64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 6.285 714 285 714 285 714 285 714 285 73 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 6.285 714 285 714 285 714 285 714 285 73(10) converted and written in 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: 0.285 714 285 714 285 714 285 714 285 73.

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) 0.285 714 285 714 285 714 285 714 285 73 × 2 = 0 + 0.571 428 571 428 571 428 571 428 571 46;
  • 2) 0.571 428 571 428 571 428 571 428 571 46 × 2 = 1 + 0.142 857 142 857 142 857 142 857 142 92;
  • 3) 0.142 857 142 857 142 857 142 857 142 92 × 2 = 0 + 0.285 714 285 714 285 714 285 714 285 84;
  • 4) 0.285 714 285 714 285 714 285 714 285 84 × 2 = 0 + 0.571 428 571 428 571 428 571 428 571 68;
  • 5) 0.571 428 571 428 571 428 571 428 571 68 × 2 = 1 + 0.142 857 142 857 142 857 142 857 143 36;
  • 6) 0.142 857 142 857 142 857 142 857 143 36 × 2 = 0 + 0.285 714 285 714 285 714 285 714 286 72;
  • 7) 0.285 714 285 714 285 714 285 714 286 72 × 2 = 0 + 0.571 428 571 428 571 428 571 428 573 44;
  • 8) 0.571 428 571 428 571 428 571 428 573 44 × 2 = 1 + 0.142 857 142 857 142 857 142 857 146 88;
  • 9) 0.142 857 142 857 142 857 142 857 146 88 × 2 = 0 + 0.285 714 285 714 285 714 285 714 293 76;
  • 10) 0.285 714 285 714 285 714 285 714 293 76 × 2 = 0 + 0.571 428 571 428 571 428 571 428 587 52;
  • 11) 0.571 428 571 428 571 428 571 428 587 52 × 2 = 1 + 0.142 857 142 857 142 857 142 857 175 04;
  • 12) 0.142 857 142 857 142 857 142 857 175 04 × 2 = 0 + 0.285 714 285 714 285 714 285 714 350 08;
  • 13) 0.285 714 285 714 285 714 285 714 350 08 × 2 = 0 + 0.571 428 571 428 571 428 571 428 700 16;
  • 14) 0.571 428 571 428 571 428 571 428 700 16 × 2 = 1 + 0.142 857 142 857 142 857 142 857 400 32;
  • 15) 0.142 857 142 857 142 857 142 857 400 32 × 2 = 0 + 0.285 714 285 714 285 714 285 714 800 64;
  • 16) 0.285 714 285 714 285 714 285 714 800 64 × 2 = 0 + 0.571 428 571 428 571 428 571 429 601 28;
  • 17) 0.571 428 571 428 571 428 571 429 601 28 × 2 = 1 + 0.142 857 142 857 142 857 142 859 202 56;
  • 18) 0.142 857 142 857 142 857 142 859 202 56 × 2 = 0 + 0.285 714 285 714 285 714 285 718 405 12;
  • 19) 0.285 714 285 714 285 714 285 718 405 12 × 2 = 0 + 0.571 428 571 428 571 428 571 436 810 24;
  • 20) 0.571 428 571 428 571 428 571 436 810 24 × 2 = 1 + 0.142 857 142 857 142 857 142 873 620 48;
  • 21) 0.142 857 142 857 142 857 142 873 620 48 × 2 = 0 + 0.285 714 285 714 285 714 285 747 240 96;
  • 22) 0.285 714 285 714 285 714 285 747 240 96 × 2 = 0 + 0.571 428 571 428 571 428 571 494 481 92;
  • 23) 0.571 428 571 428 571 428 571 494 481 92 × 2 = 1 + 0.142 857 142 857 142 857 142 988 963 84;
  • 24) 0.142 857 142 857 142 857 142 988 963 84 × 2 = 0 + 0.285 714 285 714 285 714 285 977 927 68;
  • 25) 0.285 714 285 714 285 714 285 977 927 68 × 2 = 0 + 0.571 428 571 428 571 428 571 955 855 36;
  • 26) 0.571 428 571 428 571 428 571 955 855 36 × 2 = 1 + 0.142 857 142 857 142 857 143 911 710 72;
  • 27) 0.142 857 142 857 142 857 143 911 710 72 × 2 = 0 + 0.285 714 285 714 285 714 287 823 421 44;
  • 28) 0.285 714 285 714 285 714 287 823 421 44 × 2 = 0 + 0.571 428 571 428 571 428 575 646 842 88;
  • 29) 0.571 428 571 428 571 428 575 646 842 88 × 2 = 1 + 0.142 857 142 857 142 857 151 293 685 76;
  • 30) 0.142 857 142 857 142 857 151 293 685 76 × 2 = 0 + 0.285 714 285 714 285 714 302 587 371 52;
  • 31) 0.285 714 285 714 285 714 302 587 371 52 × 2 = 0 + 0.571 428 571 428 571 428 605 174 743 04;
  • 32) 0.571 428 571 428 571 428 605 174 743 04 × 2 = 1 + 0.142 857 142 857 142 857 210 349 486 08;
  • 33) 0.142 857 142 857 142 857 210 349 486 08 × 2 = 0 + 0.285 714 285 714 285 714 420 698 972 16;
  • 34) 0.285 714 285 714 285 714 420 698 972 16 × 2 = 0 + 0.571 428 571 428 571 428 841 397 944 32;
  • 35) 0.571 428 571 428 571 428 841 397 944 32 × 2 = 1 + 0.142 857 142 857 142 857 682 795 888 64;
  • 36) 0.142 857 142 857 142 857 682 795 888 64 × 2 = 0 + 0.285 714 285 714 285 715 365 591 777 28;
  • 37) 0.285 714 285 714 285 715 365 591 777 28 × 2 = 0 + 0.571 428 571 428 571 430 731 183 554 56;
  • 38) 0.571 428 571 428 571 430 731 183 554 56 × 2 = 1 + 0.142 857 142 857 142 861 462 367 109 12;
  • 39) 0.142 857 142 857 142 861 462 367 109 12 × 2 = 0 + 0.285 714 285 714 285 722 924 734 218 24;
  • 40) 0.285 714 285 714 285 722 924 734 218 24 × 2 = 0 + 0.571 428 571 428 571 445 849 468 436 48;
  • 41) 0.571 428 571 428 571 445 849 468 436 48 × 2 = 1 + 0.142 857 142 857 142 891 698 936 872 96;
  • 42) 0.142 857 142 857 142 891 698 936 872 96 × 2 = 0 + 0.285 714 285 714 285 783 397 873 745 92;
  • 43) 0.285 714 285 714 285 783 397 873 745 92 × 2 = 0 + 0.571 428 571 428 571 566 795 747 491 84;
  • 44) 0.571 428 571 428 571 566 795 747 491 84 × 2 = 1 + 0.142 857 142 857 143 133 591 494 983 68;
  • 45) 0.142 857 142 857 143 133 591 494 983 68 × 2 = 0 + 0.285 714 285 714 286 267 182 989 967 36;
  • 46) 0.285 714 285 714 286 267 182 989 967 36 × 2 = 0 + 0.571 428 571 428 572 534 365 979 934 72;
  • 47) 0.571 428 571 428 572 534 365 979 934 72 × 2 = 1 + 0.142 857 142 857 145 068 731 959 869 44;
  • 48) 0.142 857 142 857 145 068 731 959 869 44 × 2 = 0 + 0.285 714 285 714 290 137 463 919 738 88;
  • 49) 0.285 714 285 714 290 137 463 919 738 88 × 2 = 0 + 0.571 428 571 428 580 274 927 839 477 76;
  • 50) 0.571 428 571 428 580 274 927 839 477 76 × 2 = 1 + 0.142 857 142 857 160 549 855 678 955 52;
  • 51) 0.142 857 142 857 160 549 855 678 955 52 × 2 = 0 + 0.285 714 285 714 321 099 711 357 911 04;
  • 52) 0.285 714 285 714 321 099 711 357 911 04 × 2 = 0 + 0.571 428 571 428 642 199 422 715 822 08;
  • 53) 0.571 428 571 428 642 199 422 715 822 08 × 2 = 1 + 0.142 857 142 857 284 398 845 431 644 16;

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...)


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:


0.285 714 285 714 285 714 285 714 285 73(10) =

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


5. Positive number before normalization:

6.285 714 285 714 285 714 285 714 285 73(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:


6.285 714 285 714 285 714 285 714 285 73(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


The base ten decimal number 6.285 714 285 714 285 714 285 714 285 73 converted and written in 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

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 6.285 714 285 714 285 714 285 714 285 72 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 6.285 714 285 714 285 714 285 714 285 74 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

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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