2.714 285 714 301 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.714 285 714 301 7(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
2.714 285 714 301 7(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: 0.714 285 714 301 7.

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) 0.714 285 714 301 7 × 2 = 1 + 0.428 571 428 603 4;
  • 2) 0.428 571 428 603 4 × 2 = 0 + 0.857 142 857 206 8;
  • 3) 0.857 142 857 206 8 × 2 = 1 + 0.714 285 714 413 6;
  • 4) 0.714 285 714 413 6 × 2 = 1 + 0.428 571 428 827 2;
  • 5) 0.428 571 428 827 2 × 2 = 0 + 0.857 142 857 654 4;
  • 6) 0.857 142 857 654 4 × 2 = 1 + 0.714 285 715 308 8;
  • 7) 0.714 285 715 308 8 × 2 = 1 + 0.428 571 430 617 6;
  • 8) 0.428 571 430 617 6 × 2 = 0 + 0.857 142 861 235 2;
  • 9) 0.857 142 861 235 2 × 2 = 1 + 0.714 285 722 470 4;
  • 10) 0.714 285 722 470 4 × 2 = 1 + 0.428 571 444 940 8;
  • 11) 0.428 571 444 940 8 × 2 = 0 + 0.857 142 889 881 6;
  • 12) 0.857 142 889 881 6 × 2 = 1 + 0.714 285 779 763 2;
  • 13) 0.714 285 779 763 2 × 2 = 1 + 0.428 571 559 526 4;
  • 14) 0.428 571 559 526 4 × 2 = 0 + 0.857 143 119 052 8;
  • 15) 0.857 143 119 052 8 × 2 = 1 + 0.714 286 238 105 6;
  • 16) 0.714 286 238 105 6 × 2 = 1 + 0.428 572 476 211 2;
  • 17) 0.428 572 476 211 2 × 2 = 0 + 0.857 144 952 422 4;
  • 18) 0.857 144 952 422 4 × 2 = 1 + 0.714 289 904 844 8;
  • 19) 0.714 289 904 844 8 × 2 = 1 + 0.428 579 809 689 6;
  • 20) 0.428 579 809 689 6 × 2 = 0 + 0.857 159 619 379 2;
  • 21) 0.857 159 619 379 2 × 2 = 1 + 0.714 319 238 758 4;
  • 22) 0.714 319 238 758 4 × 2 = 1 + 0.428 638 477 516 8;
  • 23) 0.428 638 477 516 8 × 2 = 0 + 0.857 276 955 033 6;
  • 24) 0.857 276 955 033 6 × 2 = 1 + 0.714 553 910 067 2;
  • 25) 0.714 553 910 067 2 × 2 = 1 + 0.429 107 820 134 4;
  • 26) 0.429 107 820 134 4 × 2 = 0 + 0.858 215 640 268 8;
  • 27) 0.858 215 640 268 8 × 2 = 1 + 0.716 431 280 537 6;
  • 28) 0.716 431 280 537 6 × 2 = 1 + 0.432 862 561 075 2;
  • 29) 0.432 862 561 075 2 × 2 = 0 + 0.865 725 122 150 4;
  • 30) 0.865 725 122 150 4 × 2 = 1 + 0.731 450 244 300 8;
  • 31) 0.731 450 244 300 8 × 2 = 1 + 0.462 900 488 601 6;
  • 32) 0.462 900 488 601 6 × 2 = 0 + 0.925 800 977 203 2;
  • 33) 0.925 800 977 203 2 × 2 = 1 + 0.851 601 954 406 4;
  • 34) 0.851 601 954 406 4 × 2 = 1 + 0.703 203 908 812 8;
  • 35) 0.703 203 908 812 8 × 2 = 1 + 0.406 407 817 625 6;
  • 36) 0.406 407 817 625 6 × 2 = 0 + 0.812 815 635 251 2;
  • 37) 0.812 815 635 251 2 × 2 = 1 + 0.625 631 270 502 4;
  • 38) 0.625 631 270 502 4 × 2 = 1 + 0.251 262 541 004 8;
  • 39) 0.251 262 541 004 8 × 2 = 0 + 0.502 525 082 009 6;
  • 40) 0.502 525 082 009 6 × 2 = 1 + 0.005 050 164 019 2;
  • 41) 0.005 050 164 019 2 × 2 = 0 + 0.010 100 328 038 4;
  • 42) 0.010 100 328 038 4 × 2 = 0 + 0.020 200 656 076 8;
  • 43) 0.020 200 656 076 8 × 2 = 0 + 0.040 401 312 153 6;
  • 44) 0.040 401 312 153 6 × 2 = 0 + 0.080 802 624 307 2;
  • 45) 0.080 802 624 307 2 × 2 = 0 + 0.161 605 248 614 4;
  • 46) 0.161 605 248 614 4 × 2 = 0 + 0.323 210 497 228 8;
  • 47) 0.323 210 497 228 8 × 2 = 0 + 0.646 420 994 457 6;
  • 48) 0.646 420 994 457 6 × 2 = 1 + 0.292 841 988 915 2;
  • 49) 0.292 841 988 915 2 × 2 = 0 + 0.585 683 977 830 4;
  • 50) 0.585 683 977 830 4 × 2 = 1 + 0.171 367 955 660 8;
  • 51) 0.171 367 955 660 8 × 2 = 0 + 0.342 735 911 321 6;
  • 52) 0.342 735 911 321 6 × 2 = 0 + 0.685 471 822 643 2;
  • 53) 0.685 471 822 643 2 × 2 = 1 + 0.370 943 645 286 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:


0.714 285 714 301 7(10) =

0.1011 0110 1101 1011 0110 1101 1011 0110 1110 1101 0000 0001 0100 1(2)

5. Positive number before normalization:

2.714 285 714 301 7(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1110 1101 0000 0001 0100 1(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:


2.714 285 714 301 7(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1110 1101 0000 0001 0100 1(2) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1110 1101 0000 0001 0100 1(2) × 20 =

1.0101 1011 0110 1101 1011 0110 1101 1011 0111 0110 1000 0000 1010 01(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 0110 1000 0000 1010 01


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 0110 1000 0000 1010 01 =

0101 1011 0110 1101 1011 0110 1101 1011 0111 0110 1000 0000 1010


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 0110 1000 0000 1010


Decimal number 2.714 285 714 301 7 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 0110 1000 0000 1010

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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