6.285 714 285 714 277 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 277 4(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
6.285 714 285 714 277 4(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: 0.285 714 285 714 277 4.

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) 0.285 714 285 714 277 4 × 2 = 0 + 0.571 428 571 428 554 8;
  • 2) 0.571 428 571 428 554 8 × 2 = 1 + 0.142 857 142 857 109 6;
  • 3) 0.142 857 142 857 109 6 × 2 = 0 + 0.285 714 285 714 219 2;
  • 4) 0.285 714 285 714 219 2 × 2 = 0 + 0.571 428 571 428 438 4;
  • 5) 0.571 428 571 428 438 4 × 2 = 1 + 0.142 857 142 856 876 8;
  • 6) 0.142 857 142 856 876 8 × 2 = 0 + 0.285 714 285 713 753 6;
  • 7) 0.285 714 285 713 753 6 × 2 = 0 + 0.571 428 571 427 507 2;
  • 8) 0.571 428 571 427 507 2 × 2 = 1 + 0.142 857 142 855 014 4;
  • 9) 0.142 857 142 855 014 4 × 2 = 0 + 0.285 714 285 710 028 8;
  • 10) 0.285 714 285 710 028 8 × 2 = 0 + 0.571 428 571 420 057 6;
  • 11) 0.571 428 571 420 057 6 × 2 = 1 + 0.142 857 142 840 115 2;
  • 12) 0.142 857 142 840 115 2 × 2 = 0 + 0.285 714 285 680 230 4;
  • 13) 0.285 714 285 680 230 4 × 2 = 0 + 0.571 428 571 360 460 8;
  • 14) 0.571 428 571 360 460 8 × 2 = 1 + 0.142 857 142 720 921 6;
  • 15) 0.142 857 142 720 921 6 × 2 = 0 + 0.285 714 285 441 843 2;
  • 16) 0.285 714 285 441 843 2 × 2 = 0 + 0.571 428 570 883 686 4;
  • 17) 0.571 428 570 883 686 4 × 2 = 1 + 0.142 857 141 767 372 8;
  • 18) 0.142 857 141 767 372 8 × 2 = 0 + 0.285 714 283 534 745 6;
  • 19) 0.285 714 283 534 745 6 × 2 = 0 + 0.571 428 567 069 491 2;
  • 20) 0.571 428 567 069 491 2 × 2 = 1 + 0.142 857 134 138 982 4;
  • 21) 0.142 857 134 138 982 4 × 2 = 0 + 0.285 714 268 277 964 8;
  • 22) 0.285 714 268 277 964 8 × 2 = 0 + 0.571 428 536 555 929 6;
  • 23) 0.571 428 536 555 929 6 × 2 = 1 + 0.142 857 073 111 859 2;
  • 24) 0.142 857 073 111 859 2 × 2 = 0 + 0.285 714 146 223 718 4;
  • 25) 0.285 714 146 223 718 4 × 2 = 0 + 0.571 428 292 447 436 8;
  • 26) 0.571 428 292 447 436 8 × 2 = 1 + 0.142 856 584 894 873 6;
  • 27) 0.142 856 584 894 873 6 × 2 = 0 + 0.285 713 169 789 747 2;
  • 28) 0.285 713 169 789 747 2 × 2 = 0 + 0.571 426 339 579 494 4;
  • 29) 0.571 426 339 579 494 4 × 2 = 1 + 0.142 852 679 158 988 8;
  • 30) 0.142 852 679 158 988 8 × 2 = 0 + 0.285 705 358 317 977 6;
  • 31) 0.285 705 358 317 977 6 × 2 = 0 + 0.571 410 716 635 955 2;
  • 32) 0.571 410 716 635 955 2 × 2 = 1 + 0.142 821 433 271 910 4;
  • 33) 0.142 821 433 271 910 4 × 2 = 0 + 0.285 642 866 543 820 8;
  • 34) 0.285 642 866 543 820 8 × 2 = 0 + 0.571 285 733 087 641 6;
  • 35) 0.571 285 733 087 641 6 × 2 = 1 + 0.142 571 466 175 283 2;
  • 36) 0.142 571 466 175 283 2 × 2 = 0 + 0.285 142 932 350 566 4;
  • 37) 0.285 142 932 350 566 4 × 2 = 0 + 0.570 285 864 701 132 8;
  • 38) 0.570 285 864 701 132 8 × 2 = 1 + 0.140 571 729 402 265 6;
  • 39) 0.140 571 729 402 265 6 × 2 = 0 + 0.281 143 458 804 531 2;
  • 40) 0.281 143 458 804 531 2 × 2 = 0 + 0.562 286 917 609 062 4;
  • 41) 0.562 286 917 609 062 4 × 2 = 1 + 0.124 573 835 218 124 8;
  • 42) 0.124 573 835 218 124 8 × 2 = 0 + 0.249 147 670 436 249 6;
  • 43) 0.249 147 670 436 249 6 × 2 = 0 + 0.498 295 340 872 499 2;
  • 44) 0.498 295 340 872 499 2 × 2 = 0 + 0.996 590 681 744 998 4;
  • 45) 0.996 590 681 744 998 4 × 2 = 1 + 0.993 181 363 489 996 8;
  • 46) 0.993 181 363 489 996 8 × 2 = 1 + 0.986 362 726 979 993 6;
  • 47) 0.986 362 726 979 993 6 × 2 = 1 + 0.972 725 453 959 987 2;
  • 48) 0.972 725 453 959 987 2 × 2 = 1 + 0.945 450 907 919 974 4;
  • 49) 0.945 450 907 919 974 4 × 2 = 1 + 0.890 901 815 839 948 8;
  • 50) 0.890 901 815 839 948 8 × 2 = 1 + 0.781 803 631 679 897 6;
  • 51) 0.781 803 631 679 897 6 × 2 = 1 + 0.563 607 263 359 795 2;
  • 52) 0.563 607 263 359 795 2 × 2 = 1 + 0.127 214 526 719 590 4;
  • 53) 0.127 214 526 719 590 4 × 2 = 0 + 0.254 429 053 439 180 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:


0.285 714 285 714 277 4(10) =

0.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1000 1111 1111 0(2)

5. Positive number before normalization:

6.285 714 285 714 277 4(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1000 1111 1111 0(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:


6.285 714 285 714 277 4(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1000 1111 1111 0(2) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1000 1111 1111 0(2) × 20 =

1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0011 1111 110(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 0011 1111 110


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 0011 1111 110 =

1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0011 1111


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 0011 1111


Decimal number 6.285 714 285 714 277 4 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 0011 1111

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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