2.718 281 828 459 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.718 281 828 459 8(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
2.718 281 828 459 8(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: 0.718 281 828 459 8.

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) 0.718 281 828 459 8 × 2 = 1 + 0.436 563 656 919 6;
  • 2) 0.436 563 656 919 6 × 2 = 0 + 0.873 127 313 839 2;
  • 3) 0.873 127 313 839 2 × 2 = 1 + 0.746 254 627 678 4;
  • 4) 0.746 254 627 678 4 × 2 = 1 + 0.492 509 255 356 8;
  • 5) 0.492 509 255 356 8 × 2 = 0 + 0.985 018 510 713 6;
  • 6) 0.985 018 510 713 6 × 2 = 1 + 0.970 037 021 427 2;
  • 7) 0.970 037 021 427 2 × 2 = 1 + 0.940 074 042 854 4;
  • 8) 0.940 074 042 854 4 × 2 = 1 + 0.880 148 085 708 8;
  • 9) 0.880 148 085 708 8 × 2 = 1 + 0.760 296 171 417 6;
  • 10) 0.760 296 171 417 6 × 2 = 1 + 0.520 592 342 835 2;
  • 11) 0.520 592 342 835 2 × 2 = 1 + 0.041 184 685 670 4;
  • 12) 0.041 184 685 670 4 × 2 = 0 + 0.082 369 371 340 8;
  • 13) 0.082 369 371 340 8 × 2 = 0 + 0.164 738 742 681 6;
  • 14) 0.164 738 742 681 6 × 2 = 0 + 0.329 477 485 363 2;
  • 15) 0.329 477 485 363 2 × 2 = 0 + 0.658 954 970 726 4;
  • 16) 0.658 954 970 726 4 × 2 = 1 + 0.317 909 941 452 8;
  • 17) 0.317 909 941 452 8 × 2 = 0 + 0.635 819 882 905 6;
  • 18) 0.635 819 882 905 6 × 2 = 1 + 0.271 639 765 811 2;
  • 19) 0.271 639 765 811 2 × 2 = 0 + 0.543 279 531 622 4;
  • 20) 0.543 279 531 622 4 × 2 = 1 + 0.086 559 063 244 8;
  • 21) 0.086 559 063 244 8 × 2 = 0 + 0.173 118 126 489 6;
  • 22) 0.173 118 126 489 6 × 2 = 0 + 0.346 236 252 979 2;
  • 23) 0.346 236 252 979 2 × 2 = 0 + 0.692 472 505 958 4;
  • 24) 0.692 472 505 958 4 × 2 = 1 + 0.384 945 011 916 8;
  • 25) 0.384 945 011 916 8 × 2 = 0 + 0.769 890 023 833 6;
  • 26) 0.769 890 023 833 6 × 2 = 1 + 0.539 780 047 667 2;
  • 27) 0.539 780 047 667 2 × 2 = 1 + 0.079 560 095 334 4;
  • 28) 0.079 560 095 334 4 × 2 = 0 + 0.159 120 190 668 8;
  • 29) 0.159 120 190 668 8 × 2 = 0 + 0.318 240 381 337 6;
  • 30) 0.318 240 381 337 6 × 2 = 0 + 0.636 480 762 675 2;
  • 31) 0.636 480 762 675 2 × 2 = 1 + 0.272 961 525 350 4;
  • 32) 0.272 961 525 350 4 × 2 = 0 + 0.545 923 050 700 8;
  • 33) 0.545 923 050 700 8 × 2 = 1 + 0.091 846 101 401 6;
  • 34) 0.091 846 101 401 6 × 2 = 0 + 0.183 692 202 803 2;
  • 35) 0.183 692 202 803 2 × 2 = 0 + 0.367 384 405 606 4;
  • 36) 0.367 384 405 606 4 × 2 = 0 + 0.734 768 811 212 8;
  • 37) 0.734 768 811 212 8 × 2 = 1 + 0.469 537 622 425 6;
  • 38) 0.469 537 622 425 6 × 2 = 0 + 0.939 075 244 851 2;
  • 39) 0.939 075 244 851 2 × 2 = 1 + 0.878 150 489 702 4;
  • 40) 0.878 150 489 702 4 × 2 = 1 + 0.756 300 979 404 8;
  • 41) 0.756 300 979 404 8 × 2 = 1 + 0.512 601 958 809 6;
  • 42) 0.512 601 958 809 6 × 2 = 1 + 0.025 203 917 619 2;
  • 43) 0.025 203 917 619 2 × 2 = 0 + 0.050 407 835 238 4;
  • 44) 0.050 407 835 238 4 × 2 = 0 + 0.100 815 670 476 8;
  • 45) 0.100 815 670 476 8 × 2 = 0 + 0.201 631 340 953 6;
  • 46) 0.201 631 340 953 6 × 2 = 0 + 0.403 262 681 907 2;
  • 47) 0.403 262 681 907 2 × 2 = 0 + 0.806 525 363 814 4;
  • 48) 0.806 525 363 814 4 × 2 = 1 + 0.613 050 727 628 8;
  • 49) 0.613 050 727 628 8 × 2 = 1 + 0.226 101 455 257 6;
  • 50) 0.226 101 455 257 6 × 2 = 0 + 0.452 202 910 515 2;
  • 51) 0.452 202 910 515 2 × 2 = 0 + 0.904 405 821 030 4;
  • 52) 0.904 405 821 030 4 × 2 = 1 + 0.808 811 642 060 8;
  • 53) 0.808 811 642 060 8 × 2 = 1 + 0.617 623 284 121 6;

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:


0.718 281 828 459 8(10) =

0.1011 0111 1110 0001 0101 0001 0110 0010 1000 1011 1100 0001 1001 1(2)

5. Positive number before normalization:

2.718 281 828 459 8(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1011 1100 0001 1001 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:


2.718 281 828 459 8(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1011 1100 0001 1001 1(2) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1011 1100 0001 1001 1(2) × 20 =

1.0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 1110 0000 1100 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 0101 1110 0000 1100 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 0101 1110 0000 1100 11 =

0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 1110 0000 1100


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 0101 1110 0000 1100


Decimal number 2.718 281 828 459 8 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 0101 1110 0000 1100

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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