2.333 333 333 334 87 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.333 333 333 334 87(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
2.333 333 333 334 87(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: 0.333 333 333 334 87.

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) 0.333 333 333 334 87 × 2 = 0 + 0.666 666 666 669 74;
  • 2) 0.666 666 666 669 74 × 2 = 1 + 0.333 333 333 339 48;
  • 3) 0.333 333 333 339 48 × 2 = 0 + 0.666 666 666 678 96;
  • 4) 0.666 666 666 678 96 × 2 = 1 + 0.333 333 333 357 92;
  • 5) 0.333 333 333 357 92 × 2 = 0 + 0.666 666 666 715 84;
  • 6) 0.666 666 666 715 84 × 2 = 1 + 0.333 333 333 431 68;
  • 7) 0.333 333 333 431 68 × 2 = 0 + 0.666 666 666 863 36;
  • 8) 0.666 666 666 863 36 × 2 = 1 + 0.333 333 333 726 72;
  • 9) 0.333 333 333 726 72 × 2 = 0 + 0.666 666 667 453 44;
  • 10) 0.666 666 667 453 44 × 2 = 1 + 0.333 333 334 906 88;
  • 11) 0.333 333 334 906 88 × 2 = 0 + 0.666 666 669 813 76;
  • 12) 0.666 666 669 813 76 × 2 = 1 + 0.333 333 339 627 52;
  • 13) 0.333 333 339 627 52 × 2 = 0 + 0.666 666 679 255 04;
  • 14) 0.666 666 679 255 04 × 2 = 1 + 0.333 333 358 510 08;
  • 15) 0.333 333 358 510 08 × 2 = 0 + 0.666 666 717 020 16;
  • 16) 0.666 666 717 020 16 × 2 = 1 + 0.333 333 434 040 32;
  • 17) 0.333 333 434 040 32 × 2 = 0 + 0.666 666 868 080 64;
  • 18) 0.666 666 868 080 64 × 2 = 1 + 0.333 333 736 161 28;
  • 19) 0.333 333 736 161 28 × 2 = 0 + 0.666 667 472 322 56;
  • 20) 0.666 667 472 322 56 × 2 = 1 + 0.333 334 944 645 12;
  • 21) 0.333 334 944 645 12 × 2 = 0 + 0.666 669 889 290 24;
  • 22) 0.666 669 889 290 24 × 2 = 1 + 0.333 339 778 580 48;
  • 23) 0.333 339 778 580 48 × 2 = 0 + 0.666 679 557 160 96;
  • 24) 0.666 679 557 160 96 × 2 = 1 + 0.333 359 114 321 92;
  • 25) 0.333 359 114 321 92 × 2 = 0 + 0.666 718 228 643 84;
  • 26) 0.666 718 228 643 84 × 2 = 1 + 0.333 436 457 287 68;
  • 27) 0.333 436 457 287 68 × 2 = 0 + 0.666 872 914 575 36;
  • 28) 0.666 872 914 575 36 × 2 = 1 + 0.333 745 829 150 72;
  • 29) 0.333 745 829 150 72 × 2 = 0 + 0.667 491 658 301 44;
  • 30) 0.667 491 658 301 44 × 2 = 1 + 0.334 983 316 602 88;
  • 31) 0.334 983 316 602 88 × 2 = 0 + 0.669 966 633 205 76;
  • 32) 0.669 966 633 205 76 × 2 = 1 + 0.339 933 266 411 52;
  • 33) 0.339 933 266 411 52 × 2 = 0 + 0.679 866 532 823 04;
  • 34) 0.679 866 532 823 04 × 2 = 1 + 0.359 733 065 646 08;
  • 35) 0.359 733 065 646 08 × 2 = 0 + 0.719 466 131 292 16;
  • 36) 0.719 466 131 292 16 × 2 = 1 + 0.438 932 262 584 32;
  • 37) 0.438 932 262 584 32 × 2 = 0 + 0.877 864 525 168 64;
  • 38) 0.877 864 525 168 64 × 2 = 1 + 0.755 729 050 337 28;
  • 39) 0.755 729 050 337 28 × 2 = 1 + 0.511 458 100 674 56;
  • 40) 0.511 458 100 674 56 × 2 = 1 + 0.022 916 201 349 12;
  • 41) 0.022 916 201 349 12 × 2 = 0 + 0.045 832 402 698 24;
  • 42) 0.045 832 402 698 24 × 2 = 0 + 0.091 664 805 396 48;
  • 43) 0.091 664 805 396 48 × 2 = 0 + 0.183 329 610 792 96;
  • 44) 0.183 329 610 792 96 × 2 = 0 + 0.366 659 221 585 92;
  • 45) 0.366 659 221 585 92 × 2 = 0 + 0.733 318 443 171 84;
  • 46) 0.733 318 443 171 84 × 2 = 1 + 0.466 636 886 343 68;
  • 47) 0.466 636 886 343 68 × 2 = 0 + 0.933 273 772 687 36;
  • 48) 0.933 273 772 687 36 × 2 = 1 + 0.866 547 545 374 72;
  • 49) 0.866 547 545 374 72 × 2 = 1 + 0.733 095 090 749 44;
  • 50) 0.733 095 090 749 44 × 2 = 1 + 0.466 190 181 498 88;
  • 51) 0.466 190 181 498 88 × 2 = 0 + 0.932 380 362 997 76;
  • 52) 0.932 380 362 997 76 × 2 = 1 + 0.864 760 725 995 52;
  • 53) 0.864 760 725 995 52 × 2 = 1 + 0.729 521 451 991 04;

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:


0.333 333 333 334 87(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0000 0101 1101 1(2)

5. Positive number before normalization:

2.333 333 333 334 87(10) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0000 0101 1101 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:


2.333 333 333 334 87(10) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0000 0101 1101 1(2) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0000 0101 1101 1(2) × 20 =

1.0010 1010 1010 1010 1010 1010 1010 1010 1010 1011 1000 0010 1110 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):
1.0010 1010 1010 1010 1010 1010 1010 1010 1010 1011 1000 0010 1110 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. 0010 1010 1010 1010 1010 1010 1010 1010 1010 1011 1000 0010 1110 11 =

0010 1010 1010 1010 1010 1010 1010 1010 1010 1011 1000 0010 1110


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) =
0010 1010 1010 1010 1010 1010 1010 1010 1010 1011 1000 0010 1110


Decimal number 2.333 333 333 334 87 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0010 1010 1010 1010 1010 1010 1010 1010 1010 1011 1000 0010 1110

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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