2.718 281 828 456 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.718 281 828 456 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
2.718 281 828 456 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: 0.718 281 828 456 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) 0.718 281 828 456 7 × 2 = 1 + 0.436 563 656 913 4;
  • 2) 0.436 563 656 913 4 × 2 = 0 + 0.873 127 313 826 8;
  • 3) 0.873 127 313 826 8 × 2 = 1 + 0.746 254 627 653 6;
  • 4) 0.746 254 627 653 6 × 2 = 1 + 0.492 509 255 307 2;
  • 5) 0.492 509 255 307 2 × 2 = 0 + 0.985 018 510 614 4;
  • 6) 0.985 018 510 614 4 × 2 = 1 + 0.970 037 021 228 8;
  • 7) 0.970 037 021 228 8 × 2 = 1 + 0.940 074 042 457 6;
  • 8) 0.940 074 042 457 6 × 2 = 1 + 0.880 148 084 915 2;
  • 9) 0.880 148 084 915 2 × 2 = 1 + 0.760 296 169 830 4;
  • 10) 0.760 296 169 830 4 × 2 = 1 + 0.520 592 339 660 8;
  • 11) 0.520 592 339 660 8 × 2 = 1 + 0.041 184 679 321 6;
  • 12) 0.041 184 679 321 6 × 2 = 0 + 0.082 369 358 643 2;
  • 13) 0.082 369 358 643 2 × 2 = 0 + 0.164 738 717 286 4;
  • 14) 0.164 738 717 286 4 × 2 = 0 + 0.329 477 434 572 8;
  • 15) 0.329 477 434 572 8 × 2 = 0 + 0.658 954 869 145 6;
  • 16) 0.658 954 869 145 6 × 2 = 1 + 0.317 909 738 291 2;
  • 17) 0.317 909 738 291 2 × 2 = 0 + 0.635 819 476 582 4;
  • 18) 0.635 819 476 582 4 × 2 = 1 + 0.271 638 953 164 8;
  • 19) 0.271 638 953 164 8 × 2 = 0 + 0.543 277 906 329 6;
  • 20) 0.543 277 906 329 6 × 2 = 1 + 0.086 555 812 659 2;
  • 21) 0.086 555 812 659 2 × 2 = 0 + 0.173 111 625 318 4;
  • 22) 0.173 111 625 318 4 × 2 = 0 + 0.346 223 250 636 8;
  • 23) 0.346 223 250 636 8 × 2 = 0 + 0.692 446 501 273 6;
  • 24) 0.692 446 501 273 6 × 2 = 1 + 0.384 893 002 547 2;
  • 25) 0.384 893 002 547 2 × 2 = 0 + 0.769 786 005 094 4;
  • 26) 0.769 786 005 094 4 × 2 = 1 + 0.539 572 010 188 8;
  • 27) 0.539 572 010 188 8 × 2 = 1 + 0.079 144 020 377 6;
  • 28) 0.079 144 020 377 6 × 2 = 0 + 0.158 288 040 755 2;
  • 29) 0.158 288 040 755 2 × 2 = 0 + 0.316 576 081 510 4;
  • 30) 0.316 576 081 510 4 × 2 = 0 + 0.633 152 163 020 8;
  • 31) 0.633 152 163 020 8 × 2 = 1 + 0.266 304 326 041 6;
  • 32) 0.266 304 326 041 6 × 2 = 0 + 0.532 608 652 083 2;
  • 33) 0.532 608 652 083 2 × 2 = 1 + 0.065 217 304 166 4;
  • 34) 0.065 217 304 166 4 × 2 = 0 + 0.130 434 608 332 8;
  • 35) 0.130 434 608 332 8 × 2 = 0 + 0.260 869 216 665 6;
  • 36) 0.260 869 216 665 6 × 2 = 0 + 0.521 738 433 331 2;
  • 37) 0.521 738 433 331 2 × 2 = 1 + 0.043 476 866 662 4;
  • 38) 0.043 476 866 662 4 × 2 = 0 + 0.086 953 733 324 8;
  • 39) 0.086 953 733 324 8 × 2 = 0 + 0.173 907 466 649 6;
  • 40) 0.173 907 466 649 6 × 2 = 0 + 0.347 814 933 299 2;
  • 41) 0.347 814 933 299 2 × 2 = 0 + 0.695 629 866 598 4;
  • 42) 0.695 629 866 598 4 × 2 = 1 + 0.391 259 733 196 8;
  • 43) 0.391 259 733 196 8 × 2 = 0 + 0.782 519 466 393 6;
  • 44) 0.782 519 466 393 6 × 2 = 1 + 0.565 038 932 787 2;
  • 45) 0.565 038 932 787 2 × 2 = 1 + 0.130 077 865 574 4;
  • 46) 0.130 077 865 574 4 × 2 = 0 + 0.260 155 731 148 8;
  • 47) 0.260 155 731 148 8 × 2 = 0 + 0.520 311 462 297 6;
  • 48) 0.520 311 462 297 6 × 2 = 1 + 0.040 622 924 595 2;
  • 49) 0.040 622 924 595 2 × 2 = 0 + 0.081 245 849 190 4;
  • 50) 0.081 245 849 190 4 × 2 = 0 + 0.162 491 698 380 8;
  • 51) 0.162 491 698 380 8 × 2 = 0 + 0.324 983 396 761 6;
  • 52) 0.324 983 396 761 6 × 2 = 0 + 0.649 966 793 523 2;
  • 53) 0.649 966 793 523 2 × 2 = 1 + 0.299 933 587 046 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:


0.718 281 828 456 7(10) =

0.1011 0111 1110 0001 0101 0001 0110 0010 1000 1000 0101 1001 0000 1(2)

5. Positive number before normalization:

2.718 281 828 456 7(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1000 0101 1001 0000 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:


2.718 281 828 456 7(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1000 0101 1001 0000 1(2) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1000 0101 1001 0000 1(2) × 20 =

1.0101 1011 1111 0000 1010 1000 1011 0001 0100 0100 0010 1100 1000 01(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 0100 0010 1100 1000 01


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 0100 0010 1100 1000 01 =

0101 1011 1111 0000 1010 1000 1011 0001 0100 0100 0010 1100 1000


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 0100 0010 1100 1000


Decimal number 2.718 281 828 456 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 0100 0010 1100 1000

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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