 194(10) =


100 0000 0000.1111 0111 1111 0110 0010 1011 0110 1010 1110 0111 1011 1111 1111 1(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:


1 024.968 599 999 999 923 966 242 931 783 194(10) =


100 0000 0000.1111 0111 1111 0110 0010 1011 0110 1010 1110 0111 1011 1111 1111 1(2) =


100 0000 0000.1111 0111 1111 0110 0010 1011 0110 1010 1110 0111 1011 1111 1111 1(2) × 20 =


1.0000 0000 0011 1101 1111 1101 1000 1010 1101 1010 1011 1001 1110 1111 1111 111(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 1110 1111 1111 111


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 1110 111 1111 1111 =


0000 0000 0011 1101 1111 1101 1000 1010 1101 1010 1011 1001 1110


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 1110


The base ten decimal number 1 024.968 599 999 999 923 966 242 931 783 194 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 1110

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