6.285 749 999 999 999 282 351 836 881 635 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 749 999 999 999 282 351 836 881 635(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
6.285 749 999 999 999 282 351 836 881 635(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: 0.285 749 999 999 999 282 351 836 881 635.

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) 0.285 749 999 999 999 282 351 836 881 635 × 2 = 0 + 0.571 499 999 999 998 564 703 673 763 27;
  • 2) 0.571 499 999 999 998 564 703 673 763 27 × 2 = 1 + 0.142 999 999 999 997 129 407 347 526 54;
  • 3) 0.142 999 999 999 997 129 407 347 526 54 × 2 = 0 + 0.285 999 999 999 994 258 814 695 053 08;
  • 4) 0.285 999 999 999 994 258 814 695 053 08 × 2 = 0 + 0.571 999 999 999 988 517 629 390 106 16;
  • 5) 0.571 999 999 999 988 517 629 390 106 16 × 2 = 1 + 0.143 999 999 999 977 035 258 780 212 32;
  • 6) 0.143 999 999 999 977 035 258 780 212 32 × 2 = 0 + 0.287 999 999 999 954 070 517 560 424 64;
  • 7) 0.287 999 999 999 954 070 517 560 424 64 × 2 = 0 + 0.575 999 999 999 908 141 035 120 849 28;
  • 8) 0.575 999 999 999 908 141 035 120 849 28 × 2 = 1 + 0.151 999 999 999 816 282 070 241 698 56;
  • 9) 0.151 999 999 999 816 282 070 241 698 56 × 2 = 0 + 0.303 999 999 999 632 564 140 483 397 12;
  • 10) 0.303 999 999 999 632 564 140 483 397 12 × 2 = 0 + 0.607 999 999 999 265 128 280 966 794 24;
  • 11) 0.607 999 999 999 265 128 280 966 794 24 × 2 = 1 + 0.215 999 999 998 530 256 561 933 588 48;
  • 12) 0.215 999 999 998 530 256 561 933 588 48 × 2 = 0 + 0.431 999 999 997 060 513 123 867 176 96;
  • 13) 0.431 999 999 997 060 513 123 867 176 96 × 2 = 0 + 0.863 999 999 994 121 026 247 734 353 92;
  • 14) 0.863 999 999 994 121 026 247 734 353 92 × 2 = 1 + 0.727 999 999 988 242 052 495 468 707 84;
  • 15) 0.727 999 999 988 242 052 495 468 707 84 × 2 = 1 + 0.455 999 999 976 484 104 990 937 415 68;
  • 16) 0.455 999 999 976 484 104 990 937 415 68 × 2 = 0 + 0.911 999 999 952 968 209 981 874 831 36;
  • 17) 0.911 999 999 952 968 209 981 874 831 36 × 2 = 1 + 0.823 999 999 905 936 419 963 749 662 72;
  • 18) 0.823 999 999 905 936 419 963 749 662 72 × 2 = 1 + 0.647 999 999 811 872 839 927 499 325 44;
  • 19) 0.647 999 999 811 872 839 927 499 325 44 × 2 = 1 + 0.295 999 999 623 745 679 854 998 650 88;
  • 20) 0.295 999 999 623 745 679 854 998 650 88 × 2 = 0 + 0.591 999 999 247 491 359 709 997 301 76;
  • 21) 0.591 999 999 247 491 359 709 997 301 76 × 2 = 1 + 0.183 999 998 494 982 719 419 994 603 52;
  • 22) 0.183 999 998 494 982 719 419 994 603 52 × 2 = 0 + 0.367 999 996 989 965 438 839 989 207 04;
  • 23) 0.367 999 996 989 965 438 839 989 207 04 × 2 = 0 + 0.735 999 993 979 930 877 679 978 414 08;
  • 24) 0.735 999 993 979 930 877 679 978 414 08 × 2 = 1 + 0.471 999 987 959 861 755 359 956 828 16;
  • 25) 0.471 999 987 959 861 755 359 956 828 16 × 2 = 0 + 0.943 999 975 919 723 510 719 913 656 32;
  • 26) 0.943 999 975 919 723 510 719 913 656 32 × 2 = 1 + 0.887 999 951 839 447 021 439 827 312 64;
  • 27) 0.887 999 951 839 447 021 439 827 312 64 × 2 = 1 + 0.775 999 903 678 894 042 879 654 625 28;
  • 28) 0.775 999 903 678 894 042 879 654 625 28 × 2 = 1 + 0.551 999 807 357 788 085 759 309 250 56;
  • 29) 0.551 999 807 357 788 085 759 309 250 56 × 2 = 1 + 0.103 999 614 715 576 171 518 618 501 12;
  • 30) 0.103 999 614 715 576 171 518 618 501 12 × 2 = 0 + 0.207 999 229 431 152 343 037 237 002 24;
  • 31) 0.207 999 229 431 152 343 037 237 002 24 × 2 = 0 + 0.415 998 458 862 304 686 074 474 004 48;
  • 32) 0.415 998 458 862 304 686 074 474 004 48 × 2 = 0 + 0.831 996 917 724 609 372 148 948 008 96;
  • 33) 0.831 996 917 724 609 372 148 948 008 96 × 2 = 1 + 0.663 993 835 449 218 744 297 896 017 92;
  • 34) 0.663 993 835 449 218 744 297 896 017 92 × 2 = 1 + 0.327 987 670 898 437 488 595 792 035 84;
  • 35) 0.327 987 670 898 437 488 595 792 035 84 × 2 = 0 + 0.655 975 341 796 874 977 191 584 071 68;
  • 36) 0.655 975 341 796 874 977 191 584 071 68 × 2 = 1 + 0.311 950 683 593 749 954 383 168 143 36;
  • 37) 0.311 950 683 593 749 954 383 168 143 36 × 2 = 0 + 0.623 901 367 187 499 908 766 336 286 72;
  • 38) 0.623 901 367 187 499 908 766 336 286 72 × 2 = 1 + 0.247 802 734 374 999 817 532 672 573 44;
  • 39) 0.247 802 734 374 999 817 532 672 573 44 × 2 = 0 + 0.495 605 468 749 999 635 065 345 146 88;
  • 40) 0.495 605 468 749 999 635 065 345 146 88 × 2 = 0 + 0.991 210 937 499 999 270 130 690 293 76;
  • 41) 0.991 210 937 499 999 270 130 690 293 76 × 2 = 1 + 0.982 421 874 999 998 540 261 380 587 52;
  • 42) 0.982 421 874 999 998 540 261 380 587 52 × 2 = 1 + 0.964 843 749 999 997 080 522 761 175 04;
  • 43) 0.964 843 749 999 997 080 522 761 175 04 × 2 = 1 + 0.929 687 499 999 994 161 045 522 350 08;
  • 44) 0.929 687 499 999 994 161 045 522 350 08 × 2 = 1 + 0.859 374 999 999 988 322 091 044 700 16;
  • 45) 0.859 374 999 999 988 322 091 044 700 16 × 2 = 1 + 0.718 749 999 999 976 644 182 089 400 32;
  • 46) 0.718 749 999 999 976 644 182 089 400 32 × 2 = 1 + 0.437 499 999 999 953 288 364 178 800 64;
  • 47) 0.437 499 999 999 953 288 364 178 800 64 × 2 = 0 + 0.874 999 999 999 906 576 728 357 601 28;
  • 48) 0.874 999 999 999 906 576 728 357 601 28 × 2 = 1 + 0.749 999 999 999 813 153 456 715 202 56;
  • 49) 0.749 999 999 999 813 153 456 715 202 56 × 2 = 1 + 0.499 999 999 999 626 306 913 430 405 12;
  • 50) 0.499 999 999 999 626 306 913 430 405 12 × 2 = 0 + 0.999 999 999 999 252 613 826 860 810 24;
  • 51) 0.999 999 999 999 252 613 826 860 810 24 × 2 = 1 + 0.999 999 999 998 505 227 653 721 620 48;
  • 52) 0.999 999 999 998 505 227 653 721 620 48 × 2 = 1 + 0.999 999 999 997 010 455 307 443 240 96;
  • 53) 0.999 999 999 997 010 455 307 443 240 96 × 2 = 1 + 0.999 999 999 994 020 910 614 886 481 92;

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:


0.285 749 999 999 999 282 351 836 881 635(10) =

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

5. Positive number before normalization:

6.285 749 999 999 999 282 351 836 881 635(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:


6.285 749 999 999 999 282 351 836 881 635(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 6.285 749 999 999 999 282 351 836 881 635 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