2.718 281 828 477 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.718 281 828 477 6(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
2.718 281 828 477 6(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: 0.718 281 828 477 6.

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) 0.718 281 828 477 6 × 2 = 1 + 0.436 563 656 955 2;
  • 2) 0.436 563 656 955 2 × 2 = 0 + 0.873 127 313 910 4;
  • 3) 0.873 127 313 910 4 × 2 = 1 + 0.746 254 627 820 8;
  • 4) 0.746 254 627 820 8 × 2 = 1 + 0.492 509 255 641 6;
  • 5) 0.492 509 255 641 6 × 2 = 0 + 0.985 018 511 283 2;
  • 6) 0.985 018 511 283 2 × 2 = 1 + 0.970 037 022 566 4;
  • 7) 0.970 037 022 566 4 × 2 = 1 + 0.940 074 045 132 8;
  • 8) 0.940 074 045 132 8 × 2 = 1 + 0.880 148 090 265 6;
  • 9) 0.880 148 090 265 6 × 2 = 1 + 0.760 296 180 531 2;
  • 10) 0.760 296 180 531 2 × 2 = 1 + 0.520 592 361 062 4;
  • 11) 0.520 592 361 062 4 × 2 = 1 + 0.041 184 722 124 8;
  • 12) 0.041 184 722 124 8 × 2 = 0 + 0.082 369 444 249 6;
  • 13) 0.082 369 444 249 6 × 2 = 0 + 0.164 738 888 499 2;
  • 14) 0.164 738 888 499 2 × 2 = 0 + 0.329 477 776 998 4;
  • 15) 0.329 477 776 998 4 × 2 = 0 + 0.658 955 553 996 8;
  • 16) 0.658 955 553 996 8 × 2 = 1 + 0.317 911 107 993 6;
  • 17) 0.317 911 107 993 6 × 2 = 0 + 0.635 822 215 987 2;
  • 18) 0.635 822 215 987 2 × 2 = 1 + 0.271 644 431 974 4;
  • 19) 0.271 644 431 974 4 × 2 = 0 + 0.543 288 863 948 8;
  • 20) 0.543 288 863 948 8 × 2 = 1 + 0.086 577 727 897 6;
  • 21) 0.086 577 727 897 6 × 2 = 0 + 0.173 155 455 795 2;
  • 22) 0.173 155 455 795 2 × 2 = 0 + 0.346 310 911 590 4;
  • 23) 0.346 310 911 590 4 × 2 = 0 + 0.692 621 823 180 8;
  • 24) 0.692 621 823 180 8 × 2 = 1 + 0.385 243 646 361 6;
  • 25) 0.385 243 646 361 6 × 2 = 0 + 0.770 487 292 723 2;
  • 26) 0.770 487 292 723 2 × 2 = 1 + 0.540 974 585 446 4;
  • 27) 0.540 974 585 446 4 × 2 = 1 + 0.081 949 170 892 8;
  • 28) 0.081 949 170 892 8 × 2 = 0 + 0.163 898 341 785 6;
  • 29) 0.163 898 341 785 6 × 2 = 0 + 0.327 796 683 571 2;
  • 30) 0.327 796 683 571 2 × 2 = 0 + 0.655 593 367 142 4;
  • 31) 0.655 593 367 142 4 × 2 = 1 + 0.311 186 734 284 8;
  • 32) 0.311 186 734 284 8 × 2 = 0 + 0.622 373 468 569 6;
  • 33) 0.622 373 468 569 6 × 2 = 1 + 0.244 746 937 139 2;
  • 34) 0.244 746 937 139 2 × 2 = 0 + 0.489 493 874 278 4;
  • 35) 0.489 493 874 278 4 × 2 = 0 + 0.978 987 748 556 8;
  • 36) 0.978 987 748 556 8 × 2 = 1 + 0.957 975 497 113 6;
  • 37) 0.957 975 497 113 6 × 2 = 1 + 0.915 950 994 227 2;
  • 38) 0.915 950 994 227 2 × 2 = 1 + 0.831 901 988 454 4;
  • 39) 0.831 901 988 454 4 × 2 = 1 + 0.663 803 976 908 8;
  • 40) 0.663 803 976 908 8 × 2 = 1 + 0.327 607 953 817 6;
  • 41) 0.327 607 953 817 6 × 2 = 0 + 0.655 215 907 635 2;
  • 42) 0.655 215 907 635 2 × 2 = 1 + 0.310 431 815 270 4;
  • 43) 0.310 431 815 270 4 × 2 = 0 + 0.620 863 630 540 8;
  • 44) 0.620 863 630 540 8 × 2 = 1 + 0.241 727 261 081 6;
  • 45) 0.241 727 261 081 6 × 2 = 0 + 0.483 454 522 163 2;
  • 46) 0.483 454 522 163 2 × 2 = 0 + 0.966 909 044 326 4;
  • 47) 0.966 909 044 326 4 × 2 = 1 + 0.933 818 088 652 8;
  • 48) 0.933 818 088 652 8 × 2 = 1 + 0.867 636 177 305 6;
  • 49) 0.867 636 177 305 6 × 2 = 1 + 0.735 272 354 611 2;
  • 50) 0.735 272 354 611 2 × 2 = 1 + 0.470 544 709 222 4;
  • 51) 0.470 544 709 222 4 × 2 = 0 + 0.941 089 418 444 8;
  • 52) 0.941 089 418 444 8 × 2 = 1 + 0.882 178 836 889 6;
  • 53) 0.882 178 836 889 6 × 2 = 1 + 0.764 357 673 779 2;

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:


0.718 281 828 477 6(10) =

0.1011 0111 1110 0001 0101 0001 0110 0010 1001 1111 0101 0011 1101 1(2)

5. Positive number before normalization:

2.718 281 828 477 6(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1001 1111 0101 0011 1101 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:


2.718 281 828 477 6(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1001 1111 0101 0011 1101 1(2) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1001 1111 0101 0011 1101 1(2) × 20 =

1.0101 1011 1111 0000 1010 1000 1011 0001 0100 1111 1010 1001 1110 11(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 1111 0000 1010 1000 1011 0001 0100 1111 1010 1001 1110 11


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 1111 0000 1010 1000 1011 0001 0100 1111 1010 1001 1110 11 =

0101 1011 1111 0000 1010 1000 1011 0001 0100 1111 1010 1001 1110


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 1111 0000 1010 1000 1011 0001 0100 1111 1010 1001 1110


Decimal number 2.718 281 828 477 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 1011 1111 0000 1010 1000 1011 0001 0100 1111 1010 1001 1110

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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