2.718 281 828 470 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.718 281 828 470 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
2.718 281 828 470 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: 0.718 281 828 470 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) 0.718 281 828 470 7 × 2 = 1 + 0.436 563 656 941 4;
  • 2) 0.436 563 656 941 4 × 2 = 0 + 0.873 127 313 882 8;
  • 3) 0.873 127 313 882 8 × 2 = 1 + 0.746 254 627 765 6;
  • 4) 0.746 254 627 765 6 × 2 = 1 + 0.492 509 255 531 2;
  • 5) 0.492 509 255 531 2 × 2 = 0 + 0.985 018 511 062 4;
  • 6) 0.985 018 511 062 4 × 2 = 1 + 0.970 037 022 124 8;
  • 7) 0.970 037 022 124 8 × 2 = 1 + 0.940 074 044 249 6;
  • 8) 0.940 074 044 249 6 × 2 = 1 + 0.880 148 088 499 2;
  • 9) 0.880 148 088 499 2 × 2 = 1 + 0.760 296 176 998 4;
  • 10) 0.760 296 176 998 4 × 2 = 1 + 0.520 592 353 996 8;
  • 11) 0.520 592 353 996 8 × 2 = 1 + 0.041 184 707 993 6;
  • 12) 0.041 184 707 993 6 × 2 = 0 + 0.082 369 415 987 2;
  • 13) 0.082 369 415 987 2 × 2 = 0 + 0.164 738 831 974 4;
  • 14) 0.164 738 831 974 4 × 2 = 0 + 0.329 477 663 948 8;
  • 15) 0.329 477 663 948 8 × 2 = 0 + 0.658 955 327 897 6;
  • 16) 0.658 955 327 897 6 × 2 = 1 + 0.317 910 655 795 2;
  • 17) 0.317 910 655 795 2 × 2 = 0 + 0.635 821 311 590 4;
  • 18) 0.635 821 311 590 4 × 2 = 1 + 0.271 642 623 180 8;
  • 19) 0.271 642 623 180 8 × 2 = 0 + 0.543 285 246 361 6;
  • 20) 0.543 285 246 361 6 × 2 = 1 + 0.086 570 492 723 2;
  • 21) 0.086 570 492 723 2 × 2 = 0 + 0.173 140 985 446 4;
  • 22) 0.173 140 985 446 4 × 2 = 0 + 0.346 281 970 892 8;
  • 23) 0.346 281 970 892 8 × 2 = 0 + 0.692 563 941 785 6;
  • 24) 0.692 563 941 785 6 × 2 = 1 + 0.385 127 883 571 2;
  • 25) 0.385 127 883 571 2 × 2 = 0 + 0.770 255 767 142 4;
  • 26) 0.770 255 767 142 4 × 2 = 1 + 0.540 511 534 284 8;
  • 27) 0.540 511 534 284 8 × 2 = 1 + 0.081 023 068 569 6;
  • 28) 0.081 023 068 569 6 × 2 = 0 + 0.162 046 137 139 2;
  • 29) 0.162 046 137 139 2 × 2 = 0 + 0.324 092 274 278 4;
  • 30) 0.324 092 274 278 4 × 2 = 0 + 0.648 184 548 556 8;
  • 31) 0.648 184 548 556 8 × 2 = 1 + 0.296 369 097 113 6;
  • 32) 0.296 369 097 113 6 × 2 = 0 + 0.592 738 194 227 2;
  • 33) 0.592 738 194 227 2 × 2 = 1 + 0.185 476 388 454 4;
  • 34) 0.185 476 388 454 4 × 2 = 0 + 0.370 952 776 908 8;
  • 35) 0.370 952 776 908 8 × 2 = 0 + 0.741 905 553 817 6;
  • 36) 0.741 905 553 817 6 × 2 = 1 + 0.483 811 107 635 2;
  • 37) 0.483 811 107 635 2 × 2 = 0 + 0.967 622 215 270 4;
  • 38) 0.967 622 215 270 4 × 2 = 1 + 0.935 244 430 540 8;
  • 39) 0.935 244 430 540 8 × 2 = 1 + 0.870 488 861 081 6;
  • 40) 0.870 488 861 081 6 × 2 = 1 + 0.740 977 722 163 2;
  • 41) 0.740 977 722 163 2 × 2 = 1 + 0.481 955 444 326 4;
  • 42) 0.481 955 444 326 4 × 2 = 0 + 0.963 910 888 652 8;
  • 43) 0.963 910 888 652 8 × 2 = 1 + 0.927 821 777 305 6;
  • 44) 0.927 821 777 305 6 × 2 = 1 + 0.855 643 554 611 2;
  • 45) 0.855 643 554 611 2 × 2 = 1 + 0.711 287 109 222 4;
  • 46) 0.711 287 109 222 4 × 2 = 1 + 0.422 574 218 444 8;
  • 47) 0.422 574 218 444 8 × 2 = 0 + 0.845 148 436 889 6;
  • 48) 0.845 148 436 889 6 × 2 = 1 + 0.690 296 873 779 2;
  • 49) 0.690 296 873 779 2 × 2 = 1 + 0.380 593 747 558 4;
  • 50) 0.380 593 747 558 4 × 2 = 0 + 0.761 187 495 116 8;
  • 51) 0.761 187 495 116 8 × 2 = 1 + 0.522 374 990 233 6;
  • 52) 0.522 374 990 233 6 × 2 = 1 + 0.044 749 980 467 2;
  • 53) 0.044 749 980 467 2 × 2 = 0 + 0.089 499 960 934 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:


0.718 281 828 470 7(10) =

0.1011 0111 1110 0001 0101 0001 0110 0010 1001 0111 1011 1101 1011 0(2)

5. Positive number before normalization:

2.718 281 828 470 7(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1001 0111 1011 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:


2.718 281 828 470 7(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1001 0111 1011 1101 1011 0(2) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1001 0111 1011 1101 1011 0(2) × 20 =

1.0101 1011 1111 0000 1010 1000 1011 0001 0100 1011 1101 1110 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 1111 0000 1010 1000 1011 0001 0100 1011 1101 1110 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 1111 0000 1010 1000 1011 0001 0100 1011 1101 1110 1101 10 =

0101 1011 1111 0000 1010 1000 1011 0001 0100 1011 1101 1110 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 1111 0000 1010 1000 1011 0001 0100 1011 1101 1110 1101


Decimal number 2.718 281 828 470 7 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 1011 1101 1110 1101

» Convert decimal 2.718 281 828 477 6 to 64 bit double precision IEEE 754 binary floating point representation standard

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

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - 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