2.714 285 714 299 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.714 285 714 299 9(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
2.714 285 714 299 9(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: 0.714 285 714 299 9.

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) 0.714 285 714 299 9 × 2 = 1 + 0.428 571 428 599 8;
  • 2) 0.428 571 428 599 8 × 2 = 0 + 0.857 142 857 199 6;
  • 3) 0.857 142 857 199 6 × 2 = 1 + 0.714 285 714 399 2;
  • 4) 0.714 285 714 399 2 × 2 = 1 + 0.428 571 428 798 4;
  • 5) 0.428 571 428 798 4 × 2 = 0 + 0.857 142 857 596 8;
  • 6) 0.857 142 857 596 8 × 2 = 1 + 0.714 285 715 193 6;
  • 7) 0.714 285 715 193 6 × 2 = 1 + 0.428 571 430 387 2;
  • 8) 0.428 571 430 387 2 × 2 = 0 + 0.857 142 860 774 4;
  • 9) 0.857 142 860 774 4 × 2 = 1 + 0.714 285 721 548 8;
  • 10) 0.714 285 721 548 8 × 2 = 1 + 0.428 571 443 097 6;
  • 11) 0.428 571 443 097 6 × 2 = 0 + 0.857 142 886 195 2;
  • 12) 0.857 142 886 195 2 × 2 = 1 + 0.714 285 772 390 4;
  • 13) 0.714 285 772 390 4 × 2 = 1 + 0.428 571 544 780 8;
  • 14) 0.428 571 544 780 8 × 2 = 0 + 0.857 143 089 561 6;
  • 15) 0.857 143 089 561 6 × 2 = 1 + 0.714 286 179 123 2;
  • 16) 0.714 286 179 123 2 × 2 = 1 + 0.428 572 358 246 4;
  • 17) 0.428 572 358 246 4 × 2 = 0 + 0.857 144 716 492 8;
  • 18) 0.857 144 716 492 8 × 2 = 1 + 0.714 289 432 985 6;
  • 19) 0.714 289 432 985 6 × 2 = 1 + 0.428 578 865 971 2;
  • 20) 0.428 578 865 971 2 × 2 = 0 + 0.857 157 731 942 4;
  • 21) 0.857 157 731 942 4 × 2 = 1 + 0.714 315 463 884 8;
  • 22) 0.714 315 463 884 8 × 2 = 1 + 0.428 630 927 769 6;
  • 23) 0.428 630 927 769 6 × 2 = 0 + 0.857 261 855 539 2;
  • 24) 0.857 261 855 539 2 × 2 = 1 + 0.714 523 711 078 4;
  • 25) 0.714 523 711 078 4 × 2 = 1 + 0.429 047 422 156 8;
  • 26) 0.429 047 422 156 8 × 2 = 0 + 0.858 094 844 313 6;
  • 27) 0.858 094 844 313 6 × 2 = 1 + 0.716 189 688 627 2;
  • 28) 0.716 189 688 627 2 × 2 = 1 + 0.432 379 377 254 4;
  • 29) 0.432 379 377 254 4 × 2 = 0 + 0.864 758 754 508 8;
  • 30) 0.864 758 754 508 8 × 2 = 1 + 0.729 517 509 017 6;
  • 31) 0.729 517 509 017 6 × 2 = 1 + 0.459 035 018 035 2;
  • 32) 0.459 035 018 035 2 × 2 = 0 + 0.918 070 036 070 4;
  • 33) 0.918 070 036 070 4 × 2 = 1 + 0.836 140 072 140 8;
  • 34) 0.836 140 072 140 8 × 2 = 1 + 0.672 280 144 281 6;
  • 35) 0.672 280 144 281 6 × 2 = 1 + 0.344 560 288 563 2;
  • 36) 0.344 560 288 563 2 × 2 = 0 + 0.689 120 577 126 4;
  • 37) 0.689 120 577 126 4 × 2 = 1 + 0.378 241 154 252 8;
  • 38) 0.378 241 154 252 8 × 2 = 0 + 0.756 482 308 505 6;
  • 39) 0.756 482 308 505 6 × 2 = 1 + 0.512 964 617 011 2;
  • 40) 0.512 964 617 011 2 × 2 = 1 + 0.025 929 234 022 4;
  • 41) 0.025 929 234 022 4 × 2 = 0 + 0.051 858 468 044 8;
  • 42) 0.051 858 468 044 8 × 2 = 0 + 0.103 716 936 089 6;
  • 43) 0.103 716 936 089 6 × 2 = 0 + 0.207 433 872 179 2;
  • 44) 0.207 433 872 179 2 × 2 = 0 + 0.414 867 744 358 4;
  • 45) 0.414 867 744 358 4 × 2 = 0 + 0.829 735 488 716 8;
  • 46) 0.829 735 488 716 8 × 2 = 1 + 0.659 470 977 433 6;
  • 47) 0.659 470 977 433 6 × 2 = 1 + 0.318 941 954 867 2;
  • 48) 0.318 941 954 867 2 × 2 = 0 + 0.637 883 909 734 4;
  • 49) 0.637 883 909 734 4 × 2 = 1 + 0.275 767 819 468 8;
  • 50) 0.275 767 819 468 8 × 2 = 0 + 0.551 535 638 937 6;
  • 51) 0.551 535 638 937 6 × 2 = 1 + 0.103 071 277 875 2;
  • 52) 0.103 071 277 875 2 × 2 = 0 + 0.206 142 555 750 4;
  • 53) 0.206 142 555 750 4 × 2 = 0 + 0.412 285 111 500 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:


0.714 285 714 299 9(10) =

0.1011 0110 1101 1011 0110 1101 1011 0110 1110 1011 0000 0110 1010 0(2)

5. Positive number before normalization:

2.714 285 714 299 9(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1110 1011 0000 0110 1010 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:


2.714 285 714 299 9(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1110 1011 0000 0110 1010 0(2) =

10.1011 0110 1101 1011 0110 1101 1011 0110 1110 1011 0000 0110 1010 0(2) × 20 =

1.0101 1011 0110 1101 1011 0110 1101 1011 0111 0101 1000 0011 0101 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 0101 1000 0011 0101 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 0101 1000 0011 0101 00 =

0101 1011 0110 1101 1011 0110 1101 1011 0111 0101 1000 0011 0101


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 0101 1000 0011 0101


Decimal number 2.714 285 714 299 9 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 0101 1000 0011 0101

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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