2.718 281 828 472 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.718 281 828 472 2(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
2.718 281 828 472 2(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: 0.718 281 828 472 2.

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) 0.718 281 828 472 2 × 2 = 1 + 0.436 563 656 944 4;
  • 2) 0.436 563 656 944 4 × 2 = 0 + 0.873 127 313 888 8;
  • 3) 0.873 127 313 888 8 × 2 = 1 + 0.746 254 627 777 6;
  • 4) 0.746 254 627 777 6 × 2 = 1 + 0.492 509 255 555 2;
  • 5) 0.492 509 255 555 2 × 2 = 0 + 0.985 018 511 110 4;
  • 6) 0.985 018 511 110 4 × 2 = 1 + 0.970 037 022 220 8;
  • 7) 0.970 037 022 220 8 × 2 = 1 + 0.940 074 044 441 6;
  • 8) 0.940 074 044 441 6 × 2 = 1 + 0.880 148 088 883 2;
  • 9) 0.880 148 088 883 2 × 2 = 1 + 0.760 296 177 766 4;
  • 10) 0.760 296 177 766 4 × 2 = 1 + 0.520 592 355 532 8;
  • 11) 0.520 592 355 532 8 × 2 = 1 + 0.041 184 711 065 6;
  • 12) 0.041 184 711 065 6 × 2 = 0 + 0.082 369 422 131 2;
  • 13) 0.082 369 422 131 2 × 2 = 0 + 0.164 738 844 262 4;
  • 14) 0.164 738 844 262 4 × 2 = 0 + 0.329 477 688 524 8;
  • 15) 0.329 477 688 524 8 × 2 = 0 + 0.658 955 377 049 6;
  • 16) 0.658 955 377 049 6 × 2 = 1 + 0.317 910 754 099 2;
  • 17) 0.317 910 754 099 2 × 2 = 0 + 0.635 821 508 198 4;
  • 18) 0.635 821 508 198 4 × 2 = 1 + 0.271 643 016 396 8;
  • 19) 0.271 643 016 396 8 × 2 = 0 + 0.543 286 032 793 6;
  • 20) 0.543 286 032 793 6 × 2 = 1 + 0.086 572 065 587 2;
  • 21) 0.086 572 065 587 2 × 2 = 0 + 0.173 144 131 174 4;
  • 22) 0.173 144 131 174 4 × 2 = 0 + 0.346 288 262 348 8;
  • 23) 0.346 288 262 348 8 × 2 = 0 + 0.692 576 524 697 6;
  • 24) 0.692 576 524 697 6 × 2 = 1 + 0.385 153 049 395 2;
  • 25) 0.385 153 049 395 2 × 2 = 0 + 0.770 306 098 790 4;
  • 26) 0.770 306 098 790 4 × 2 = 1 + 0.540 612 197 580 8;
  • 27) 0.540 612 197 580 8 × 2 = 1 + 0.081 224 395 161 6;
  • 28) 0.081 224 395 161 6 × 2 = 0 + 0.162 448 790 323 2;
  • 29) 0.162 448 790 323 2 × 2 = 0 + 0.324 897 580 646 4;
  • 30) 0.324 897 580 646 4 × 2 = 0 + 0.649 795 161 292 8;
  • 31) 0.649 795 161 292 8 × 2 = 1 + 0.299 590 322 585 6;
  • 32) 0.299 590 322 585 6 × 2 = 0 + 0.599 180 645 171 2;
  • 33) 0.599 180 645 171 2 × 2 = 1 + 0.198 361 290 342 4;
  • 34) 0.198 361 290 342 4 × 2 = 0 + 0.396 722 580 684 8;
  • 35) 0.396 722 580 684 8 × 2 = 0 + 0.793 445 161 369 6;
  • 36) 0.793 445 161 369 6 × 2 = 1 + 0.586 890 322 739 2;
  • 37) 0.586 890 322 739 2 × 2 = 1 + 0.173 780 645 478 4;
  • 38) 0.173 780 645 478 4 × 2 = 0 + 0.347 561 290 956 8;
  • 39) 0.347 561 290 956 8 × 2 = 0 + 0.695 122 581 913 6;
  • 40) 0.695 122 581 913 6 × 2 = 1 + 0.390 245 163 827 2;
  • 41) 0.390 245 163 827 2 × 2 = 0 + 0.780 490 327 654 4;
  • 42) 0.780 490 327 654 4 × 2 = 1 + 0.560 980 655 308 8;
  • 43) 0.560 980 655 308 8 × 2 = 1 + 0.121 961 310 617 6;
  • 44) 0.121 961 310 617 6 × 2 = 0 + 0.243 922 621 235 2;
  • 45) 0.243 922 621 235 2 × 2 = 0 + 0.487 845 242 470 4;
  • 46) 0.487 845 242 470 4 × 2 = 0 + 0.975 690 484 940 8;
  • 47) 0.975 690 484 940 8 × 2 = 1 + 0.951 380 969 881 6;
  • 48) 0.951 380 969 881 6 × 2 = 1 + 0.902 761 939 763 2;
  • 49) 0.902 761 939 763 2 × 2 = 1 + 0.805 523 879 526 4;
  • 50) 0.805 523 879 526 4 × 2 = 1 + 0.611 047 759 052 8;
  • 51) 0.611 047 759 052 8 × 2 = 1 + 0.222 095 518 105 6;
  • 52) 0.222 095 518 105 6 × 2 = 0 + 0.444 191 036 211 2;
  • 53) 0.444 191 036 211 2 × 2 = 0 + 0.888 382 072 422 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:


0.718 281 828 472 2(10) =

0.1011 0111 1110 0001 0101 0001 0110 0010 1001 1001 0110 0011 1110 0(2)

5. Positive number before normalization:

2.718 281 828 472 2(10) =

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


2.718 281 828 472 2(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1001 1001 0110 0011 1110 0(2) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1001 1001 0110 0011 1110 0(2) × 20 =

1.0101 1011 1111 0000 1010 1000 1011 0001 0100 1100 1011 0001 1111 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 0100 1100 1011 0001 1111 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 0100 1100 1011 0001 1111 00 =

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


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 1100 1011 0001 1111


Decimal number 2.718 281 828 472 2 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 1100 1011 0001 1111

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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