0.600 000 000 000 21 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.600 000 000 000 21(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
0.600 000 000 000 21(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: 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: 0.600 000 000 000 21.

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) 0.600 000 000 000 21 × 2 = 1 + 0.200 000 000 000 42;
  • 2) 0.200 000 000 000 42 × 2 = 0 + 0.400 000 000 000 84;
  • 3) 0.400 000 000 000 84 × 2 = 0 + 0.800 000 000 001 68;
  • 4) 0.800 000 000 001 68 × 2 = 1 + 0.600 000 000 003 36;
  • 5) 0.600 000 000 003 36 × 2 = 1 + 0.200 000 000 006 72;
  • 6) 0.200 000 000 006 72 × 2 = 0 + 0.400 000 000 013 44;
  • 7) 0.400 000 000 013 44 × 2 = 0 + 0.800 000 000 026 88;
  • 8) 0.800 000 000 026 88 × 2 = 1 + 0.600 000 000 053 76;
  • 9) 0.600 000 000 053 76 × 2 = 1 + 0.200 000 000 107 52;
  • 10) 0.200 000 000 107 52 × 2 = 0 + 0.400 000 000 215 04;
  • 11) 0.400 000 000 215 04 × 2 = 0 + 0.800 000 000 430 08;
  • 12) 0.800 000 000 430 08 × 2 = 1 + 0.600 000 000 860 16;
  • 13) 0.600 000 000 860 16 × 2 = 1 + 0.200 000 001 720 32;
  • 14) 0.200 000 001 720 32 × 2 = 0 + 0.400 000 003 440 64;
  • 15) 0.400 000 003 440 64 × 2 = 0 + 0.800 000 006 881 28;
  • 16) 0.800 000 006 881 28 × 2 = 1 + 0.600 000 013 762 56;
  • 17) 0.600 000 013 762 56 × 2 = 1 + 0.200 000 027 525 12;
  • 18) 0.200 000 027 525 12 × 2 = 0 + 0.400 000 055 050 24;
  • 19) 0.400 000 055 050 24 × 2 = 0 + 0.800 000 110 100 48;
  • 20) 0.800 000 110 100 48 × 2 = 1 + 0.600 000 220 200 96;
  • 21) 0.600 000 220 200 96 × 2 = 1 + 0.200 000 440 401 92;
  • 22) 0.200 000 440 401 92 × 2 = 0 + 0.400 000 880 803 84;
  • 23) 0.400 000 880 803 84 × 2 = 0 + 0.800 001 761 607 68;
  • 24) 0.800 001 761 607 68 × 2 = 1 + 0.600 003 523 215 36;
  • 25) 0.600 003 523 215 36 × 2 = 1 + 0.200 007 046 430 72;
  • 26) 0.200 007 046 430 72 × 2 = 0 + 0.400 014 092 861 44;
  • 27) 0.400 014 092 861 44 × 2 = 0 + 0.800 028 185 722 88;
  • 28) 0.800 028 185 722 88 × 2 = 1 + 0.600 056 371 445 76;
  • 29) 0.600 056 371 445 76 × 2 = 1 + 0.200 112 742 891 52;
  • 30) 0.200 112 742 891 52 × 2 = 0 + 0.400 225 485 783 04;
  • 31) 0.400 225 485 783 04 × 2 = 0 + 0.800 450 971 566 08;
  • 32) 0.800 450 971 566 08 × 2 = 1 + 0.600 901 943 132 16;
  • 33) 0.600 901 943 132 16 × 2 = 1 + 0.201 803 886 264 32;
  • 34) 0.201 803 886 264 32 × 2 = 0 + 0.403 607 772 528 64;
  • 35) 0.403 607 772 528 64 × 2 = 0 + 0.807 215 545 057 28;
  • 36) 0.807 215 545 057 28 × 2 = 1 + 0.614 431 090 114 56;
  • 37) 0.614 431 090 114 56 × 2 = 1 + 0.228 862 180 229 12;
  • 38) 0.228 862 180 229 12 × 2 = 0 + 0.457 724 360 458 24;
  • 39) 0.457 724 360 458 24 × 2 = 0 + 0.915 448 720 916 48;
  • 40) 0.915 448 720 916 48 × 2 = 1 + 0.830 897 441 832 96;
  • 41) 0.830 897 441 832 96 × 2 = 1 + 0.661 794 883 665 92;
  • 42) 0.661 794 883 665 92 × 2 = 1 + 0.323 589 767 331 84;
  • 43) 0.323 589 767 331 84 × 2 = 0 + 0.647 179 534 663 68;
  • 44) 0.647 179 534 663 68 × 2 = 1 + 0.294 359 069 327 36;
  • 45) 0.294 359 069 327 36 × 2 = 0 + 0.588 718 138 654 72;
  • 46) 0.588 718 138 654 72 × 2 = 1 + 0.177 436 277 309 44;
  • 47) 0.177 436 277 309 44 × 2 = 0 + 0.354 872 554 618 88;
  • 48) 0.354 872 554 618 88 × 2 = 0 + 0.709 745 109 237 76;
  • 49) 0.709 745 109 237 76 × 2 = 1 + 0.419 490 218 475 52;
  • 50) 0.419 490 218 475 52 × 2 = 0 + 0.838 980 436 951 04;
  • 51) 0.838 980 436 951 04 × 2 = 1 + 0.677 960 873 902 08;
  • 52) 0.677 960 873 902 08 × 2 = 1 + 0.355 921 747 804 16;
  • 53) 0.355 921 747 804 16 × 2 = 0 + 0.711 843 495 608 32;

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:


0.600 000 000 000 21(10) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1101 0100 1011 0(2)

5. Positive number before normalization:

0.600 000 000 000 21(10) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1101 0100 1011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.600 000 000 000 21(10) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1101 0100 1011 0(2) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1101 0100 1011 0(2) × 20 =

1.0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 1010 1001 0110(2) × 2-1


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): -1

Mantissa (not normalized):
1.0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 1010 1001 0110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 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) =

1022(10) =

011 1111 1110(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. 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 1010 1001 0110 =

0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 1010 1001 0110


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 1110

Mantissa (52 bits) =
0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 1010 1001 0110


Decimal number 0.600 000 000 000 21 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 1010 1001 0110

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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