6.285 714 285 714 285 714 285 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 714 285 15(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
6.285 714 285 714 285 714 285 15(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: 0.285 714 285 714 285 714 285 15.

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) 0.285 714 285 714 285 714 285 15 × 2 = 0 + 0.571 428 571 428 571 428 570 3;
  • 2) 0.571 428 571 428 571 428 570 3 × 2 = 1 + 0.142 857 142 857 142 857 140 6;
  • 3) 0.142 857 142 857 142 857 140 6 × 2 = 0 + 0.285 714 285 714 285 714 281 2;
  • 4) 0.285 714 285 714 285 714 281 2 × 2 = 0 + 0.571 428 571 428 571 428 562 4;
  • 5) 0.571 428 571 428 571 428 562 4 × 2 = 1 + 0.142 857 142 857 142 857 124 8;
  • 6) 0.142 857 142 857 142 857 124 8 × 2 = 0 + 0.285 714 285 714 285 714 249 6;
  • 7) 0.285 714 285 714 285 714 249 6 × 2 = 0 + 0.571 428 571 428 571 428 499 2;
  • 8) 0.571 428 571 428 571 428 499 2 × 2 = 1 + 0.142 857 142 857 142 856 998 4;
  • 9) 0.142 857 142 857 142 856 998 4 × 2 = 0 + 0.285 714 285 714 285 713 996 8;
  • 10) 0.285 714 285 714 285 713 996 8 × 2 = 0 + 0.571 428 571 428 571 427 993 6;
  • 11) 0.571 428 571 428 571 427 993 6 × 2 = 1 + 0.142 857 142 857 142 855 987 2;
  • 12) 0.142 857 142 857 142 855 987 2 × 2 = 0 + 0.285 714 285 714 285 711 974 4;
  • 13) 0.285 714 285 714 285 711 974 4 × 2 = 0 + 0.571 428 571 428 571 423 948 8;
  • 14) 0.571 428 571 428 571 423 948 8 × 2 = 1 + 0.142 857 142 857 142 847 897 6;
  • 15) 0.142 857 142 857 142 847 897 6 × 2 = 0 + 0.285 714 285 714 285 695 795 2;
  • 16) 0.285 714 285 714 285 695 795 2 × 2 = 0 + 0.571 428 571 428 571 391 590 4;
  • 17) 0.571 428 571 428 571 391 590 4 × 2 = 1 + 0.142 857 142 857 142 783 180 8;
  • 18) 0.142 857 142 857 142 783 180 8 × 2 = 0 + 0.285 714 285 714 285 566 361 6;
  • 19) 0.285 714 285 714 285 566 361 6 × 2 = 0 + 0.571 428 571 428 571 132 723 2;
  • 20) 0.571 428 571 428 571 132 723 2 × 2 = 1 + 0.142 857 142 857 142 265 446 4;
  • 21) 0.142 857 142 857 142 265 446 4 × 2 = 0 + 0.285 714 285 714 284 530 892 8;
  • 22) 0.285 714 285 714 284 530 892 8 × 2 = 0 + 0.571 428 571 428 569 061 785 6;
  • 23) 0.571 428 571 428 569 061 785 6 × 2 = 1 + 0.142 857 142 857 138 123 571 2;
  • 24) 0.142 857 142 857 138 123 571 2 × 2 = 0 + 0.285 714 285 714 276 247 142 4;
  • 25) 0.285 714 285 714 276 247 142 4 × 2 = 0 + 0.571 428 571 428 552 494 284 8;
  • 26) 0.571 428 571 428 552 494 284 8 × 2 = 1 + 0.142 857 142 857 104 988 569 6;
  • 27) 0.142 857 142 857 104 988 569 6 × 2 = 0 + 0.285 714 285 714 209 977 139 2;
  • 28) 0.285 714 285 714 209 977 139 2 × 2 = 0 + 0.571 428 571 428 419 954 278 4;
  • 29) 0.571 428 571 428 419 954 278 4 × 2 = 1 + 0.142 857 142 856 839 908 556 8;
  • 30) 0.142 857 142 856 839 908 556 8 × 2 = 0 + 0.285 714 285 713 679 817 113 6;
  • 31) 0.285 714 285 713 679 817 113 6 × 2 = 0 + 0.571 428 571 427 359 634 227 2;
  • 32) 0.571 428 571 427 359 634 227 2 × 2 = 1 + 0.142 857 142 854 719 268 454 4;
  • 33) 0.142 857 142 854 719 268 454 4 × 2 = 0 + 0.285 714 285 709 438 536 908 8;
  • 34) 0.285 714 285 709 438 536 908 8 × 2 = 0 + 0.571 428 571 418 877 073 817 6;
  • 35) 0.571 428 571 418 877 073 817 6 × 2 = 1 + 0.142 857 142 837 754 147 635 2;
  • 36) 0.142 857 142 837 754 147 635 2 × 2 = 0 + 0.285 714 285 675 508 295 270 4;
  • 37) 0.285 714 285 675 508 295 270 4 × 2 = 0 + 0.571 428 571 351 016 590 540 8;
  • 38) 0.571 428 571 351 016 590 540 8 × 2 = 1 + 0.142 857 142 702 033 181 081 6;
  • 39) 0.142 857 142 702 033 181 081 6 × 2 = 0 + 0.285 714 285 404 066 362 163 2;
  • 40) 0.285 714 285 404 066 362 163 2 × 2 = 0 + 0.571 428 570 808 132 724 326 4;
  • 41) 0.571 428 570 808 132 724 326 4 × 2 = 1 + 0.142 857 141 616 265 448 652 8;
  • 42) 0.142 857 141 616 265 448 652 8 × 2 = 0 + 0.285 714 283 232 530 897 305 6;
  • 43) 0.285 714 283 232 530 897 305 6 × 2 = 0 + 0.571 428 566 465 061 794 611 2;
  • 44) 0.571 428 566 465 061 794 611 2 × 2 = 1 + 0.142 857 132 930 123 589 222 4;
  • 45) 0.142 857 132 930 123 589 222 4 × 2 = 0 + 0.285 714 265 860 247 178 444 8;
  • 46) 0.285 714 265 860 247 178 444 8 × 2 = 0 + 0.571 428 531 720 494 356 889 6;
  • 47) 0.571 428 531 720 494 356 889 6 × 2 = 1 + 0.142 857 063 440 988 713 779 2;
  • 48) 0.142 857 063 440 988 713 779 2 × 2 = 0 + 0.285 714 126 881 977 427 558 4;
  • 49) 0.285 714 126 881 977 427 558 4 × 2 = 0 + 0.571 428 253 763 954 855 116 8;
  • 50) 0.571 428 253 763 954 855 116 8 × 2 = 1 + 0.142 856 507 527 909 710 233 6;
  • 51) 0.142 856 507 527 909 710 233 6 × 2 = 0 + 0.285 713 015 055 819 420 467 2;
  • 52) 0.285 713 015 055 819 420 467 2 × 2 = 0 + 0.571 426 030 111 638 840 934 4;
  • 53) 0.571 426 030 111 638 840 934 4 × 2 = 1 + 0.142 852 060 223 277 681 868 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 - 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:


0.285 714 285 714 285 714 285 15(10) =

0.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1(2)

5. Positive number before normalization:

6.285 714 285 714 285 714 285 15(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 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:


6.285 714 285 714 285 714 285 15(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1(2) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1(2) × 20 =

1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 001(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 0010 0100 1001 0010 0100 1001 0010 0100 1001 001


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 0010 0100 1001 0010 0100 1001 0010 0100 1001 001 =

1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001


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 0010 0100 1001 0010 0100 1001 0010 0100 1001


Decimal number 6.285 714 285 714 285 714 285 15 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001

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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