20.333 333 333 333 333 334 28 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 20.333 333 333 333 333 334 28(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
20.333 333 333 333 333 334 28(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: 0.333 333 333 333 333 334 28.

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) 0.333 333 333 333 333 334 28 × 2 = 0 + 0.666 666 666 666 666 668 56;
  • 2) 0.666 666 666 666 666 668 56 × 2 = 1 + 0.333 333 333 333 333 337 12;
  • 3) 0.333 333 333 333 333 337 12 × 2 = 0 + 0.666 666 666 666 666 674 24;
  • 4) 0.666 666 666 666 666 674 24 × 2 = 1 + 0.333 333 333 333 333 348 48;
  • 5) 0.333 333 333 333 333 348 48 × 2 = 0 + 0.666 666 666 666 666 696 96;
  • 6) 0.666 666 666 666 666 696 96 × 2 = 1 + 0.333 333 333 333 333 393 92;
  • 7) 0.333 333 333 333 333 393 92 × 2 = 0 + 0.666 666 666 666 666 787 84;
  • 8) 0.666 666 666 666 666 787 84 × 2 = 1 + 0.333 333 333 333 333 575 68;
  • 9) 0.333 333 333 333 333 575 68 × 2 = 0 + 0.666 666 666 666 667 151 36;
  • 10) 0.666 666 666 666 667 151 36 × 2 = 1 + 0.333 333 333 333 334 302 72;
  • 11) 0.333 333 333 333 334 302 72 × 2 = 0 + 0.666 666 666 666 668 605 44;
  • 12) 0.666 666 666 666 668 605 44 × 2 = 1 + 0.333 333 333 333 337 210 88;
  • 13) 0.333 333 333 333 337 210 88 × 2 = 0 + 0.666 666 666 666 674 421 76;
  • 14) 0.666 666 666 666 674 421 76 × 2 = 1 + 0.333 333 333 333 348 843 52;
  • 15) 0.333 333 333 333 348 843 52 × 2 = 0 + 0.666 666 666 666 697 687 04;
  • 16) 0.666 666 666 666 697 687 04 × 2 = 1 + 0.333 333 333 333 395 374 08;
  • 17) 0.333 333 333 333 395 374 08 × 2 = 0 + 0.666 666 666 666 790 748 16;
  • 18) 0.666 666 666 666 790 748 16 × 2 = 1 + 0.333 333 333 333 581 496 32;
  • 19) 0.333 333 333 333 581 496 32 × 2 = 0 + 0.666 666 666 667 162 992 64;
  • 20) 0.666 666 666 667 162 992 64 × 2 = 1 + 0.333 333 333 334 325 985 28;
  • 21) 0.333 333 333 334 325 985 28 × 2 = 0 + 0.666 666 666 668 651 970 56;
  • 22) 0.666 666 666 668 651 970 56 × 2 = 1 + 0.333 333 333 337 303 941 12;
  • 23) 0.333 333 333 337 303 941 12 × 2 = 0 + 0.666 666 666 674 607 882 24;
  • 24) 0.666 666 666 674 607 882 24 × 2 = 1 + 0.333 333 333 349 215 764 48;
  • 25) 0.333 333 333 349 215 764 48 × 2 = 0 + 0.666 666 666 698 431 528 96;
  • 26) 0.666 666 666 698 431 528 96 × 2 = 1 + 0.333 333 333 396 863 057 92;
  • 27) 0.333 333 333 396 863 057 92 × 2 = 0 + 0.666 666 666 793 726 115 84;
  • 28) 0.666 666 666 793 726 115 84 × 2 = 1 + 0.333 333 333 587 452 231 68;
  • 29) 0.333 333 333 587 452 231 68 × 2 = 0 + 0.666 666 667 174 904 463 36;
  • 30) 0.666 666 667 174 904 463 36 × 2 = 1 + 0.333 333 334 349 808 926 72;
  • 31) 0.333 333 334 349 808 926 72 × 2 = 0 + 0.666 666 668 699 617 853 44;
  • 32) 0.666 666 668 699 617 853 44 × 2 = 1 + 0.333 333 337 399 235 706 88;
  • 33) 0.333 333 337 399 235 706 88 × 2 = 0 + 0.666 666 674 798 471 413 76;
  • 34) 0.666 666 674 798 471 413 76 × 2 = 1 + 0.333 333 349 596 942 827 52;
  • 35) 0.333 333 349 596 942 827 52 × 2 = 0 + 0.666 666 699 193 885 655 04;
  • 36) 0.666 666 699 193 885 655 04 × 2 = 1 + 0.333 333 398 387 771 310 08;
  • 37) 0.333 333 398 387 771 310 08 × 2 = 0 + 0.666 666 796 775 542 620 16;
  • 38) 0.666 666 796 775 542 620 16 × 2 = 1 + 0.333 333 593 551 085 240 32;
  • 39) 0.333 333 593 551 085 240 32 × 2 = 0 + 0.666 667 187 102 170 480 64;
  • 40) 0.666 667 187 102 170 480 64 × 2 = 1 + 0.333 334 374 204 340 961 28;
  • 41) 0.333 334 374 204 340 961 28 × 2 = 0 + 0.666 668 748 408 681 922 56;
  • 42) 0.666 668 748 408 681 922 56 × 2 = 1 + 0.333 337 496 817 363 845 12;
  • 43) 0.333 337 496 817 363 845 12 × 2 = 0 + 0.666 674 993 634 727 690 24;
  • 44) 0.666 674 993 634 727 690 24 × 2 = 1 + 0.333 349 987 269 455 380 48;
  • 45) 0.333 349 987 269 455 380 48 × 2 = 0 + 0.666 699 974 538 910 760 96;
  • 46) 0.666 699 974 538 910 760 96 × 2 = 1 + 0.333 399 949 077 821 521 92;
  • 47) 0.333 399 949 077 821 521 92 × 2 = 0 + 0.666 799 898 155 643 043 84;
  • 48) 0.666 799 898 155 643 043 84 × 2 = 1 + 0.333 599 796 311 286 087 68;
  • 49) 0.333 599 796 311 286 087 68 × 2 = 0 + 0.667 199 592 622 572 175 36;
  • 50) 0.667 199 592 622 572 175 36 × 2 = 1 + 0.334 399 185 245 144 350 72;
  • 51) 0.334 399 185 245 144 350 72 × 2 = 0 + 0.668 798 370 490 288 701 44;
  • 52) 0.668 798 370 490 288 701 44 × 2 = 1 + 0.337 596 740 980 577 402 88;
  • 53) 0.337 596 740 980 577 402 88 × 2 = 0 + 0.675 193 481 961 154 805 76;

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:


0.333 333 333 333 333 334 28(10) =

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

5. Positive number before normalization:

20.333 333 333 333 333 334 28(10) =

1 0100.0101 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 4 positions to the left, so that only one non zero digit remains to the left of it:


20.333 333 333 333 333 334 28(10) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) × 20 =

1.0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 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 0101 0101 0101 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 0101 0101 0 1010 =

0100 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) =
100 0000 0011

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


Decimal number 20.333 333 333 333 333 334 28 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 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