64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -0.333 333 333 337 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -0.333 333 333 337(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.333 333 333 337| = 0.333 333 333 337

2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.


0(10) =


0(2)


4. Convert to binary (base 2) the fractional part: 0.333 333 333 337.

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) 0.333 333 333 337 × 2 = 0 + 0.666 666 666 674;
  • 2) 0.666 666 666 674 × 2 = 1 + 0.333 333 333 348;
  • 3) 0.333 333 333 348 × 2 = 0 + 0.666 666 666 696;
  • 4) 0.666 666 666 696 × 2 = 1 + 0.333 333 333 392;
  • 5) 0.333 333 333 392 × 2 = 0 + 0.666 666 666 784;
  • 6) 0.666 666 666 784 × 2 = 1 + 0.333 333 333 568;
  • 7) 0.333 333 333 568 × 2 = 0 + 0.666 666 667 136;
  • 8) 0.666 666 667 136 × 2 = 1 + 0.333 333 334 272;
  • 9) 0.333 333 334 272 × 2 = 0 + 0.666 666 668 544;
  • 10) 0.666 666 668 544 × 2 = 1 + 0.333 333 337 088;
  • 11) 0.333 333 337 088 × 2 = 0 + 0.666 666 674 176;
  • 12) 0.666 666 674 176 × 2 = 1 + 0.333 333 348 352;
  • 13) 0.333 333 348 352 × 2 = 0 + 0.666 666 696 704;
  • 14) 0.666 666 696 704 × 2 = 1 + 0.333 333 393 408;
  • 15) 0.333 333 393 408 × 2 = 0 + 0.666 666 786 816;
  • 16) 0.666 666 786 816 × 2 = 1 + 0.333 333 573 632;
  • 17) 0.333 333 573 632 × 2 = 0 + 0.666 667 147 264;
  • 18) 0.666 667 147 264 × 2 = 1 + 0.333 334 294 528;
  • 19) 0.333 334 294 528 × 2 = 0 + 0.666 668 589 056;
  • 20) 0.666 668 589 056 × 2 = 1 + 0.333 337 178 112;
  • 21) 0.333 337 178 112 × 2 = 0 + 0.666 674 356 224;
  • 22) 0.666 674 356 224 × 2 = 1 + 0.333 348 712 448;
  • 23) 0.333 348 712 448 × 2 = 0 + 0.666 697 424 896;
  • 24) 0.666 697 424 896 × 2 = 1 + 0.333 394 849 792;
  • 25) 0.333 394 849 792 × 2 = 0 + 0.666 789 699 584;
  • 26) 0.666 789 699 584 × 2 = 1 + 0.333 579 399 168;
  • 27) 0.333 579 399 168 × 2 = 0 + 0.667 158 798 336;
  • 28) 0.667 158 798 336 × 2 = 1 + 0.334 317 596 672;
  • 29) 0.334 317 596 672 × 2 = 0 + 0.668 635 193 344;
  • 30) 0.668 635 193 344 × 2 = 1 + 0.337 270 386 688;
  • 31) 0.337 270 386 688 × 2 = 0 + 0.674 540 773 376;
  • 32) 0.674 540 773 376 × 2 = 1 + 0.349 081 546 752;
  • 33) 0.349 081 546 752 × 2 = 0 + 0.698 163 093 504;
  • 34) 0.698 163 093 504 × 2 = 1 + 0.396 326 187 008;
  • 35) 0.396 326 187 008 × 2 = 0 + 0.792 652 374 016;
  • 36) 0.792 652 374 016 × 2 = 1 + 0.585 304 748 032;
  • 37) 0.585 304 748 032 × 2 = 1 + 0.170 609 496 064;
  • 38) 0.170 609 496 064 × 2 = 0 + 0.341 218 992 128;
  • 39) 0.341 218 992 128 × 2 = 0 + 0.682 437 984 256;
  • 40) 0.682 437 984 256 × 2 = 1 + 0.364 875 968 512;
  • 41) 0.364 875 968 512 × 2 = 0 + 0.729 751 937 024;
  • 42) 0.729 751 937 024 × 2 = 1 + 0.459 503 874 048;
  • 43) 0.459 503 874 048 × 2 = 0 + 0.919 007 748 096;
  • 44) 0.919 007 748 096 × 2 = 1 + 0.838 015 496 192;
  • 45) 0.838 015 496 192 × 2 = 1 + 0.676 030 992 384;
  • 46) 0.676 030 992 384 × 2 = 1 + 0.352 061 984 768;
  • 47) 0.352 061 984 768 × 2 = 0 + 0.704 123 969 536;
  • 48) 0.704 123 969 536 × 2 = 1 + 0.408 247 939 072;
  • 49) 0.408 247 939 072 × 2 = 0 + 0.816 495 878 144;
  • 50) 0.816 495 878 144 × 2 = 1 + 0.632 991 756 288;
  • 51) 0.632 991 756 288 × 2 = 1 + 0.265 983 512 576;
  • 52) 0.265 983 512 576 × 2 = 0 + 0.531 967 025 152;
  • 53) 0.531 967 025 152 × 2 = 1 + 0.063 934 050 304;
  • 54) 0.063 934 050 304 × 2 = 0 + 0.127 868 100 608;

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...)


5. 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:


0.333 333 333 337(10) =


0.0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0101 1101 0110 10(2)


6. Positive number before normalization:

0.333 333 333 337(10) =


0.0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0101 1101 0110 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.333 333 333 337(10) =


0.0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0101 1101 0110 10(2) =


0.0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0101 1101 0110 10(2) × 20 =


1.0101 0101 0101 0101 0101 0101 0101 0101 0110 0101 0111 0101 1010(2) × 2-2


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): -2


Mantissa (not normalized):
1.0101 0101 0101 0101 0101 0101 0101 0101 0110 0101 0111 0101 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


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


(-2 + 1 023)(10) =


1 021(10)


10. 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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


1021(10) =


011 1111 1101(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =


1. 0101 0101 0101 0101 0101 0101 0101 0101 0110 0101 0111 0101 1010 =


0101 0101 0101 0101 0101 0101 0101 0101 0110 0101 0111 0101 1010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)


Exponent (11 bits) =
011 1111 1101


Mantissa (52 bits) =
0101 0101 0101 0101 0101 0101 0101 0101 0110 0101 0111 0101 1010


The base ten decimal number -0.333 333 333 337 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 011 1111 1101 - 0101 0101 0101 0101 0101 0101 0101 0101 0110 0101 0111 0101 1010

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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