3.133 333 333 333 333 328 22 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.133 333 333 333 333 328 22(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
3.133 333 333 333 333 328 22(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.133 333 333 333 333 328 22.

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) 0.133 333 333 333 333 328 22 × 2 = 0 + 0.266 666 666 666 666 656 44;
  • 2) 0.266 666 666 666 666 656 44 × 2 = 0 + 0.533 333 333 333 333 312 88;
  • 3) 0.533 333 333 333 333 312 88 × 2 = 1 + 0.066 666 666 666 666 625 76;
  • 4) 0.066 666 666 666 666 625 76 × 2 = 0 + 0.133 333 333 333 333 251 52;
  • 5) 0.133 333 333 333 333 251 52 × 2 = 0 + 0.266 666 666 666 666 503 04;
  • 6) 0.266 666 666 666 666 503 04 × 2 = 0 + 0.533 333 333 333 333 006 08;
  • 7) 0.533 333 333 333 333 006 08 × 2 = 1 + 0.066 666 666 666 666 012 16;
  • 8) 0.066 666 666 666 666 012 16 × 2 = 0 + 0.133 333 333 333 332 024 32;
  • 9) 0.133 333 333 333 332 024 32 × 2 = 0 + 0.266 666 666 666 664 048 64;
  • 10) 0.266 666 666 666 664 048 64 × 2 = 0 + 0.533 333 333 333 328 097 28;
  • 11) 0.533 333 333 333 328 097 28 × 2 = 1 + 0.066 666 666 666 656 194 56;
  • 12) 0.066 666 666 666 656 194 56 × 2 = 0 + 0.133 333 333 333 312 389 12;
  • 13) 0.133 333 333 333 312 389 12 × 2 = 0 + 0.266 666 666 666 624 778 24;
  • 14) 0.266 666 666 666 624 778 24 × 2 = 0 + 0.533 333 333 333 249 556 48;
  • 15) 0.533 333 333 333 249 556 48 × 2 = 1 + 0.066 666 666 666 499 112 96;
  • 16) 0.066 666 666 666 499 112 96 × 2 = 0 + 0.133 333 333 332 998 225 92;
  • 17) 0.133 333 333 332 998 225 92 × 2 = 0 + 0.266 666 666 665 996 451 84;
  • 18) 0.266 666 666 665 996 451 84 × 2 = 0 + 0.533 333 333 331 992 903 68;
  • 19) 0.533 333 333 331 992 903 68 × 2 = 1 + 0.066 666 666 663 985 807 36;
  • 20) 0.066 666 666 663 985 807 36 × 2 = 0 + 0.133 333 333 327 971 614 72;
  • 21) 0.133 333 333 327 971 614 72 × 2 = 0 + 0.266 666 666 655 943 229 44;
  • 22) 0.266 666 666 655 943 229 44 × 2 = 0 + 0.533 333 333 311 886 458 88;
  • 23) 0.533 333 333 311 886 458 88 × 2 = 1 + 0.066 666 666 623 772 917 76;
  • 24) 0.066 666 666 623 772 917 76 × 2 = 0 + 0.133 333 333 247 545 835 52;
  • 25) 0.133 333 333 247 545 835 52 × 2 = 0 + 0.266 666 666 495 091 671 04;
  • 26) 0.266 666 666 495 091 671 04 × 2 = 0 + 0.533 333 332 990 183 342 08;
  • 27) 0.533 333 332 990 183 342 08 × 2 = 1 + 0.066 666 665 980 366 684 16;
  • 28) 0.066 666 665 980 366 684 16 × 2 = 0 + 0.133 333 331 960 733 368 32;
  • 29) 0.133 333 331 960 733 368 32 × 2 = 0 + 0.266 666 663 921 466 736 64;
  • 30) 0.266 666 663 921 466 736 64 × 2 = 0 + 0.533 333 327 842 933 473 28;
  • 31) 0.533 333 327 842 933 473 28 × 2 = 1 + 0.066 666 655 685 866 946 56;
  • 32) 0.066 666 655 685 866 946 56 × 2 = 0 + 0.133 333 311 371 733 893 12;
  • 33) 0.133 333 311 371 733 893 12 × 2 = 0 + 0.266 666 622 743 467 786 24;
  • 34) 0.266 666 622 743 467 786 24 × 2 = 0 + 0.533 333 245 486 935 572 48;
  • 35) 0.533 333 245 486 935 572 48 × 2 = 1 + 0.066 666 490 973 871 144 96;
  • 36) 0.066 666 490 973 871 144 96 × 2 = 0 + 0.133 332 981 947 742 289 92;
  • 37) 0.133 332 981 947 742 289 92 × 2 = 0 + 0.266 665 963 895 484 579 84;
  • 38) 0.266 665 963 895 484 579 84 × 2 = 0 + 0.533 331 927 790 969 159 68;
  • 39) 0.533 331 927 790 969 159 68 × 2 = 1 + 0.066 663 855 581 938 319 36;
  • 40) 0.066 663 855 581 938 319 36 × 2 = 0 + 0.133 327 711 163 876 638 72;
  • 41) 0.133 327 711 163 876 638 72 × 2 = 0 + 0.266 655 422 327 753 277 44;
  • 42) 0.266 655 422 327 753 277 44 × 2 = 0 + 0.533 310 844 655 506 554 88;
  • 43) 0.533 310 844 655 506 554 88 × 2 = 1 + 0.066 621 689 311 013 109 76;
  • 44) 0.066 621 689 311 013 109 76 × 2 = 0 + 0.133 243 378 622 026 219 52;
  • 45) 0.133 243 378 622 026 219 52 × 2 = 0 + 0.266 486 757 244 052 439 04;
  • 46) 0.266 486 757 244 052 439 04 × 2 = 0 + 0.532 973 514 488 104 878 08;
  • 47) 0.532 973 514 488 104 878 08 × 2 = 1 + 0.065 947 028 976 209 756 16;
  • 48) 0.065 947 028 976 209 756 16 × 2 = 0 + 0.131 894 057 952 419 512 32;
  • 49) 0.131 894 057 952 419 512 32 × 2 = 0 + 0.263 788 115 904 839 024 64;
  • 50) 0.263 788 115 904 839 024 64 × 2 = 0 + 0.527 576 231 809 678 049 28;
  • 51) 0.527 576 231 809 678 049 28 × 2 = 1 + 0.055 152 463 619 356 098 56;
  • 52) 0.055 152 463 619 356 098 56 × 2 = 0 + 0.110 304 927 238 712 197 12;
  • 53) 0.110 304 927 238 712 197 12 × 2 = 0 + 0.220 609 854 477 424 394 24;

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:


0.133 333 333 333 333 328 22(10) =

0.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0(2)

5. Positive number before normalization:

3.133 333 333 333 333 328 22(10) =

11.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 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:


3.133 333 333 333 333 328 22(10) =

11.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0(2) =

11.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0(2) × 20 =

1.1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 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):
1.1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 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. 1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 00 =

1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001


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) =
1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001


Decimal number 3.133 333 333 333 333 328 22 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001

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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