Convert 1 982.000 12 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

How to convert the decimal number 1 982.000 12(10)
to
64 bit double precision IEEE 754 binary floating point
(1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (base 2) the integer part: 1 982. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 1 982 ÷ 2 = 991 + 0;
  • 991 ÷ 2 = 495 + 1;
  • 495 ÷ 2 = 247 + 1;
  • 247 ÷ 2 = 123 + 1;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number, by taking all the remainders starting from the bottom of the list constructed above:

1 982(10) =


111 1011 1110(2)

3. Convert to binary (base 2) the fractional part: 0.000 12. Multiply it 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) 0.000 12 × 2 = 0 + 0.000 24;
  • 2) 0.000 24 × 2 = 0 + 0.000 48;
  • 3) 0.000 48 × 2 = 0 + 0.000 96;
  • 4) 0.000 96 × 2 = 0 + 0.001 92;
  • 5) 0.001 92 × 2 = 0 + 0.003 84;
  • 6) 0.003 84 × 2 = 0 + 0.007 68;
  • 7) 0.007 68 × 2 = 0 + 0.015 36;
  • 8) 0.015 36 × 2 = 0 + 0.030 72;
  • 9) 0.030 72 × 2 = 0 + 0.061 44;
  • 10) 0.061 44 × 2 = 0 + 0.122 88;
  • 11) 0.122 88 × 2 = 0 + 0.245 76;
  • 12) 0.245 76 × 2 = 0 + 0.491 52;
  • 13) 0.491 52 × 2 = 0 + 0.983 04;
  • 14) 0.983 04 × 2 = 1 + 0.966 08;
  • 15) 0.966 08 × 2 = 1 + 0.932 16;
  • 16) 0.932 16 × 2 = 1 + 0.864 32;
  • 17) 0.864 32 × 2 = 1 + 0.728 64;
  • 18) 0.728 64 × 2 = 1 + 0.457 28;
  • 19) 0.457 28 × 2 = 0 + 0.914 56;
  • 20) 0.914 56 × 2 = 1 + 0.829 12;
  • 21) 0.829 12 × 2 = 1 + 0.658 24;
  • 22) 0.658 24 × 2 = 1 + 0.316 48;
  • 23) 0.316 48 × 2 = 0 + 0.632 96;
  • 24) 0.632 96 × 2 = 1 + 0.265 92;
  • 25) 0.265 92 × 2 = 0 + 0.531 84;
  • 26) 0.531 84 × 2 = 1 + 0.063 68;
  • 27) 0.063 68 × 2 = 0 + 0.127 36;
  • 28) 0.127 36 × 2 = 0 + 0.254 72;
  • 29) 0.254 72 × 2 = 0 + 0.509 44;
  • 30) 0.509 44 × 2 = 1 + 0.018 88;
  • 31) 0.018 88 × 2 = 0 + 0.037 76;
  • 32) 0.037 76 × 2 = 0 + 0.075 52;
  • 33) 0.075 52 × 2 = 0 + 0.151 04;
  • 34) 0.151 04 × 2 = 0 + 0.302 08;
  • 35) 0.302 08 × 2 = 0 + 0.604 16;
  • 36) 0.604 16 × 2 = 1 + 0.208 32;
  • 37) 0.208 32 × 2 = 0 + 0.416 64;
  • 38) 0.416 64 × 2 = 0 + 0.833 28;
  • 39) 0.833 28 × 2 = 1 + 0.666 56;
  • 40) 0.666 56 × 2 = 1 + 0.333 12;
  • 41) 0.333 12 × 2 = 0 + 0.666 24;
  • 42) 0.666 24 × 2 = 1 + 0.332 48;
  • 43) 0.332 48 × 2 = 0 + 0.664 96;
  • 44) 0.664 96 × 2 = 1 + 0.329 92;
  • 45) 0.329 92 × 2 = 0 + 0.659 84;
  • 46) 0.659 84 × 2 = 1 + 0.319 68;
  • 47) 0.319 68 × 2 = 0 + 0.639 36;
  • 48) 0.639 36 × 2 = 1 + 0.278 72;
  • 49) 0.278 72 × 2 = 0 + 0.557 44;
  • 50) 0.557 44 × 2 = 1 + 0.114 88;
  • 51) 0.114 88 × 2 = 0 + 0.229 76;
  • 52) 0.229 76 × 2 = 0 + 0.459 52;
  • 53) 0.459 52 × 2 = 0 + 0.919 04;

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...)

4. 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 constructed list above:

0.000 12(10) =


0.0000 0000 0000 0111 1101 1101 0100 0100 0001 0011 0101 0101 0100 0(2)

Positive number before normalization:

1 982.000 12(10) =


111 1011 1110.0000 0000 0000 0111 1101 1101 0100 0100 0001 0011 0101 0101 0100 0(2)

5. Normalize the binary representation of the number, shifting the decimal mark 10 positions to the left so that only one non zero digit remains to the left of it:

1 982.000 12(10) =


111 1011 1110.0000 0000 0000 0111 1101 1101 0100 0100 0001 0011 0101 0101 0100 0(2) =


111 1011 1110.0000 0000 0000 0111 1101 1101 0100 0100 0001 0011 0101 0101 0100 0(2) × 20 =


1.1110 1111 1000 0000 0000 0001 1111 0111 0101 0001 0000 0100 1101 0101 0101 000(2) × 210

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): 10


Mantissa (not normalized): 1.1110 1111 1000 0000 0000 0001 1111 0111 0101 0001 0000 0100 1101 0101 0101 000

6. 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:

Exponent (adjusted) =


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


10 + 2(11-1) - 1 =


(10 + 1 023)(10) =


1 033(10)


  • division = quotient + remainder;
  • 1 033 ÷ 2 = 516 + 1;
  • 516 ÷ 2 = 258 + 0;
  • 258 ÷ 2 = 129 + 0;
  • 129 ÷ 2 = 64 + 1;
  • 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;

Exponent (adjusted) =


1033(10) =


100 0000 1001(2)

7. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then 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. 1110 1111 1000 0000 0000 0001 1111 0111 0101 0001 0000 0100 1101 010 1010 1000 =


1110 1111 1000 0000 0000 0001 1111 0111 0101 0001 0000 0100 1101

Conclusion:

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 1001


Mantissa (52 bits) =
1110 1111 1000 0000 0000 0001 1111 0111 0101 0001 0000 0100 1101

Number 1 982.000 12 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 1001 - 1110 1111 1000 0000 0000 0001 1111 0111 0101 0001 0000 0100 1101

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 1

      55
    • 0

      54
    • 0

      53
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 1

      50
    • 1

      49
    • 0

      48
    • 1

      47
    • 1

      46
    • 1

      45
    • 1

      44
    • 1

      43
    • 0

      42
    • 0

      41
    • 0

      40
    • 0

      39
    • 0

      38
    • 0

      37
    • 0

      36
    • 0

      35
    • 0

      34
    • 0

      33
    • 0

      32
    • 0

      31
    • 0

      30
    • 0

      29
    • 1

      28
    • 1

      27
    • 1

      26
    • 1

      25
    • 1

      24
    • 0

      23
    • 1

      22
    • 1

      21
    • 1

      20
    • 0

      19
    • 1

      18
    • 0

      17
    • 1

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 0

      4
    • 1

      3
    • 1

      2
    • 0

      1
    • 1

      0

1 982.000 11 = ? ... 1 982.000 13 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

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