2.718 281 834 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.718 281 834 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
2.718 281 834 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: 0.718 281 834 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) 0.718 281 834 6 × 2 = 1 + 0.436 563 669 2;
  • 2) 0.436 563 669 2 × 2 = 0 + 0.873 127 338 4;
  • 3) 0.873 127 338 4 × 2 = 1 + 0.746 254 676 8;
  • 4) 0.746 254 676 8 × 2 = 1 + 0.492 509 353 6;
  • 5) 0.492 509 353 6 × 2 = 0 + 0.985 018 707 2;
  • 6) 0.985 018 707 2 × 2 = 1 + 0.970 037 414 4;
  • 7) 0.970 037 414 4 × 2 = 1 + 0.940 074 828 8;
  • 8) 0.940 074 828 8 × 2 = 1 + 0.880 149 657 6;
  • 9) 0.880 149 657 6 × 2 = 1 + 0.760 299 315 2;
  • 10) 0.760 299 315 2 × 2 = 1 + 0.520 598 630 4;
  • 11) 0.520 598 630 4 × 2 = 1 + 0.041 197 260 8;
  • 12) 0.041 197 260 8 × 2 = 0 + 0.082 394 521 6;
  • 13) 0.082 394 521 6 × 2 = 0 + 0.164 789 043 2;
  • 14) 0.164 789 043 2 × 2 = 0 + 0.329 578 086 4;
  • 15) 0.329 578 086 4 × 2 = 0 + 0.659 156 172 8;
  • 16) 0.659 156 172 8 × 2 = 1 + 0.318 312 345 6;
  • 17) 0.318 312 345 6 × 2 = 0 + 0.636 624 691 2;
  • 18) 0.636 624 691 2 × 2 = 1 + 0.273 249 382 4;
  • 19) 0.273 249 382 4 × 2 = 0 + 0.546 498 764 8;
  • 20) 0.546 498 764 8 × 2 = 1 + 0.092 997 529 6;
  • 21) 0.092 997 529 6 × 2 = 0 + 0.185 995 059 2;
  • 22) 0.185 995 059 2 × 2 = 0 + 0.371 990 118 4;
  • 23) 0.371 990 118 4 × 2 = 0 + 0.743 980 236 8;
  • 24) 0.743 980 236 8 × 2 = 1 + 0.487 960 473 6;
  • 25) 0.487 960 473 6 × 2 = 0 + 0.975 920 947 2;
  • 26) 0.975 920 947 2 × 2 = 1 + 0.951 841 894 4;
  • 27) 0.951 841 894 4 × 2 = 1 + 0.903 683 788 8;
  • 28) 0.903 683 788 8 × 2 = 1 + 0.807 367 577 6;
  • 29) 0.807 367 577 6 × 2 = 1 + 0.614 735 155 2;
  • 30) 0.614 735 155 2 × 2 = 1 + 0.229 470 310 4;
  • 31) 0.229 470 310 4 × 2 = 0 + 0.458 940 620 8;
  • 32) 0.458 940 620 8 × 2 = 0 + 0.917 881 241 6;
  • 33) 0.917 881 241 6 × 2 = 1 + 0.835 762 483 2;
  • 34) 0.835 762 483 2 × 2 = 1 + 0.671 524 966 4;
  • 35) 0.671 524 966 4 × 2 = 1 + 0.343 049 932 8;
  • 36) 0.343 049 932 8 × 2 = 0 + 0.686 099 865 6;
  • 37) 0.686 099 865 6 × 2 = 1 + 0.372 199 731 2;
  • 38) 0.372 199 731 2 × 2 = 0 + 0.744 399 462 4;
  • 39) 0.744 399 462 4 × 2 = 1 + 0.488 798 924 8;
  • 40) 0.488 798 924 8 × 2 = 0 + 0.977 597 849 6;
  • 41) 0.977 597 849 6 × 2 = 1 + 0.955 195 699 2;
  • 42) 0.955 195 699 2 × 2 = 1 + 0.910 391 398 4;
  • 43) 0.910 391 398 4 × 2 = 1 + 0.820 782 796 8;
  • 44) 0.820 782 796 8 × 2 = 1 + 0.641 565 593 6;
  • 45) 0.641 565 593 6 × 2 = 1 + 0.283 131 187 2;
  • 46) 0.283 131 187 2 × 2 = 0 + 0.566 262 374 4;
  • 47) 0.566 262 374 4 × 2 = 1 + 0.132 524 748 8;
  • 48) 0.132 524 748 8 × 2 = 0 + 0.265 049 497 6;
  • 49) 0.265 049 497 6 × 2 = 0 + 0.530 098 995 2;
  • 50) 0.530 098 995 2 × 2 = 1 + 0.060 197 990 4;
  • 51) 0.060 197 990 4 × 2 = 0 + 0.120 395 980 8;
  • 52) 0.120 395 980 8 × 2 = 0 + 0.240 791 961 6;
  • 53) 0.240 791 961 6 × 2 = 0 + 0.481 583 923 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:


0.718 281 834 6(10) =

0.1011 0111 1110 0001 0101 0001 0111 1100 1110 1010 1111 1010 0100 0(2)

5. Positive number before normalization:

2.718 281 834 6(10) =

10.1011 0111 1110 0001 0101 0001 0111 1100 1110 1010 1111 1010 0100 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:


2.718 281 834 6(10) =

10.1011 0111 1110 0001 0101 0001 0111 1100 1110 1010 1111 1010 0100 0(2) =

10.1011 0111 1110 0001 0101 0001 0111 1100 1110 1010 1111 1010 0100 0(2) × 20 =

1.0101 1011 1111 0000 1010 1000 1011 1110 0111 0101 0111 1101 0010 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 1110 0111 0101 0111 1101 0010 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 1110 0111 0101 0111 1101 0010 00 =

0101 1011 1111 0000 1010 1000 1011 1110 0111 0101 0111 1101 0010


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 1110 0111 0101 0111 1101 0010


Decimal number 2.718 281 834 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 1011 1111 0000 1010 1000 1011 1110 0111 0101 0111 1101 0010

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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