2.666 666 666 666 666 669 17 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.666 666 666 666 666 669 17(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
2.666 666 666 666 666 669 17(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: 0.666 666 666 666 666 669 17.

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) 0.666 666 666 666 666 669 17 × 2 = 1 + 0.333 333 333 333 333 338 34;
  • 2) 0.333 333 333 333 333 338 34 × 2 = 0 + 0.666 666 666 666 666 676 68;
  • 3) 0.666 666 666 666 666 676 68 × 2 = 1 + 0.333 333 333 333 333 353 36;
  • 4) 0.333 333 333 333 333 353 36 × 2 = 0 + 0.666 666 666 666 666 706 72;
  • 5) 0.666 666 666 666 666 706 72 × 2 = 1 + 0.333 333 333 333 333 413 44;
  • 6) 0.333 333 333 333 333 413 44 × 2 = 0 + 0.666 666 666 666 666 826 88;
  • 7) 0.666 666 666 666 666 826 88 × 2 = 1 + 0.333 333 333 333 333 653 76;
  • 8) 0.333 333 333 333 333 653 76 × 2 = 0 + 0.666 666 666 666 667 307 52;
  • 9) 0.666 666 666 666 667 307 52 × 2 = 1 + 0.333 333 333 333 334 615 04;
  • 10) 0.333 333 333 333 334 615 04 × 2 = 0 + 0.666 666 666 666 669 230 08;
  • 11) 0.666 666 666 666 669 230 08 × 2 = 1 + 0.333 333 333 333 338 460 16;
  • 12) 0.333 333 333 333 338 460 16 × 2 = 0 + 0.666 666 666 666 676 920 32;
  • 13) 0.666 666 666 666 676 920 32 × 2 = 1 + 0.333 333 333 333 353 840 64;
  • 14) 0.333 333 333 333 353 840 64 × 2 = 0 + 0.666 666 666 666 707 681 28;
  • 15) 0.666 666 666 666 707 681 28 × 2 = 1 + 0.333 333 333 333 415 362 56;
  • 16) 0.333 333 333 333 415 362 56 × 2 = 0 + 0.666 666 666 666 830 725 12;
  • 17) 0.666 666 666 666 830 725 12 × 2 = 1 + 0.333 333 333 333 661 450 24;
  • 18) 0.333 333 333 333 661 450 24 × 2 = 0 + 0.666 666 666 667 322 900 48;
  • 19) 0.666 666 666 667 322 900 48 × 2 = 1 + 0.333 333 333 334 645 800 96;
  • 20) 0.333 333 333 334 645 800 96 × 2 = 0 + 0.666 666 666 669 291 601 92;
  • 21) 0.666 666 666 669 291 601 92 × 2 = 1 + 0.333 333 333 338 583 203 84;
  • 22) 0.333 333 333 338 583 203 84 × 2 = 0 + 0.666 666 666 677 166 407 68;
  • 23) 0.666 666 666 677 166 407 68 × 2 = 1 + 0.333 333 333 354 332 815 36;
  • 24) 0.333 333 333 354 332 815 36 × 2 = 0 + 0.666 666 666 708 665 630 72;
  • 25) 0.666 666 666 708 665 630 72 × 2 = 1 + 0.333 333 333 417 331 261 44;
  • 26) 0.333 333 333 417 331 261 44 × 2 = 0 + 0.666 666 666 834 662 522 88;
  • 27) 0.666 666 666 834 662 522 88 × 2 = 1 + 0.333 333 333 669 325 045 76;
  • 28) 0.333 333 333 669 325 045 76 × 2 = 0 + 0.666 666 667 338 650 091 52;
  • 29) 0.666 666 667 338 650 091 52 × 2 = 1 + 0.333 333 334 677 300 183 04;
  • 30) 0.333 333 334 677 300 183 04 × 2 = 0 + 0.666 666 669 354 600 366 08;
  • 31) 0.666 666 669 354 600 366 08 × 2 = 1 + 0.333 333 338 709 200 732 16;
  • 32) 0.333 333 338 709 200 732 16 × 2 = 0 + 0.666 666 677 418 401 464 32;
  • 33) 0.666 666 677 418 401 464 32 × 2 = 1 + 0.333 333 354 836 802 928 64;
  • 34) 0.333 333 354 836 802 928 64 × 2 = 0 + 0.666 666 709 673 605 857 28;
  • 35) 0.666 666 709 673 605 857 28 × 2 = 1 + 0.333 333 419 347 211 714 56;
  • 36) 0.333 333 419 347 211 714 56 × 2 = 0 + 0.666 666 838 694 423 429 12;
  • 37) 0.666 666 838 694 423 429 12 × 2 = 1 + 0.333 333 677 388 846 858 24;
  • 38) 0.333 333 677 388 846 858 24 × 2 = 0 + 0.666 667 354 777 693 716 48;
  • 39) 0.666 667 354 777 693 716 48 × 2 = 1 + 0.333 334 709 555 387 432 96;
  • 40) 0.333 334 709 555 387 432 96 × 2 = 0 + 0.666 669 419 110 774 865 92;
  • 41) 0.666 669 419 110 774 865 92 × 2 = 1 + 0.333 338 838 221 549 731 84;
  • 42) 0.333 338 838 221 549 731 84 × 2 = 0 + 0.666 677 676 443 099 463 68;
  • 43) 0.666 677 676 443 099 463 68 × 2 = 1 + 0.333 355 352 886 198 927 36;
  • 44) 0.333 355 352 886 198 927 36 × 2 = 0 + 0.666 710 705 772 397 854 72;
  • 45) 0.666 710 705 772 397 854 72 × 2 = 1 + 0.333 421 411 544 795 709 44;
  • 46) 0.333 421 411 544 795 709 44 × 2 = 0 + 0.666 842 823 089 591 418 88;
  • 47) 0.666 842 823 089 591 418 88 × 2 = 1 + 0.333 685 646 179 182 837 76;
  • 48) 0.333 685 646 179 182 837 76 × 2 = 0 + 0.667 371 292 358 365 675 52;
  • 49) 0.667 371 292 358 365 675 52 × 2 = 1 + 0.334 742 584 716 731 351 04;
  • 50) 0.334 742 584 716 731 351 04 × 2 = 0 + 0.669 485 169 433 462 702 08;
  • 51) 0.669 485 169 433 462 702 08 × 2 = 1 + 0.338 970 338 866 925 404 16;
  • 52) 0.338 970 338 866 925 404 16 × 2 = 0 + 0.677 940 677 733 850 808 32;
  • 53) 0.677 940 677 733 850 808 32 × 2 = 1 + 0.355 881 355 467 701 616 64;

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:


0.666 666 666 666 666 669 17(10) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1(2)

5. Positive number before normalization:

2.666 666 666 666 666 669 17(10) =

10.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 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:


2.666 666 666 666 666 669 17(10) =

10.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1(2) =

10.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1(2) × 20 =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 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 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 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 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01 =

0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


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 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


Decimal number 2.666 666 666 666 666 669 17 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101

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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