0.142 838 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.142 838 8(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
0.142 838 8(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: 0.142 838 8.

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) 0.142 838 8 × 2 = 0 + 0.285 677 6;
  • 2) 0.285 677 6 × 2 = 0 + 0.571 355 2;
  • 3) 0.571 355 2 × 2 = 1 + 0.142 710 4;
  • 4) 0.142 710 4 × 2 = 0 + 0.285 420 8;
  • 5) 0.285 420 8 × 2 = 0 + 0.570 841 6;
  • 6) 0.570 841 6 × 2 = 1 + 0.141 683 2;
  • 7) 0.141 683 2 × 2 = 0 + 0.283 366 4;
  • 8) 0.283 366 4 × 2 = 0 + 0.566 732 8;
  • 9) 0.566 732 8 × 2 = 1 + 0.133 465 6;
  • 10) 0.133 465 6 × 2 = 0 + 0.266 931 2;
  • 11) 0.266 931 2 × 2 = 0 + 0.533 862 4;
  • 12) 0.533 862 4 × 2 = 1 + 0.067 724 8;
  • 13) 0.067 724 8 × 2 = 0 + 0.135 449 6;
  • 14) 0.135 449 6 × 2 = 0 + 0.270 899 2;
  • 15) 0.270 899 2 × 2 = 0 + 0.541 798 4;
  • 16) 0.541 798 4 × 2 = 1 + 0.083 596 8;
  • 17) 0.083 596 8 × 2 = 0 + 0.167 193 6;
  • 18) 0.167 193 6 × 2 = 0 + 0.334 387 2;
  • 19) 0.334 387 2 × 2 = 0 + 0.668 774 4;
  • 20) 0.668 774 4 × 2 = 1 + 0.337 548 8;
  • 21) 0.337 548 8 × 2 = 0 + 0.675 097 6;
  • 22) 0.675 097 6 × 2 = 1 + 0.350 195 2;
  • 23) 0.350 195 2 × 2 = 0 + 0.700 390 4;
  • 24) 0.700 390 4 × 2 = 1 + 0.400 780 8;
  • 25) 0.400 780 8 × 2 = 0 + 0.801 561 6;
  • 26) 0.801 561 6 × 2 = 1 + 0.603 123 2;
  • 27) 0.603 123 2 × 2 = 1 + 0.206 246 4;
  • 28) 0.206 246 4 × 2 = 0 + 0.412 492 8;
  • 29) 0.412 492 8 × 2 = 0 + 0.824 985 6;
  • 30) 0.824 985 6 × 2 = 1 + 0.649 971 2;
  • 31) 0.649 971 2 × 2 = 1 + 0.299 942 4;
  • 32) 0.299 942 4 × 2 = 0 + 0.599 884 8;
  • 33) 0.599 884 8 × 2 = 1 + 0.199 769 6;
  • 34) 0.199 769 6 × 2 = 0 + 0.399 539 2;
  • 35) 0.399 539 2 × 2 = 0 + 0.799 078 4;
  • 36) 0.799 078 4 × 2 = 1 + 0.598 156 8;
  • 37) 0.598 156 8 × 2 = 1 + 0.196 313 6;
  • 38) 0.196 313 6 × 2 = 0 + 0.392 627 2;
  • 39) 0.392 627 2 × 2 = 0 + 0.785 254 4;
  • 40) 0.785 254 4 × 2 = 1 + 0.570 508 8;
  • 41) 0.570 508 8 × 2 = 1 + 0.141 017 6;
  • 42) 0.141 017 6 × 2 = 0 + 0.282 035 2;
  • 43) 0.282 035 2 × 2 = 0 + 0.564 070 4;
  • 44) 0.564 070 4 × 2 = 1 + 0.128 140 8;
  • 45) 0.128 140 8 × 2 = 0 + 0.256 281 6;
  • 46) 0.256 281 6 × 2 = 0 + 0.512 563 2;
  • 47) 0.512 563 2 × 2 = 1 + 0.025 126 4;
  • 48) 0.025 126 4 × 2 = 0 + 0.050 252 8;
  • 49) 0.050 252 8 × 2 = 0 + 0.100 505 6;
  • 50) 0.100 505 6 × 2 = 0 + 0.201 011 2;
  • 51) 0.201 011 2 × 2 = 0 + 0.402 022 4;
  • 52) 0.402 022 4 × 2 = 0 + 0.804 044 8;
  • 53) 0.804 044 8 × 2 = 1 + 0.608 089 6;
  • 54) 0.608 089 6 × 2 = 1 + 0.216 179 2;
  • 55) 0.216 179 2 × 2 = 0 + 0.432 358 4;

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:


0.142 838 8(10) =

0.0010 0100 1001 0001 0001 0101 0110 0110 1001 1001 1001 0010 0000 110(2)

5. Positive number before normalization:

0.142 838 8(10) =

0.0010 0100 1001 0001 0001 0101 0110 0110 1001 1001 1001 0010 0000 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.142 838 8(10) =

0.0010 0100 1001 0001 0001 0101 0110 0110 1001 1001 1001 0010 0000 110(2) =

0.0010 0100 1001 0001 0001 0101 0110 0110 1001 1001 1001 0010 0000 110(2) × 20 =

1.0010 0100 1000 1000 1010 1011 0011 0100 1100 1100 1001 0000 0110(2) × 2-3


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): -3

Mantissa (not normalized):
1.0010 0100 1000 1000 1010 1011 0011 0100 1100 1100 1001 0000 0110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 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) =

1020(10) =

011 1111 1100(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. 0010 0100 1000 1000 1010 1011 0011 0100 1100 1100 1001 0000 0110 =

0010 0100 1000 1000 1010 1011 0011 0100 1100 1100 1001 0000 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 1100

Mantissa (52 bits) =
0010 0100 1000 1000 1010 1011 0011 0100 1100 1100 1001 0000 0110


Decimal number 0.142 838 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1100 - 0010 0100 1000 1000 1010 1011 0011 0100 1100 1100 1001 0000 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