64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.333 333 333 5 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 5(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. 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;

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.


0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.333 333 333 5.

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 5 × 2 = 0 + 0.666 666 667;
  • 2) 0.666 666 667 × 2 = 1 + 0.333 333 334;
  • 3) 0.333 333 334 × 2 = 0 + 0.666 666 668;
  • 4) 0.666 666 668 × 2 = 1 + 0.333 333 336;
  • 5) 0.333 333 336 × 2 = 0 + 0.666 666 672;
  • 6) 0.666 666 672 × 2 = 1 + 0.333 333 344;
  • 7) 0.333 333 344 × 2 = 0 + 0.666 666 688;
  • 8) 0.666 666 688 × 2 = 1 + 0.333 333 376;
  • 9) 0.333 333 376 × 2 = 0 + 0.666 666 752;
  • 10) 0.666 666 752 × 2 = 1 + 0.333 333 504;
  • 11) 0.333 333 504 × 2 = 0 + 0.666 667 008;
  • 12) 0.666 667 008 × 2 = 1 + 0.333 334 016;
  • 13) 0.333 334 016 × 2 = 0 + 0.666 668 032;
  • 14) 0.666 668 032 × 2 = 1 + 0.333 336 064;
  • 15) 0.333 336 064 × 2 = 0 + 0.666 672 128;
  • 16) 0.666 672 128 × 2 = 1 + 0.333 344 256;
  • 17) 0.333 344 256 × 2 = 0 + 0.666 688 512;
  • 18) 0.666 688 512 × 2 = 1 + 0.333 377 024;
  • 19) 0.333 377 024 × 2 = 0 + 0.666 754 048;
  • 20) 0.666 754 048 × 2 = 1 + 0.333 508 096;
  • 21) 0.333 508 096 × 2 = 0 + 0.667 016 192;
  • 22) 0.667 016 192 × 2 = 1 + 0.334 032 384;
  • 23) 0.334 032 384 × 2 = 0 + 0.668 064 768;
  • 24) 0.668 064 768 × 2 = 1 + 0.336 129 536;
  • 25) 0.336 129 536 × 2 = 0 + 0.672 259 072;
  • 26) 0.672 259 072 × 2 = 1 + 0.344 518 144;
  • 27) 0.344 518 144 × 2 = 0 + 0.689 036 288;
  • 28) 0.689 036 288 × 2 = 1 + 0.378 072 576;
  • 29) 0.378 072 576 × 2 = 0 + 0.756 145 152;
  • 30) 0.756 145 152 × 2 = 1 + 0.512 290 304;
  • 31) 0.512 290 304 × 2 = 1 + 0.024 580 608;
  • 32) 0.024 580 608 × 2 = 0 + 0.049 161 216;
  • 33) 0.049 161 216 × 2 = 0 + 0.098 322 432;
  • 34) 0.098 322 432 × 2 = 0 + 0.196 644 864;
  • 35) 0.196 644 864 × 2 = 0 + 0.393 289 728;
  • 36) 0.393 289 728 × 2 = 0 + 0.786 579 456;
  • 37) 0.786 579 456 × 2 = 1 + 0.573 158 912;
  • 38) 0.573 158 912 × 2 = 1 + 0.146 317 824;
  • 39) 0.146 317 824 × 2 = 0 + 0.292 635 648;
  • 40) 0.292 635 648 × 2 = 0 + 0.585 271 296;
  • 41) 0.585 271 296 × 2 = 1 + 0.170 542 592;
  • 42) 0.170 542 592 × 2 = 0 + 0.341 085 184;
  • 43) 0.341 085 184 × 2 = 0 + 0.682 170 368;
  • 44) 0.682 170 368 × 2 = 1 + 0.364 340 736;
  • 45) 0.364 340 736 × 2 = 0 + 0.728 681 472;
  • 46) 0.728 681 472 × 2 = 1 + 0.457 362 944;
  • 47) 0.457 362 944 × 2 = 0 + 0.914 725 888;
  • 48) 0.914 725 888 × 2 = 1 + 0.829 451 776;
  • 49) 0.829 451 776 × 2 = 1 + 0.658 903 552;
  • 50) 0.658 903 552 × 2 = 1 + 0.317 807 104;
  • 51) 0.317 807 104 × 2 = 0 + 0.635 614 208;
  • 52) 0.635 614 208 × 2 = 1 + 0.271 228 416;
  • 53) 0.271 228 416 × 2 = 0 + 0.542 456 832;
  • 54) 0.542 456 832 × 2 = 1 + 0.084 913 664;

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...)


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 5(10) =

0.0101 0101 0101 0101 0101 0101 0101 0110 0000 1100 1001 0101 1101 01(2)


5. Positive number before normalization:

0.333 333 333 5(10) =

0.0101 0101 0101 0101 0101 0101 0101 0110 0000 1100 1001 0101 1101 01(2)

6. 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 5(10) =

0.0101 0101 0101 0101 0101 0101 0101 0110 0000 1100 1001 0101 1101 01(2) =

0.0101 0101 0101 0101 0101 0101 0101 0110 0000 1100 1001 0101 1101 01(2) × 20 =

1.0101 0101 0101 0101 0101 0101 0101 1000 0011 0010 0101 0111 0101(2) × 2-2


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): -2

Mantissa (not normalized):
1.0101 0101 0101 0101 0101 0101 0101 1000 0011 0010 0101 0111 0101


8. 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)


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;
  • 1 021 ÷ 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;

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) =

1021(10) =

011 1111 1101(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0101 0101 0101 0101 0101 0101 0101 1000 0011 0010 0101 0111 0101 =

0101 0101 0101 0101 0101 0101 0101 1000 0011 0010 0101 0111 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 1101

Mantissa (52 bits) =
0101 0101 0101 0101 0101 0101 0101 1000 0011 0010 0101 0111 0101


The base ten decimal number 0.333 333 333 5 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1101 - 0101 0101 0101 0101 0101 0101 0101 1000 0011 0010 0101 0111 0101

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 0.333 333 333 4 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 0.333 333 333 6 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

Number 361 518 965 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 15 18:59 UTC (GMT)
Number -25.001 024 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 15 18:59 UTC (GMT)
Number -13.648 752 6 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 15 18:59 UTC (GMT)
Number 9 927 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 15 18:59 UTC (GMT)
Number -557 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 15 18:59 UTC (GMT)
Number 19 283 475 674 839 293 847 565 473 829 384 756 473 852 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 15 18:59 UTC (GMT)
Number 88 771 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 15 18:59 UTC (GMT)
Number 1 669 320 456 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 15 18:59 UTC (GMT)
Number 227 524 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 15 18:59 UTC (GMT)
Number -0.000 000 019 557 774 1 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 15 18:59 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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