4.166 666 676 24 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 4.166 666 676 24(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
4.166 666 676 24(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: 4.
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;
  • 4 ÷ 2 = 2 + 0;
  • 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.

4(10) =

100(2)


3. Convert to binary (base 2) the fractional part: 0.166 666 676 24.

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) 0.166 666 676 24 × 2 = 0 + 0.333 333 352 48;
  • 2) 0.333 333 352 48 × 2 = 0 + 0.666 666 704 96;
  • 3) 0.666 666 704 96 × 2 = 1 + 0.333 333 409 92;
  • 4) 0.333 333 409 92 × 2 = 0 + 0.666 666 819 84;
  • 5) 0.666 666 819 84 × 2 = 1 + 0.333 333 639 68;
  • 6) 0.333 333 639 68 × 2 = 0 + 0.666 667 279 36;
  • 7) 0.666 667 279 36 × 2 = 1 + 0.333 334 558 72;
  • 8) 0.333 334 558 72 × 2 = 0 + 0.666 669 117 44;
  • 9) 0.666 669 117 44 × 2 = 1 + 0.333 338 234 88;
  • 10) 0.333 338 234 88 × 2 = 0 + 0.666 676 469 76;
  • 11) 0.666 676 469 76 × 2 = 1 + 0.333 352 939 52;
  • 12) 0.333 352 939 52 × 2 = 0 + 0.666 705 879 04;
  • 13) 0.666 705 879 04 × 2 = 1 + 0.333 411 758 08;
  • 14) 0.333 411 758 08 × 2 = 0 + 0.666 823 516 16;
  • 15) 0.666 823 516 16 × 2 = 1 + 0.333 647 032 32;
  • 16) 0.333 647 032 32 × 2 = 0 + 0.667 294 064 64;
  • 17) 0.667 294 064 64 × 2 = 1 + 0.334 588 129 28;
  • 18) 0.334 588 129 28 × 2 = 0 + 0.669 176 258 56;
  • 19) 0.669 176 258 56 × 2 = 1 + 0.338 352 517 12;
  • 20) 0.338 352 517 12 × 2 = 0 + 0.676 705 034 24;
  • 21) 0.676 705 034 24 × 2 = 1 + 0.353 410 068 48;
  • 22) 0.353 410 068 48 × 2 = 0 + 0.706 820 136 96;
  • 23) 0.706 820 136 96 × 2 = 1 + 0.413 640 273 92;
  • 24) 0.413 640 273 92 × 2 = 0 + 0.827 280 547 84;
  • 25) 0.827 280 547 84 × 2 = 1 + 0.654 561 095 68;
  • 26) 0.654 561 095 68 × 2 = 1 + 0.309 122 191 36;
  • 27) 0.309 122 191 36 × 2 = 0 + 0.618 244 382 72;
  • 28) 0.618 244 382 72 × 2 = 1 + 0.236 488 765 44;
  • 29) 0.236 488 765 44 × 2 = 0 + 0.472 977 530 88;
  • 30) 0.472 977 530 88 × 2 = 0 + 0.945 955 061 76;
  • 31) 0.945 955 061 76 × 2 = 1 + 0.891 910 123 52;
  • 32) 0.891 910 123 52 × 2 = 1 + 0.783 820 247 04;
  • 33) 0.783 820 247 04 × 2 = 1 + 0.567 640 494 08;
  • 34) 0.567 640 494 08 × 2 = 1 + 0.135 280 988 16;
  • 35) 0.135 280 988 16 × 2 = 0 + 0.270 561 976 32;
  • 36) 0.270 561 976 32 × 2 = 0 + 0.541 123 952 64;
  • 37) 0.541 123 952 64 × 2 = 1 + 0.082 247 905 28;
  • 38) 0.082 247 905 28 × 2 = 0 + 0.164 495 810 56;
  • 39) 0.164 495 810 56 × 2 = 0 + 0.328 991 621 12;
  • 40) 0.328 991 621 12 × 2 = 0 + 0.657 983 242 24;
  • 41) 0.657 983 242 24 × 2 = 1 + 0.315 966 484 48;
  • 42) 0.315 966 484 48 × 2 = 0 + 0.631 932 968 96;
  • 43) 0.631 932 968 96 × 2 = 1 + 0.263 865 937 92;
  • 44) 0.263 865 937 92 × 2 = 0 + 0.527 731 875 84;
  • 45) 0.527 731 875 84 × 2 = 1 + 0.055 463 751 68;
  • 46) 0.055 463 751 68 × 2 = 0 + 0.110 927 503 36;
  • 47) 0.110 927 503 36 × 2 = 0 + 0.221 855 006 72;
  • 48) 0.221 855 006 72 × 2 = 0 + 0.443 710 013 44;
  • 49) 0.443 710 013 44 × 2 = 0 + 0.887 420 026 88;
  • 50) 0.887 420 026 88 × 2 = 1 + 0.774 840 053 76;
  • 51) 0.774 840 053 76 × 2 = 1 + 0.549 680 107 52;
  • 52) 0.549 680 107 52 × 2 = 1 + 0.099 360 215 04;
  • 53) 0.099 360 215 04 × 2 = 0 + 0.198 720 430 08;

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:


0.166 666 676 24(10) =

0.0010 1010 1010 1010 1010 1010 1101 0011 1100 1000 1010 1000 0111 0(2)

5. Positive number before normalization:

4.166 666 676 24(10) =

100.0010 1010 1010 1010 1010 1010 1101 0011 1100 1000 1010 1000 0111 0(2)

6. Normalize the binary representation of the number.

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


4.166 666 676 24(10) =

100.0010 1010 1010 1010 1010 1010 1101 0011 1100 1000 1010 1000 0111 0(2) =

100.0010 1010 1010 1010 1010 1010 1101 0011 1100 1000 1010 1000 0111 0(2) × 20 =

1.0000 1010 1010 1010 1010 1010 1011 0100 1111 0010 0010 1010 0001 110(2) × 22


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):
1.0000 1010 1010 1010 1010 1010 1011 0100 1111 0010 0010 1010 0001 110


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) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 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) =

1025(10) =

100 0000 0001(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. 0000 1010 1010 1010 1010 1010 1011 0100 1111 0010 0010 1010 0001 110 =

0000 1010 1010 1010 1010 1010 1011 0100 1111 0010 0010 1010 0001


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 0001

Mantissa (52 bits) =
0000 1010 1010 1010 1010 1010 1011 0100 1111 0010 0010 1010 0001


Decimal number 4.166 666 676 24 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 0000 1010 1010 1010 1010 1010 1011 0100 1111 0010 0010 1010 0001

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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