Decimal to 64 Bit IEEE 754 Binary: Convert Number 2.714 285 714 285 714 285 714 288 57 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 2.714 285 714 285 714 285 714 288 57(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.714 285 714 285 714 285 714 288 57.

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) 0.714 285 714 285 714 285 714 288 57 × 2 = 1 + 0.428 571 428 571 428 571 428 577 14;
  • 2) 0.428 571 428 571 428 571 428 577 14 × 2 = 0 + 0.857 142 857 142 857 142 857 154 28;
  • 3) 0.857 142 857 142 857 142 857 154 28 × 2 = 1 + 0.714 285 714 285 714 285 714 308 56;
  • 4) 0.714 285 714 285 714 285 714 308 56 × 2 = 1 + 0.428 571 428 571 428 571 428 617 12;
  • 5) 0.428 571 428 571 428 571 428 617 12 × 2 = 0 + 0.857 142 857 142 857 142 857 234 24;
  • 6) 0.857 142 857 142 857 142 857 234 24 × 2 = 1 + 0.714 285 714 285 714 285 714 468 48;
  • 7) 0.714 285 714 285 714 285 714 468 48 × 2 = 1 + 0.428 571 428 571 428 571 428 936 96;
  • 8) 0.428 571 428 571 428 571 428 936 96 × 2 = 0 + 0.857 142 857 142 857 142 857 873 92;
  • 9) 0.857 142 857 142 857 142 857 873 92 × 2 = 1 + 0.714 285 714 285 714 285 715 747 84;
  • 10) 0.714 285 714 285 714 285 715 747 84 × 2 = 1 + 0.428 571 428 571 428 571 431 495 68;
  • 11) 0.428 571 428 571 428 571 431 495 68 × 2 = 0 + 0.857 142 857 142 857 142 862 991 36;
  • 12) 0.857 142 857 142 857 142 862 991 36 × 2 = 1 + 0.714 285 714 285 714 285 725 982 72;
  • 13) 0.714 285 714 285 714 285 725 982 72 × 2 = 1 + 0.428 571 428 571 428 571 451 965 44;
  • 14) 0.428 571 428 571 428 571 451 965 44 × 2 = 0 + 0.857 142 857 142 857 142 903 930 88;
  • 15) 0.857 142 857 142 857 142 903 930 88 × 2 = 1 + 0.714 285 714 285 714 285 807 861 76;
  • 16) 0.714 285 714 285 714 285 807 861 76 × 2 = 1 + 0.428 571 428 571 428 571 615 723 52;
  • 17) 0.428 571 428 571 428 571 615 723 52 × 2 = 0 + 0.857 142 857 142 857 143 231 447 04;
  • 18) 0.857 142 857 142 857 143 231 447 04 × 2 = 1 + 0.714 285 714 285 714 286 462 894 08;
  • 19) 0.714 285 714 285 714 286 462 894 08 × 2 = 1 + 0.428 571 428 571 428 572 925 788 16;
  • 20) 0.428 571 428 571 428 572 925 788 16 × 2 = 0 + 0.857 142 857 142 857 145 851 576 32;
  • 21) 0.857 142 857 142 857 145 851 576 32 × 2 = 1 + 0.714 285 714 285 714 291 703 152 64;
  • 22) 0.714 285 714 285 714 291 703 152 64 × 2 = 1 + 0.428 571 428 571 428 583 406 305 28;
  • 23) 0.428 571 428 571 428 583 406 305 28 × 2 = 0 + 0.857 142 857 142 857 166 812 610 56;
  • 24) 0.857 142 857 142 857 166 812 610 56 × 2 = 1 + 0.714 285 714 285 714 333 625 221 12;
  • 25) 0.714 285 714 285 714 333 625 221 12 × 2 = 1 + 0.428 571 428 571 428 667 250 442 24;
  • 26) 0.428 571 428 571 428 667 250 442 24 × 2 = 0 + 0.857 142 857 142 857 334 500 884 48;
  • 27) 0.857 142 857 142 857 334 500 884 48 × 2 = 1 + 0.714 285 714 285 714 669 001 768 96;
  • 28) 0.714 285 714 285 714 669 001 768 96 × 2 = 1 + 0.428 571 428 571 429 338 003 537 92;
  • 29) 0.428 571 428 571 429 338 003 537 92 × 2 = 0 + 0.857 142 857 142 858 676 007 075 84;
  • 30) 0.857 142 857 142 858 676 007 075 84 × 2 = 1 + 0.714 285 714 285 717 352 014 151 68;
  • 31) 0.714 285 714 285 717 352 014 151 68 × 2 = 1 + 0.428 571 428 571 434 704 028 303 36;
  • 32) 0.428 571 428 571 434 704 028 303 36 × 2 = 0 + 0.857 142 857 142 869 408 056 606 72;
  • 33) 0.857 142 857 142 869 408 056 606 72 × 2 = 1 + 0.714 285 714 285 738 816 113 213 44;
  • 34) 0.714 285 714 285 738 816 113 213 44 × 2 = 1 + 0.428 571 428 571 477 632 226 426 88;
  • 35) 0.428 571 428 571 477 632 226 426 88 × 2 = 0 + 0.857 142 857 142 955 264 452 853 76;
  • 36) 0.857 142 857 142 955 264 452 853 76 × 2 = 1 + 0.714 285 714 285 910 528 905 707 52;
  • 37) 0.714 285 714 285 910 528 905 707 52 × 2 = 1 + 0.428 571 428 571 821 057 811 415 04;
  • 38) 0.428 571 428 571 821 057 811 415 04 × 2 = 0 + 0.857 142 857 143 642 115 622 830 08;
  • 39) 0.857 142 857 143 642 115 622 830 08 × 2 = 1 + 0.714 285 714 287 284 231 245 660 16;
  • 40) 0.714 285 714 287 284 231 245 660 16 × 2 = 1 + 0.428 571 428 574 568 462 491 320 32;
  • 41) 0.428 571 428 574 568 462 491 320 32 × 2 = 0 + 0.857 142 857 149 136 924 982 640 64;
  • 42) 0.857 142 857 149 136 924 982 640 64 × 2 = 1 + 0.714 285 714 298 273 849 965 281 28;
  • 43) 0.714 285 714 298 273 849 965 281 28 × 2 = 1 + 0.428 571 428 596 547 699 930 562 56;
  • 44) 0.428 571 428 596 547 699 930 562 56 × 2 = 0 + 0.857 142 857 193 095 399 861 125 12;
  • 45) 0.857 142 857 193 095 399 861 125 12 × 2 = 1 + 0.714 285 714 386 190 799 722 250 24;
  • 46) 0.714 285 714 386 190 799 722 250 24 × 2 = 1 + 0.428 571 428 772 381 599 444 500 48;
  • 47) 0.428 571 428 772 381 599 444 500 48 × 2 = 0 + 0.857 142 857 544 763 198 889 000 96;
  • 48) 0.857 142 857 544 763 198 889 000 96 × 2 = 1 + 0.714 285 715 089 526 397 778 001 92;
  • 49) 0.714 285 715 089 526 397 778 001 92 × 2 = 1 + 0.428 571 430 179 052 795 556 003 84;
  • 50) 0.428 571 430 179 052 795 556 003 84 × 2 = 0 + 0.857 142 860 358 105 591 112 007 68;
  • 51) 0.857 142 860 358 105 591 112 007 68 × 2 = 1 + 0.714 285 720 716 211 182 224 015 36;
  • 52) 0.714 285 720 716 211 182 224 015 36 × 2 = 1 + 0.428 571 441 432 422 364 448 030 72;
  • 53) 0.428 571 441 432 422 364 448 030 72 × 2 = 0 + 0.857 142 882 864 844 728 896 061 44;

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:


0.714 285 714 285 714 285 714 288 57(10) =

0.1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0(2)

5. Positive number before normalization:

2.714 285 714 285 714 285 714 288 57(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.714 285 714 285 714 285 714 288 57(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0(2) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0(2) × 20 =

1.0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 10(2) × 21


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): 1

Mantissa (not normalized):
1.0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 10


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 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) =

1024(10) =

100 0000 0000(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. 0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101 10 =

0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101


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 0000

Mantissa (52 bits) =
0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101


The base ten decimal number 2.714 285 714 285 714 285 714 288 57 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 1011 0110 1101 1011 0110 1101 1011 0110 1101 1011 0110 1101

» Decimal number in base ten 2.714 285 714 285 714 285 714 288 7 converted and written as 64 bit double precision IEEE 754 binary floating point representation standard

» New: All the Calculations Performed by Our Visitors: Decimal Numbers Converted and Written as 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 04, 2025 [April]: Decimal Numbers Converted and Written as 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Monthly Calculations performed during the month of: April - by our visitors

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