6.285 749 999 999 999 282 351 836 881 549 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 749 999 999 999 282 351 836 881 549(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
6.285 749 999 999 999 282 351 836 881 549(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: 6.
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;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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.

6(10) =

110(2)


3. Convert to binary (base 2) the fractional part: 0.285 749 999 999 999 282 351 836 881 549.

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) 0.285 749 999 999 999 282 351 836 881 549 × 2 = 0 + 0.571 499 999 999 998 564 703 673 763 098;
  • 2) 0.571 499 999 999 998 564 703 673 763 098 × 2 = 1 + 0.142 999 999 999 997 129 407 347 526 196;
  • 3) 0.142 999 999 999 997 129 407 347 526 196 × 2 = 0 + 0.285 999 999 999 994 258 814 695 052 392;
  • 4) 0.285 999 999 999 994 258 814 695 052 392 × 2 = 0 + 0.571 999 999 999 988 517 629 390 104 784;
  • 5) 0.571 999 999 999 988 517 629 390 104 784 × 2 = 1 + 0.143 999 999 999 977 035 258 780 209 568;
  • 6) 0.143 999 999 999 977 035 258 780 209 568 × 2 = 0 + 0.287 999 999 999 954 070 517 560 419 136;
  • 7) 0.287 999 999 999 954 070 517 560 419 136 × 2 = 0 + 0.575 999 999 999 908 141 035 120 838 272;
  • 8) 0.575 999 999 999 908 141 035 120 838 272 × 2 = 1 + 0.151 999 999 999 816 282 070 241 676 544;
  • 9) 0.151 999 999 999 816 282 070 241 676 544 × 2 = 0 + 0.303 999 999 999 632 564 140 483 353 088;
  • 10) 0.303 999 999 999 632 564 140 483 353 088 × 2 = 0 + 0.607 999 999 999 265 128 280 966 706 176;
  • 11) 0.607 999 999 999 265 128 280 966 706 176 × 2 = 1 + 0.215 999 999 998 530 256 561 933 412 352;
  • 12) 0.215 999 999 998 530 256 561 933 412 352 × 2 = 0 + 0.431 999 999 997 060 513 123 866 824 704;
  • 13) 0.431 999 999 997 060 513 123 866 824 704 × 2 = 0 + 0.863 999 999 994 121 026 247 733 649 408;
  • 14) 0.863 999 999 994 121 026 247 733 649 408 × 2 = 1 + 0.727 999 999 988 242 052 495 467 298 816;
  • 15) 0.727 999 999 988 242 052 495 467 298 816 × 2 = 1 + 0.455 999 999 976 484 104 990 934 597 632;
  • 16) 0.455 999 999 976 484 104 990 934 597 632 × 2 = 0 + 0.911 999 999 952 968 209 981 869 195 264;
  • 17) 0.911 999 999 952 968 209 981 869 195 264 × 2 = 1 + 0.823 999 999 905 936 419 963 738 390 528;
  • 18) 0.823 999 999 905 936 419 963 738 390 528 × 2 = 1 + 0.647 999 999 811 872 839 927 476 781 056;
  • 19) 0.647 999 999 811 872 839 927 476 781 056 × 2 = 1 + 0.295 999 999 623 745 679 854 953 562 112;
  • 20) 0.295 999 999 623 745 679 854 953 562 112 × 2 = 0 + 0.591 999 999 247 491 359 709 907 124 224;
  • 21) 0.591 999 999 247 491 359 709 907 124 224 × 2 = 1 + 0.183 999 998 494 982 719 419 814 248 448;
  • 22) 0.183 999 998 494 982 719 419 814 248 448 × 2 = 0 + 0.367 999 996 989 965 438 839 628 496 896;
  • 23) 0.367 999 996 989 965 438 839 628 496 896 × 2 = 0 + 0.735 999 993 979 930 877 679 256 993 792;
  • 24) 0.735 999 993 979 930 877 679 256 993 792 × 2 = 1 + 0.471 999 987 959 861 755 358 513 987 584;
  • 25) 0.471 999 987 959 861 755 358 513 987 584 × 2 = 0 + 0.943 999 975 919 723 510 717 027 975 168;
  • 26) 0.943 999 975 919 723 510 717 027 975 168 × 2 = 1 + 0.887 999 951 839 447 021 434 055 950 336;
  • 27) 0.887 999 951 839 447 021 434 055 950 336 × 2 = 1 + 0.775 999 903 678 894 042 868 111 900 672;
  • 28) 0.775 999 903 678 894 042 868 111 900 672 × 2 = 1 + 0.551 999 807 357 788 085 736 223 801 344;
  • 29) 0.551 999 807 357 788 085 736 223 801 344 × 2 = 1 + 0.103 999 614 715 576 171 472 447 602 688;
  • 30) 0.103 999 614 715 576 171 472 447 602 688 × 2 = 0 + 0.207 999 229 431 152 342 944 895 205 376;
  • 31) 0.207 999 229 431 152 342 944 895 205 376 × 2 = 0 + 0.415 998 458 862 304 685 889 790 410 752;
  • 32) 0.415 998 458 862 304 685 889 790 410 752 × 2 = 0 + 0.831 996 917 724 609 371 779 580 821 504;
  • 33) 0.831 996 917 724 609 371 779 580 821 504 × 2 = 1 + 0.663 993 835 449 218 743 559 161 643 008;
  • 34) 0.663 993 835 449 218 743 559 161 643 008 × 2 = 1 + 0.327 987 670 898 437 487 118 323 286 016;
  • 35) 0.327 987 670 898 437 487 118 323 286 016 × 2 = 0 + 0.655 975 341 796 874 974 236 646 572 032;
  • 36) 0.655 975 341 796 874 974 236 646 572 032 × 2 = 1 + 0.311 950 683 593 749 948 473 293 144 064;
  • 37) 0.311 950 683 593 749 948 473 293 144 064 × 2 = 0 + 0.623 901 367 187 499 896 946 586 288 128;
  • 38) 0.623 901 367 187 499 896 946 586 288 128 × 2 = 1 + 0.247 802 734 374 999 793 893 172 576 256;
  • 39) 0.247 802 734 374 999 793 893 172 576 256 × 2 = 0 + 0.495 605 468 749 999 587 786 345 152 512;
  • 40) 0.495 605 468 749 999 587 786 345 152 512 × 2 = 0 + 0.991 210 937 499 999 175 572 690 305 024;
  • 41) 0.991 210 937 499 999 175 572 690 305 024 × 2 = 1 + 0.982 421 874 999 998 351 145 380 610 048;
  • 42) 0.982 421 874 999 998 351 145 380 610 048 × 2 = 1 + 0.964 843 749 999 996 702 290 761 220 096;
  • 43) 0.964 843 749 999 996 702 290 761 220 096 × 2 = 1 + 0.929 687 499 999 993 404 581 522 440 192;
  • 44) 0.929 687 499 999 993 404 581 522 440 192 × 2 = 1 + 0.859 374 999 999 986 809 163 044 880 384;
  • 45) 0.859 374 999 999 986 809 163 044 880 384 × 2 = 1 + 0.718 749 999 999 973 618 326 089 760 768;
  • 46) 0.718 749 999 999 973 618 326 089 760 768 × 2 = 1 + 0.437 499 999 999 947 236 652 179 521 536;
  • 47) 0.437 499 999 999 947 236 652 179 521 536 × 2 = 0 + 0.874 999 999 999 894 473 304 359 043 072;
  • 48) 0.874 999 999 999 894 473 304 359 043 072 × 2 = 1 + 0.749 999 999 999 788 946 608 718 086 144;
  • 49) 0.749 999 999 999 788 946 608 718 086 144 × 2 = 1 + 0.499 999 999 999 577 893 217 436 172 288;
  • 50) 0.499 999 999 999 577 893 217 436 172 288 × 2 = 0 + 0.999 999 999 999 155 786 434 872 344 576;
  • 51) 0.999 999 999 999 155 786 434 872 344 576 × 2 = 1 + 0.999 999 999 998 311 572 869 744 689 152;
  • 52) 0.999 999 999 998 311 572 869 744 689 152 × 2 = 1 + 0.999 999 999 996 623 145 739 489 378 304;
  • 53) 0.999 999 999 996 623 145 739 489 378 304 × 2 = 1 + 0.999 999 999 993 246 291 478 978 756 608;

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:


0.285 749 999 999 999 282 351 836 881 549(10) =

0.0100 1001 0010 0110 1110 1001 0111 1000 1101 0100 1111 1101 1011 1(2)

5. Positive number before normalization:

6.285 749 999 999 999 282 351 836 881 549(10) =

110.0100 1001 0010 0110 1110 1001 0111 1000 1101 0100 1111 1101 1011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


6.285 749 999 999 999 282 351 836 881 549(10) =

110.0100 1001 0010 0110 1110 1001 0111 1000 1101 0100 1111 1101 1011 1(2) =

110.0100 1001 0010 0110 1110 1001 0111 1000 1101 0100 1111 1101 1011 1(2) × 20 =

1.1001 0010 0100 1001 1011 1010 0101 1110 0011 0101 0011 1111 0110 111(2) × 22


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): 2

Mantissa (not normalized):
1.1001 0010 0100 1001 1011 1010 0101 1110 0011 0101 0011 1111 0110 111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 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;

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) =

1025(10) =

100 0000 0001(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. 1001 0010 0100 1001 1011 1010 0101 1110 0011 0101 0011 1111 0110 111 =

1001 0010 0100 1001 1011 1010 0101 1110 0011 0101 0011 1111 0110


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 0001

Mantissa (52 bits) =
1001 0010 0100 1001 1011 1010 0101 1110 0011 0101 0011 1111 0110


Decimal number 6.285 749 999 999 999 282 351 836 881 549 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1001 0010 0100 1001 1011 1010 0101 1110 0011 0101 0011 1111 0110

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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