2.718 281 828 481 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.718 281 828 481(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
2.718 281 828 481(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: 0.718 281 828 481.

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) 0.718 281 828 481 × 2 = 1 + 0.436 563 656 962;
  • 2) 0.436 563 656 962 × 2 = 0 + 0.873 127 313 924;
  • 3) 0.873 127 313 924 × 2 = 1 + 0.746 254 627 848;
  • 4) 0.746 254 627 848 × 2 = 1 + 0.492 509 255 696;
  • 5) 0.492 509 255 696 × 2 = 0 + 0.985 018 511 392;
  • 6) 0.985 018 511 392 × 2 = 1 + 0.970 037 022 784;
  • 7) 0.970 037 022 784 × 2 = 1 + 0.940 074 045 568;
  • 8) 0.940 074 045 568 × 2 = 1 + 0.880 148 091 136;
  • 9) 0.880 148 091 136 × 2 = 1 + 0.760 296 182 272;
  • 10) 0.760 296 182 272 × 2 = 1 + 0.520 592 364 544;
  • 11) 0.520 592 364 544 × 2 = 1 + 0.041 184 729 088;
  • 12) 0.041 184 729 088 × 2 = 0 + 0.082 369 458 176;
  • 13) 0.082 369 458 176 × 2 = 0 + 0.164 738 916 352;
  • 14) 0.164 738 916 352 × 2 = 0 + 0.329 477 832 704;
  • 15) 0.329 477 832 704 × 2 = 0 + 0.658 955 665 408;
  • 16) 0.658 955 665 408 × 2 = 1 + 0.317 911 330 816;
  • 17) 0.317 911 330 816 × 2 = 0 + 0.635 822 661 632;
  • 18) 0.635 822 661 632 × 2 = 1 + 0.271 645 323 264;
  • 19) 0.271 645 323 264 × 2 = 0 + 0.543 290 646 528;
  • 20) 0.543 290 646 528 × 2 = 1 + 0.086 581 293 056;
  • 21) 0.086 581 293 056 × 2 = 0 + 0.173 162 586 112;
  • 22) 0.173 162 586 112 × 2 = 0 + 0.346 325 172 224;
  • 23) 0.346 325 172 224 × 2 = 0 + 0.692 650 344 448;
  • 24) 0.692 650 344 448 × 2 = 1 + 0.385 300 688 896;
  • 25) 0.385 300 688 896 × 2 = 0 + 0.770 601 377 792;
  • 26) 0.770 601 377 792 × 2 = 1 + 0.541 202 755 584;
  • 27) 0.541 202 755 584 × 2 = 1 + 0.082 405 511 168;
  • 28) 0.082 405 511 168 × 2 = 0 + 0.164 811 022 336;
  • 29) 0.164 811 022 336 × 2 = 0 + 0.329 622 044 672;
  • 30) 0.329 622 044 672 × 2 = 0 + 0.659 244 089 344;
  • 31) 0.659 244 089 344 × 2 = 1 + 0.318 488 178 688;
  • 32) 0.318 488 178 688 × 2 = 0 + 0.636 976 357 376;
  • 33) 0.636 976 357 376 × 2 = 1 + 0.273 952 714 752;
  • 34) 0.273 952 714 752 × 2 = 0 + 0.547 905 429 504;
  • 35) 0.547 905 429 504 × 2 = 1 + 0.095 810 859 008;
  • 36) 0.095 810 859 008 × 2 = 0 + 0.191 621 718 016;
  • 37) 0.191 621 718 016 × 2 = 0 + 0.383 243 436 032;
  • 38) 0.383 243 436 032 × 2 = 0 + 0.766 486 872 064;
  • 39) 0.766 486 872 064 × 2 = 1 + 0.532 973 744 128;
  • 40) 0.532 973 744 128 × 2 = 1 + 0.065 947 488 256;
  • 41) 0.065 947 488 256 × 2 = 0 + 0.131 894 976 512;
  • 42) 0.131 894 976 512 × 2 = 0 + 0.263 789 953 024;
  • 43) 0.263 789 953 024 × 2 = 0 + 0.527 579 906 048;
  • 44) 0.527 579 906 048 × 2 = 1 + 0.055 159 812 096;
  • 45) 0.055 159 812 096 × 2 = 0 + 0.110 319 624 192;
  • 46) 0.110 319 624 192 × 2 = 0 + 0.220 639 248 384;
  • 47) 0.220 639 248 384 × 2 = 0 + 0.441 278 496 768;
  • 48) 0.441 278 496 768 × 2 = 0 + 0.882 556 993 536;
  • 49) 0.882 556 993 536 × 2 = 1 + 0.765 113 987 072;
  • 50) 0.765 113 987 072 × 2 = 1 + 0.530 227 974 144;
  • 51) 0.530 227 974 144 × 2 = 1 + 0.060 455 948 288;
  • 52) 0.060 455 948 288 × 2 = 0 + 0.120 911 896 576;
  • 53) 0.120 911 896 576 × 2 = 0 + 0.241 823 793 152;

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:


0.718 281 828 481(10) =

0.1011 0111 1110 0001 0101 0001 0110 0010 1010 0011 0001 0000 1110 0(2)

5. Positive number before normalization:

2.718 281 828 481(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1010 0011 0001 0000 1110 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:


2.718 281 828 481(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1010 0011 0001 0000 1110 0(2) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1010 0011 0001 0000 1110 0(2) × 20 =

1.0101 1011 1111 0000 1010 1000 1011 0001 0101 0001 1000 1000 0111 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 1111 0000 1010 1000 1011 0001 0101 0001 1000 1000 0111 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 1111 0000 1010 1000 1011 0001 0101 0001 1000 1000 0111 00 =

0101 1011 1111 0000 1010 1000 1011 0001 0101 0001 1000 1000 0111


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 0101 0001 1000 1000 0111


Decimal number 2.718 281 828 481 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 1011 1111 0000 1010 1000 1011 0001 0101 0001 1000 1000 0111

» Convert decimal 2.718 281 828 427 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