Convert Decimal 10 000.555 555 555 555 555 555 468 545 468 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 10 000.555 555 555 555 555 555 468 545 468(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
10 000.555 555 555 555 555 555 468 545 468(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: 10 000.
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;
  • 10 000 ÷ 2 = 5 000 + 0;
  • 5 000 ÷ 2 = 2 500 + 0;
  • 2 500 ÷ 2 = 1 250 + 0;
  • 1 250 ÷ 2 = 625 + 0;
  • 625 ÷ 2 = 312 + 1;
  • 312 ÷ 2 = 156 + 0;
  • 156 ÷ 2 = 78 + 0;
  • 78 ÷ 2 = 39 + 0;
  • 39 ÷ 2 = 19 + 1;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 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.

10 000(10) =

10 0111 0001 0000(2)


3. Convert to binary (base 2) the fractional part: 0.555 555 555 555 555 555 468 545 468.

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) 0.555 555 555 555 555 555 468 545 468 × 2 = 1 + 0.111 111 111 111 111 110 937 090 936;
  • 2) 0.111 111 111 111 111 110 937 090 936 × 2 = 0 + 0.222 222 222 222 222 221 874 181 872;
  • 3) 0.222 222 222 222 222 221 874 181 872 × 2 = 0 + 0.444 444 444 444 444 443 748 363 744;
  • 4) 0.444 444 444 444 444 443 748 363 744 × 2 = 0 + 0.888 888 888 888 888 887 496 727 488;
  • 5) 0.888 888 888 888 888 887 496 727 488 × 2 = 1 + 0.777 777 777 777 777 774 993 454 976;
  • 6) 0.777 777 777 777 777 774 993 454 976 × 2 = 1 + 0.555 555 555 555 555 549 986 909 952;
  • 7) 0.555 555 555 555 555 549 986 909 952 × 2 = 1 + 0.111 111 111 111 111 099 973 819 904;
  • 8) 0.111 111 111 111 111 099 973 819 904 × 2 = 0 + 0.222 222 222 222 222 199 947 639 808;
  • 9) 0.222 222 222 222 222 199 947 639 808 × 2 = 0 + 0.444 444 444 444 444 399 895 279 616;
  • 10) 0.444 444 444 444 444 399 895 279 616 × 2 = 0 + 0.888 888 888 888 888 799 790 559 232;
  • 11) 0.888 888 888 888 888 799 790 559 232 × 2 = 1 + 0.777 777 777 777 777 599 581 118 464;
  • 12) 0.777 777 777 777 777 599 581 118 464 × 2 = 1 + 0.555 555 555 555 555 199 162 236 928;
  • 13) 0.555 555 555 555 555 199 162 236 928 × 2 = 1 + 0.111 111 111 111 110 398 324 473 856;
  • 14) 0.111 111 111 111 110 398 324 473 856 × 2 = 0 + 0.222 222 222 222 220 796 648 947 712;
  • 15) 0.222 222 222 222 220 796 648 947 712 × 2 = 0 + 0.444 444 444 444 441 593 297 895 424;
  • 16) 0.444 444 444 444 441 593 297 895 424 × 2 = 0 + 0.888 888 888 888 883 186 595 790 848;
  • 17) 0.888 888 888 888 883 186 595 790 848 × 2 = 1 + 0.777 777 777 777 766 373 191 581 696;
  • 18) 0.777 777 777 777 766 373 191 581 696 × 2 = 1 + 0.555 555 555 555 532 746 383 163 392;
  • 19) 0.555 555 555 555 532 746 383 163 392 × 2 = 1 + 0.111 111 111 111 065 492 766 326 784;
  • 20) 0.111 111 111 111 065 492 766 326 784 × 2 = 0 + 0.222 222 222 222 130 985 532 653 568;
  • 21) 0.222 222 222 222 130 985 532 653 568 × 2 = 0 + 0.444 444 444 444 261 971 065 307 136;
  • 22) 0.444 444 444 444 261 971 065 307 136 × 2 = 0 + 0.888 888 888 888 523 942 130 614 272;
  • 23) 0.888 888 888 888 523 942 130 614 272 × 2 = 1 + 0.777 777 777 777 047 884 261 228 544;
  • 24) 0.777 777 777 777 047 884 261 228 544 × 2 = 1 + 0.555 555 555 554 095 768 522 457 088;
  • 25) 0.555 555 555 554 095 768 522 457 088 × 2 = 1 + 0.111 111 111 108 191 537 044 914 176;
  • 26) 0.111 111 111 108 191 537 044 914 176 × 2 = 0 + 0.222 222 222 216 383 074 089 828 352;
  • 27) 0.222 222 222 216 383 074 089 828 352 × 2 = 0 + 0.444 444 444 432 766 148 179 656 704;
  • 28) 0.444 444 444 432 766 148 179 656 704 × 2 = 0 + 0.888 888 888 865 532 296 359 313 408;
  • 29) 0.888 888 888 865 532 296 359 313 408 × 2 = 1 + 0.777 777 777 731 064 592 718 626 816;
  • 30) 0.777 777 777 731 064 592 718 626 816 × 2 = 1 + 0.555 555 555 462 129 185 437 253 632;
  • 31) 0.555 555 555 462 129 185 437 253 632 × 2 = 1 + 0.111 111 110 924 258 370 874 507 264;
  • 32) 0.111 111 110 924 258 370 874 507 264 × 2 = 0 + 0.222 222 221 848 516 741 749 014 528;
  • 33) 0.222 222 221 848 516 741 749 014 528 × 2 = 0 + 0.444 444 443 697 033 483 498 029 056;
  • 34) 0.444 444 443 697 033 483 498 029 056 × 2 = 0 + 0.888 888 887 394 066 966 996 058 112;
  • 35) 0.888 888 887 394 066 966 996 058 112 × 2 = 1 + 0.777 777 774 788 133 933 992 116 224;
  • 36) 0.777 777 774 788 133 933 992 116 224 × 2 = 1 + 0.555 555 549 576 267 867 984 232 448;
  • 37) 0.555 555 549 576 267 867 984 232 448 × 2 = 1 + 0.111 111 099 152 535 735 968 464 896;
  • 38) 0.111 111 099 152 535 735 968 464 896 × 2 = 0 + 0.222 222 198 305 071 471 936 929 792;
  • 39) 0.222 222 198 305 071 471 936 929 792 × 2 = 0 + 0.444 444 396 610 142 943 873 859 584;
  • 40) 0.444 444 396 610 142 943 873 859 584 × 2 = 0 + 0.888 888 793 220 285 887 747 719 168;
  • 41) 0.888 888 793 220 285 887 747 719 168 × 2 = 1 + 0.777 777 586 440 571 775 495 438 336;
  • 42) 0.777 777 586 440 571 775 495 438 336 × 2 = 1 + 0.555 555 172 881 143 550 990 876 672;
  • 43) 0.555 555 172 881 143 550 990 876 672 × 2 = 1 + 0.111 110 345 762 287 101 981 753 344;
  • 44) 0.111 110 345 762 287 101 981 753 344 × 2 = 0 + 0.222 220 691 524 574 203 963 506 688;
  • 45) 0.222 220 691 524 574 203 963 506 688 × 2 = 0 + 0.444 441 383 049 148 407 927 013 376;
  • 46) 0.444 441 383 049 148 407 927 013 376 × 2 = 0 + 0.888 882 766 098 296 815 854 026 752;
  • 47) 0.888 882 766 098 296 815 854 026 752 × 2 = 1 + 0.777 765 532 196 593 631 708 053 504;
  • 48) 0.777 765 532 196 593 631 708 053 504 × 2 = 1 + 0.555 531 064 393 187 263 416 107 008;
  • 49) 0.555 531 064 393 187 263 416 107 008 × 2 = 1 + 0.111 062 128 786 374 526 832 214 016;
  • 50) 0.111 062 128 786 374 526 832 214 016 × 2 = 0 + 0.222 124 257 572 749 053 664 428 032;
  • 51) 0.222 124 257 572 749 053 664 428 032 × 2 = 0 + 0.444 248 515 145 498 107 328 856 064;
  • 52) 0.444 248 515 145 498 107 328 856 064 × 2 = 0 + 0.888 497 030 290 996 214 657 712 128;
  • 53) 0.888 497 030 290 996 214 657 712 128 × 2 = 1 + 0.776 994 060 581 992 429 315 424 256;

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:


0.555 555 555 555 555 555 468 545 468(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2)

5. Positive number before normalization:

10 000.555 555 555 555 555 555 468 545 468(10) =

10 0111 0001 0000.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the left, so that only one non zero digit remains to the left of it:


10 000.555 555 555 555 555 555 468 545 468(10) =

10 0111 0001 0000.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2) =

10 0111 0001 0000.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1(2) × 20 =

1.0011 1000 1000 0100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 01(2) × 213


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): 13

Mantissa (not normalized):
1.0011 1000 1000 0100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 01


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

13 + 2(11-1) - 1 =

(13 + 1 023)(10) =

1 036(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;
  • 1 036 ÷ 2 = 518 + 0;
  • 518 ÷ 2 = 259 + 0;
  • 259 ÷ 2 = 129 + 1;
  • 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) =

1036(10) =

100 0000 1100(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. 0011 1000 1000 0100 0111 0001 1100 0111 0001 1100 0111 0001 1100 01 1100 0111 0001 =

0011 1000 1000 0100 0111 0001 1100 0111 0001 1100 0111 0001 1100


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 1100

Mantissa (52 bits) =
0011 1000 1000 0100 0111 0001 1100 0111 0001 1100 0111 0001 1100


Decimal number 10 000.555 555 555 555 555 555 468 545 468 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 1100 - 0011 1000 1000 0100 0111 0001 1100 0111 0001 1100 0111 0001 1100

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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