2.718 281 828 459 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.718 281 828 459 1(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 1(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 1.

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 1 × 2 = 1 + 0.436 563 656 918 2;
  • 2) 0.436 563 656 918 2 × 2 = 0 + 0.873 127 313 836 4;
  • 3) 0.873 127 313 836 4 × 2 = 1 + 0.746 254 627 672 8;
  • 4) 0.746 254 627 672 8 × 2 = 1 + 0.492 509 255 345 6;
  • 5) 0.492 509 255 345 6 × 2 = 0 + 0.985 018 510 691 2;
  • 6) 0.985 018 510 691 2 × 2 = 1 + 0.970 037 021 382 4;
  • 7) 0.970 037 021 382 4 × 2 = 1 + 0.940 074 042 764 8;
  • 8) 0.940 074 042 764 8 × 2 = 1 + 0.880 148 085 529 6;
  • 9) 0.880 148 085 529 6 × 2 = 1 + 0.760 296 171 059 2;
  • 10) 0.760 296 171 059 2 × 2 = 1 + 0.520 592 342 118 4;
  • 11) 0.520 592 342 118 4 × 2 = 1 + 0.041 184 684 236 8;
  • 12) 0.041 184 684 236 8 × 2 = 0 + 0.082 369 368 473 6;
  • 13) 0.082 369 368 473 6 × 2 = 0 + 0.164 738 736 947 2;
  • 14) 0.164 738 736 947 2 × 2 = 0 + 0.329 477 473 894 4;
  • 15) 0.329 477 473 894 4 × 2 = 0 + 0.658 954 947 788 8;
  • 16) 0.658 954 947 788 8 × 2 = 1 + 0.317 909 895 577 6;
  • 17) 0.317 909 895 577 6 × 2 = 0 + 0.635 819 791 155 2;
  • 18) 0.635 819 791 155 2 × 2 = 1 + 0.271 639 582 310 4;
  • 19) 0.271 639 582 310 4 × 2 = 0 + 0.543 279 164 620 8;
  • 20) 0.543 279 164 620 8 × 2 = 1 + 0.086 558 329 241 6;
  • 21) 0.086 558 329 241 6 × 2 = 0 + 0.173 116 658 483 2;
  • 22) 0.173 116 658 483 2 × 2 = 0 + 0.346 233 316 966 4;
  • 23) 0.346 233 316 966 4 × 2 = 0 + 0.692 466 633 932 8;
  • 24) 0.692 466 633 932 8 × 2 = 1 + 0.384 933 267 865 6;
  • 25) 0.384 933 267 865 6 × 2 = 0 + 0.769 866 535 731 2;
  • 26) 0.769 866 535 731 2 × 2 = 1 + 0.539 733 071 462 4;
  • 27) 0.539 733 071 462 4 × 2 = 1 + 0.079 466 142 924 8;
  • 28) 0.079 466 142 924 8 × 2 = 0 + 0.158 932 285 849 6;
  • 29) 0.158 932 285 849 6 × 2 = 0 + 0.317 864 571 699 2;
  • 30) 0.317 864 571 699 2 × 2 = 0 + 0.635 729 143 398 4;
  • 31) 0.635 729 143 398 4 × 2 = 1 + 0.271 458 286 796 8;
  • 32) 0.271 458 286 796 8 × 2 = 0 + 0.542 916 573 593 6;
  • 33) 0.542 916 573 593 6 × 2 = 1 + 0.085 833 147 187 2;
  • 34) 0.085 833 147 187 2 × 2 = 0 + 0.171 666 294 374 4;
  • 35) 0.171 666 294 374 4 × 2 = 0 + 0.343 332 588 748 8;
  • 36) 0.343 332 588 748 8 × 2 = 0 + 0.686 665 177 497 6;
  • 37) 0.686 665 177 497 6 × 2 = 1 + 0.373 330 354 995 2;
  • 38) 0.373 330 354 995 2 × 2 = 0 + 0.746 660 709 990 4;
  • 39) 0.746 660 709 990 4 × 2 = 1 + 0.493 321 419 980 8;
  • 40) 0.493 321 419 980 8 × 2 = 0 + 0.986 642 839 961 6;
  • 41) 0.986 642 839 961 6 × 2 = 1 + 0.973 285 679 923 2;
  • 42) 0.973 285 679 923 2 × 2 = 1 + 0.946 571 359 846 4;
  • 43) 0.946 571 359 846 4 × 2 = 1 + 0.893 142 719 692 8;
  • 44) 0.893 142 719 692 8 × 2 = 1 + 0.786 285 439 385 6;
  • 45) 0.786 285 439 385 6 × 2 = 1 + 0.572 570 878 771 2;
  • 46) 0.572 570 878 771 2 × 2 = 1 + 0.145 141 757 542 4;
  • 47) 0.145 141 757 542 4 × 2 = 0 + 0.290 283 515 084 8;
  • 48) 0.290 283 515 084 8 × 2 = 0 + 0.580 567 030 169 6;
  • 49) 0.580 567 030 169 6 × 2 = 1 + 0.161 134 060 339 2;
  • 50) 0.161 134 060 339 2 × 2 = 0 + 0.322 268 120 678 4;
  • 51) 0.322 268 120 678 4 × 2 = 0 + 0.644 536 241 356 8;
  • 52) 0.644 536 241 356 8 × 2 = 1 + 0.289 072 482 713 6;
  • 53) 0.289 072 482 713 6 × 2 = 0 + 0.578 144 965 427 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:


0.718 281 828 459 1(10) =

0.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1111 1100 1001 0(2)

5. Positive number before normalization:

2.718 281 828 459 1(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1111 1100 1001 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:


2.718 281 828 459 1(10) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1111 1100 1001 0(2) =

10.1011 0111 1110 0001 0101 0001 0110 0010 1000 1010 1111 1100 1001 0(2) × 20 =

1.0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0111 1110 0100 10(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 0111 1110 0100 10


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 0111 1110 0100 10 =

0101 1011 1111 0000 1010 1000 1011 0001 0100 0101 0111 1110 0100


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 0111 1110 0100


Decimal number 2.718 281 828 459 1 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 0111 1110 0100

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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