2.714 285 714 292 95 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.714 285 714 292 95(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
2.714 285 714 292 95(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: 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: 0.714 285 714 292 95.

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) 0.714 285 714 292 95 × 2 = 1 + 0.428 571 428 585 9;
  • 2) 0.428 571 428 585 9 × 2 = 0 + 0.857 142 857 171 8;
  • 3) 0.857 142 857 171 8 × 2 = 1 + 0.714 285 714 343 6;
  • 4) 0.714 285 714 343 6 × 2 = 1 + 0.428 571 428 687 2;
  • 5) 0.428 571 428 687 2 × 2 = 0 + 0.857 142 857 374 4;
  • 6) 0.857 142 857 374 4 × 2 = 1 + 0.714 285 714 748 8;
  • 7) 0.714 285 714 748 8 × 2 = 1 + 0.428 571 429 497 6;
  • 8) 0.428 571 429 497 6 × 2 = 0 + 0.857 142 858 995 2;
  • 9) 0.857 142 858 995 2 × 2 = 1 + 0.714 285 717 990 4;
  • 10) 0.714 285 717 990 4 × 2 = 1 + 0.428 571 435 980 8;
  • 11) 0.428 571 435 980 8 × 2 = 0 + 0.857 142 871 961 6;
  • 12) 0.857 142 871 961 6 × 2 = 1 + 0.714 285 743 923 2;
  • 13) 0.714 285 743 923 2 × 2 = 1 + 0.428 571 487 846 4;
  • 14) 0.428 571 487 846 4 × 2 = 0 + 0.857 142 975 692 8;
  • 15) 0.857 142 975 692 8 × 2 = 1 + 0.714 285 951 385 6;
  • 16) 0.714 285 951 385 6 × 2 = 1 + 0.428 571 902 771 2;
  • 17) 0.428 571 902 771 2 × 2 = 0 + 0.857 143 805 542 4;
  • 18) 0.857 143 805 542 4 × 2 = 1 + 0.714 287 611 084 8;
  • 19) 0.714 287 611 084 8 × 2 = 1 + 0.428 575 222 169 6;
  • 20) 0.428 575 222 169 6 × 2 = 0 + 0.857 150 444 339 2;
  • 21) 0.857 150 444 339 2 × 2 = 1 + 0.714 300 888 678 4;
  • 22) 0.714 300 888 678 4 × 2 = 1 + 0.428 601 777 356 8;
  • 23) 0.428 601 777 356 8 × 2 = 0 + 0.857 203 554 713 6;
  • 24) 0.857 203 554 713 6 × 2 = 1 + 0.714 407 109 427 2;
  • 25) 0.714 407 109 427 2 × 2 = 1 + 0.428 814 218 854 4;
  • 26) 0.428 814 218 854 4 × 2 = 0 + 0.857 628 437 708 8;
  • 27) 0.857 628 437 708 8 × 2 = 1 + 0.715 256 875 417 6;
  • 28) 0.715 256 875 417 6 × 2 = 1 + 0.430 513 750 835 2;
  • 29) 0.430 513 750 835 2 × 2 = 0 + 0.861 027 501 670 4;
  • 30) 0.861 027 501 670 4 × 2 = 1 + 0.722 055 003 340 8;
  • 31) 0.722 055 003 340 8 × 2 = 1 + 0.444 110 006 681 6;
  • 32) 0.444 110 006 681 6 × 2 = 0 + 0.888 220 013 363 2;
  • 33) 0.888 220 013 363 2 × 2 = 1 + 0.776 440 026 726 4;
  • 34) 0.776 440 026 726 4 × 2 = 1 + 0.552 880 053 452 8;
  • 35) 0.552 880 053 452 8 × 2 = 1 + 0.105 760 106 905 6;
  • 36) 0.105 760 106 905 6 × 2 = 0 + 0.211 520 213 811 2;
  • 37) 0.211 520 213 811 2 × 2 = 0 + 0.423 040 427 622 4;
  • 38) 0.423 040 427 622 4 × 2 = 0 + 0.846 080 855 244 8;
  • 39) 0.846 080 855 244 8 × 2 = 1 + 0.692 161 710 489 6;
  • 40) 0.692 161 710 489 6 × 2 = 1 + 0.384 323 420 979 2;
  • 41) 0.384 323 420 979 2 × 2 = 0 + 0.768 646 841 958 4;
  • 42) 0.768 646 841 958 4 × 2 = 1 + 0.537 293 683 916 8;
  • 43) 0.537 293 683 916 8 × 2 = 1 + 0.074 587 367 833 6;
  • 44) 0.074 587 367 833 6 × 2 = 0 + 0.149 174 735 667 2;
  • 45) 0.149 174 735 667 2 × 2 = 0 + 0.298 349 471 334 4;
  • 46) 0.298 349 471 334 4 × 2 = 0 + 0.596 698 942 668 8;
  • 47) 0.596 698 942 668 8 × 2 = 1 + 0.193 397 885 337 6;
  • 48) 0.193 397 885 337 6 × 2 = 0 + 0.386 795 770 675 2;
  • 49) 0.386 795 770 675 2 × 2 = 0 + 0.773 591 541 350 4;
  • 50) 0.773 591 541 350 4 × 2 = 1 + 0.547 183 082 700 8;
  • 51) 0.547 183 082 700 8 × 2 = 1 + 0.094 366 165 401 6;
  • 52) 0.094 366 165 401 6 × 2 = 0 + 0.188 732 330 803 2;
  • 53) 0.188 732 330 803 2 × 2 = 0 + 0.377 464 661 606 4;

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:


0.714 285 714 292 95(10) =

0.1011 0110 1101 1011 0110 1101 1011 0110 1110 0011 0110 0010 0110 0(2)

5. Positive number before normalization:

2.714 285 714 292 95(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1110 0011 0110 0010 0110 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:


2.714 285 714 292 95(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1110 0011 0110 0010 0110 0(2) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1110 0011 0110 0010 0110 0(2) × 20 =

1.0101 1011 0110 1101 1011 0110 1101 1011 0111 0001 1011 0001 0011 00(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 0111 0001 1011 0001 0011 00


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 0111 0001 1011 0001 0011 00 =

0101 1011 0110 1101 1011 0110 1101 1011 0111 0001 1011 0001 0011


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 0111 0001 1011 0001 0011


Decimal number 2.714 285 714 292 95 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 1011 0110 1101 1011 0110 1101 1011 0111 0001 1011 0001 0011

» Convert decimal 2.714 285 714 293 51 to 64 bit double precision IEEE 754 binary floating point representation standard

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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