20.333 333 333 333 556 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 20.333 333 333 333 556(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
20.333 333 333 333 556(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: 20.
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;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 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.

20(10) =

1 0100(2)


3. Convert to binary (base 2) the fractional part: 0.333 333 333 333 556.

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) 0.333 333 333 333 556 × 2 = 0 + 0.666 666 666 667 112;
  • 2) 0.666 666 666 667 112 × 2 = 1 + 0.333 333 333 334 224;
  • 3) 0.333 333 333 334 224 × 2 = 0 + 0.666 666 666 668 448;
  • 4) 0.666 666 666 668 448 × 2 = 1 + 0.333 333 333 336 896;
  • 5) 0.333 333 333 336 896 × 2 = 0 + 0.666 666 666 673 792;
  • 6) 0.666 666 666 673 792 × 2 = 1 + 0.333 333 333 347 584;
  • 7) 0.333 333 333 347 584 × 2 = 0 + 0.666 666 666 695 168;
  • 8) 0.666 666 666 695 168 × 2 = 1 + 0.333 333 333 390 336;
  • 9) 0.333 333 333 390 336 × 2 = 0 + 0.666 666 666 780 672;
  • 10) 0.666 666 666 780 672 × 2 = 1 + 0.333 333 333 561 344;
  • 11) 0.333 333 333 561 344 × 2 = 0 + 0.666 666 667 122 688;
  • 12) 0.666 666 667 122 688 × 2 = 1 + 0.333 333 334 245 376;
  • 13) 0.333 333 334 245 376 × 2 = 0 + 0.666 666 668 490 752;
  • 14) 0.666 666 668 490 752 × 2 = 1 + 0.333 333 336 981 504;
  • 15) 0.333 333 336 981 504 × 2 = 0 + 0.666 666 673 963 008;
  • 16) 0.666 666 673 963 008 × 2 = 1 + 0.333 333 347 926 016;
  • 17) 0.333 333 347 926 016 × 2 = 0 + 0.666 666 695 852 032;
  • 18) 0.666 666 695 852 032 × 2 = 1 + 0.333 333 391 704 064;
  • 19) 0.333 333 391 704 064 × 2 = 0 + 0.666 666 783 408 128;
  • 20) 0.666 666 783 408 128 × 2 = 1 + 0.333 333 566 816 256;
  • 21) 0.333 333 566 816 256 × 2 = 0 + 0.666 667 133 632 512;
  • 22) 0.666 667 133 632 512 × 2 = 1 + 0.333 334 267 265 024;
  • 23) 0.333 334 267 265 024 × 2 = 0 + 0.666 668 534 530 048;
  • 24) 0.666 668 534 530 048 × 2 = 1 + 0.333 337 069 060 096;
  • 25) 0.333 337 069 060 096 × 2 = 0 + 0.666 674 138 120 192;
  • 26) 0.666 674 138 120 192 × 2 = 1 + 0.333 348 276 240 384;
  • 27) 0.333 348 276 240 384 × 2 = 0 + 0.666 696 552 480 768;
  • 28) 0.666 696 552 480 768 × 2 = 1 + 0.333 393 104 961 536;
  • 29) 0.333 393 104 961 536 × 2 = 0 + 0.666 786 209 923 072;
  • 30) 0.666 786 209 923 072 × 2 = 1 + 0.333 572 419 846 144;
  • 31) 0.333 572 419 846 144 × 2 = 0 + 0.667 144 839 692 288;
  • 32) 0.667 144 839 692 288 × 2 = 1 + 0.334 289 679 384 576;
  • 33) 0.334 289 679 384 576 × 2 = 0 + 0.668 579 358 769 152;
  • 34) 0.668 579 358 769 152 × 2 = 1 + 0.337 158 717 538 304;
  • 35) 0.337 158 717 538 304 × 2 = 0 + 0.674 317 435 076 608;
  • 36) 0.674 317 435 076 608 × 2 = 1 + 0.348 634 870 153 216;
  • 37) 0.348 634 870 153 216 × 2 = 0 + 0.697 269 740 306 432;
  • 38) 0.697 269 740 306 432 × 2 = 1 + 0.394 539 480 612 864;
  • 39) 0.394 539 480 612 864 × 2 = 0 + 0.789 078 961 225 728;
  • 40) 0.789 078 961 225 728 × 2 = 1 + 0.578 157 922 451 456;
  • 41) 0.578 157 922 451 456 × 2 = 1 + 0.156 315 844 902 912;
  • 42) 0.156 315 844 902 912 × 2 = 0 + 0.312 631 689 805 824;
  • 43) 0.312 631 689 805 824 × 2 = 0 + 0.625 263 379 611 648;
  • 44) 0.625 263 379 611 648 × 2 = 1 + 0.250 526 759 223 296;
  • 45) 0.250 526 759 223 296 × 2 = 0 + 0.501 053 518 446 592;
  • 46) 0.501 053 518 446 592 × 2 = 1 + 0.002 107 036 893 184;
  • 47) 0.002 107 036 893 184 × 2 = 0 + 0.004 214 073 786 368;
  • 48) 0.004 214 073 786 368 × 2 = 0 + 0.008 428 147 572 736;
  • 49) 0.008 428 147 572 736 × 2 = 0 + 0.016 856 295 145 472;
  • 50) 0.016 856 295 145 472 × 2 = 0 + 0.033 712 590 290 944;
  • 51) 0.033 712 590 290 944 × 2 = 0 + 0.067 425 180 581 888;
  • 52) 0.067 425 180 581 888 × 2 = 0 + 0.134 850 361 163 776;
  • 53) 0.134 850 361 163 776 × 2 = 0 + 0.269 700 722 327 552;

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:


0.333 333 333 333 556(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0100 0000 0(2)

5. Positive number before normalization:

20.333 333 333 333 556(10) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0100 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


20.333 333 333 333 556(10) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0100 0000 0(2) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0100 0000 0(2) × 20 =

1.0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0100 0000 0(2) × 24


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): 4

Mantissa (not normalized):
1.0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0100 0000 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

4 + 2(11-1) - 1 =

(4 + 1 023)(10) =

1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 513 ÷ 2 = 256 + 1;
  • 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) =

1027(10) =

100 0000 0011(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. 0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0100 0 0000 =

0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0100


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 0011

Mantissa (52 bits) =
0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0100


Decimal number 20.333 333 333 333 556 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0011 - 0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0100

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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