1.333 333 333 333 333 259 318 465 025 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.333 333 333 333 333 259 318 465 025(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
1.333 333 333 333 333 259 318 465 025(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.333 333 333 333 333 259 318 465 025.

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) 0.333 333 333 333 333 259 318 465 025 × 2 = 0 + 0.666 666 666 666 666 518 636 930 05;
  • 2) 0.666 666 666 666 666 518 636 930 05 × 2 = 1 + 0.333 333 333 333 333 037 273 860 1;
  • 3) 0.333 333 333 333 333 037 273 860 1 × 2 = 0 + 0.666 666 666 666 666 074 547 720 2;
  • 4) 0.666 666 666 666 666 074 547 720 2 × 2 = 1 + 0.333 333 333 333 332 149 095 440 4;
  • 5) 0.333 333 333 333 332 149 095 440 4 × 2 = 0 + 0.666 666 666 666 664 298 190 880 8;
  • 6) 0.666 666 666 666 664 298 190 880 8 × 2 = 1 + 0.333 333 333 333 328 596 381 761 6;
  • 7) 0.333 333 333 333 328 596 381 761 6 × 2 = 0 + 0.666 666 666 666 657 192 763 523 2;
  • 8) 0.666 666 666 666 657 192 763 523 2 × 2 = 1 + 0.333 333 333 333 314 385 527 046 4;
  • 9) 0.333 333 333 333 314 385 527 046 4 × 2 = 0 + 0.666 666 666 666 628 771 054 092 8;
  • 10) 0.666 666 666 666 628 771 054 092 8 × 2 = 1 + 0.333 333 333 333 257 542 108 185 6;
  • 11) 0.333 333 333 333 257 542 108 185 6 × 2 = 0 + 0.666 666 666 666 515 084 216 371 2;
  • 12) 0.666 666 666 666 515 084 216 371 2 × 2 = 1 + 0.333 333 333 333 030 168 432 742 4;
  • 13) 0.333 333 333 333 030 168 432 742 4 × 2 = 0 + 0.666 666 666 666 060 336 865 484 8;
  • 14) 0.666 666 666 666 060 336 865 484 8 × 2 = 1 + 0.333 333 333 332 120 673 730 969 6;
  • 15) 0.333 333 333 332 120 673 730 969 6 × 2 = 0 + 0.666 666 666 664 241 347 461 939 2;
  • 16) 0.666 666 666 664 241 347 461 939 2 × 2 = 1 + 0.333 333 333 328 482 694 923 878 4;
  • 17) 0.333 333 333 328 482 694 923 878 4 × 2 = 0 + 0.666 666 666 656 965 389 847 756 8;
  • 18) 0.666 666 666 656 965 389 847 756 8 × 2 = 1 + 0.333 333 333 313 930 779 695 513 6;
  • 19) 0.333 333 333 313 930 779 695 513 6 × 2 = 0 + 0.666 666 666 627 861 559 391 027 2;
  • 20) 0.666 666 666 627 861 559 391 027 2 × 2 = 1 + 0.333 333 333 255 723 118 782 054 4;
  • 21) 0.333 333 333 255 723 118 782 054 4 × 2 = 0 + 0.666 666 666 511 446 237 564 108 8;
  • 22) 0.666 666 666 511 446 237 564 108 8 × 2 = 1 + 0.333 333 333 022 892 475 128 217 6;
  • 23) 0.333 333 333 022 892 475 128 217 6 × 2 = 0 + 0.666 666 666 045 784 950 256 435 2;
  • 24) 0.666 666 666 045 784 950 256 435 2 × 2 = 1 + 0.333 333 332 091 569 900 512 870 4;
  • 25) 0.333 333 332 091 569 900 512 870 4 × 2 = 0 + 0.666 666 664 183 139 801 025 740 8;
  • 26) 0.666 666 664 183 139 801 025 740 8 × 2 = 1 + 0.333 333 328 366 279 602 051 481 6;
  • 27) 0.333 333 328 366 279 602 051 481 6 × 2 = 0 + 0.666 666 656 732 559 204 102 963 2;
  • 28) 0.666 666 656 732 559 204 102 963 2 × 2 = 1 + 0.333 333 313 465 118 408 205 926 4;
  • 29) 0.333 333 313 465 118 408 205 926 4 × 2 = 0 + 0.666 666 626 930 236 816 411 852 8;
  • 30) 0.666 666 626 930 236 816 411 852 8 × 2 = 1 + 0.333 333 253 860 473 632 823 705 6;
  • 31) 0.333 333 253 860 473 632 823 705 6 × 2 = 0 + 0.666 666 507 720 947 265 647 411 2;
  • 32) 0.666 666 507 720 947 265 647 411 2 × 2 = 1 + 0.333 333 015 441 894 531 294 822 4;
  • 33) 0.333 333 015 441 894 531 294 822 4 × 2 = 0 + 0.666 666 030 883 789 062 589 644 8;
  • 34) 0.666 666 030 883 789 062 589 644 8 × 2 = 1 + 0.333 332 061 767 578 125 179 289 6;
  • 35) 0.333 332 061 767 578 125 179 289 6 × 2 = 0 + 0.666 664 123 535 156 250 358 579 2;
  • 36) 0.666 664 123 535 156 250 358 579 2 × 2 = 1 + 0.333 328 247 070 312 500 717 158 4;
  • 37) 0.333 328 247 070 312 500 717 158 4 × 2 = 0 + 0.666 656 494 140 625 001 434 316 8;
  • 38) 0.666 656 494 140 625 001 434 316 8 × 2 = 1 + 0.333 312 988 281 250 002 868 633 6;
  • 39) 0.333 312 988 281 250 002 868 633 6 × 2 = 0 + 0.666 625 976 562 500 005 737 267 2;
  • 40) 0.666 625 976 562 500 005 737 267 2 × 2 = 1 + 0.333 251 953 125 000 011 474 534 4;
  • 41) 0.333 251 953 125 000 011 474 534 4 × 2 = 0 + 0.666 503 906 250 000 022 949 068 8;
  • 42) 0.666 503 906 250 000 022 949 068 8 × 2 = 1 + 0.333 007 812 500 000 045 898 137 6;
  • 43) 0.333 007 812 500 000 045 898 137 6 × 2 = 0 + 0.666 015 625 000 000 091 796 275 2;
  • 44) 0.666 015 625 000 000 091 796 275 2 × 2 = 1 + 0.332 031 250 000 000 183 592 550 4;
  • 45) 0.332 031 250 000 000 183 592 550 4 × 2 = 0 + 0.664 062 500 000 000 367 185 100 8;
  • 46) 0.664 062 500 000 000 367 185 100 8 × 2 = 1 + 0.328 125 000 000 000 734 370 201 6;
  • 47) 0.328 125 000 000 000 734 370 201 6 × 2 = 0 + 0.656 250 000 000 001 468 740 403 2;
  • 48) 0.656 250 000 000 001 468 740 403 2 × 2 = 1 + 0.312 500 000 000 002 937 480 806 4;
  • 49) 0.312 500 000 000 002 937 480 806 4 × 2 = 0 + 0.625 000 000 000 005 874 961 612 8;
  • 50) 0.625 000 000 000 005 874 961 612 8 × 2 = 1 + 0.250 000 000 000 011 749 923 225 6;
  • 51) 0.250 000 000 000 011 749 923 225 6 × 2 = 0 + 0.500 000 000 000 023 499 846 451 2;
  • 52) 0.500 000 000 000 023 499 846 451 2 × 2 = 1 + 0.000 000 000 000 046 999 692 902 4;
  • 53) 0.000 000 000 000 046 999 692 902 4 × 2 = 0 + 0.000 000 000 000 093 999 385 804 8;

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:


0.333 333 333 333 333 259 318 465 025(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2)

5. Positive number before normalization:

1.333 333 333 333 333 259 318 465 025(10) =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.333 333 333 333 333 259 318 465 025(10) =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) × 20


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): 0

Mantissa (not normalized):
1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1023(10) =

011 1111 1111(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 0 =

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) =
011 1111 1111

Mantissa (52 bits) =
0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


Decimal number 1.333 333 333 333 333 259 318 465 025 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 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