Convert 0.070 38 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 0.070 38(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 the binary (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 the binary (base 2) the fractional part: 0.070 38.

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) 0.070 38 × 2 = 0 + 0.140 76;
  • 2) 0.140 76 × 2 = 0 + 0.281 52;
  • 3) 0.281 52 × 2 = 0 + 0.563 04;
  • 4) 0.563 04 × 2 = 1 + 0.126 08;
  • 5) 0.126 08 × 2 = 0 + 0.252 16;
  • 6) 0.252 16 × 2 = 0 + 0.504 32;
  • 7) 0.504 32 × 2 = 1 + 0.008 64;
  • 8) 0.008 64 × 2 = 0 + 0.017 28;
  • 9) 0.017 28 × 2 = 0 + 0.034 56;
  • 10) 0.034 56 × 2 = 0 + 0.069 12;
  • 11) 0.069 12 × 2 = 0 + 0.138 24;
  • 12) 0.138 24 × 2 = 0 + 0.276 48;
  • 13) 0.276 48 × 2 = 0 + 0.552 96;
  • 14) 0.552 96 × 2 = 1 + 0.105 92;
  • 15) 0.105 92 × 2 = 0 + 0.211 84;
  • 16) 0.211 84 × 2 = 0 + 0.423 68;
  • 17) 0.423 68 × 2 = 0 + 0.847 36;
  • 18) 0.847 36 × 2 = 1 + 0.694 72;
  • 19) 0.694 72 × 2 = 1 + 0.389 44;
  • 20) 0.389 44 × 2 = 0 + 0.778 88;
  • 21) 0.778 88 × 2 = 1 + 0.557 76;
  • 22) 0.557 76 × 2 = 1 + 0.115 52;
  • 23) 0.115 52 × 2 = 0 + 0.231 04;
  • 24) 0.231 04 × 2 = 0 + 0.462 08;
  • 25) 0.462 08 × 2 = 0 + 0.924 16;
  • 26) 0.924 16 × 2 = 1 + 0.848 32;
  • 27) 0.848 32 × 2 = 1 + 0.696 64;
  • 28) 0.696 64 × 2 = 1 + 0.393 28;
  • 29) 0.393 28 × 2 = 0 + 0.786 56;
  • 30) 0.786 56 × 2 = 1 + 0.573 12;
  • 31) 0.573 12 × 2 = 1 + 0.146 24;
  • 32) 0.146 24 × 2 = 0 + 0.292 48;
  • 33) 0.292 48 × 2 = 0 + 0.584 96;
  • 34) 0.584 96 × 2 = 1 + 0.169 92;
  • 35) 0.169 92 × 2 = 0 + 0.339 84;
  • 36) 0.339 84 × 2 = 0 + 0.679 68;
  • 37) 0.679 68 × 2 = 1 + 0.359 36;
  • 38) 0.359 36 × 2 = 0 + 0.718 72;
  • 39) 0.718 72 × 2 = 1 + 0.437 44;
  • 40) 0.437 44 × 2 = 0 + 0.874 88;
  • 41) 0.874 88 × 2 = 1 + 0.749 76;
  • 42) 0.749 76 × 2 = 1 + 0.499 52;
  • 43) 0.499 52 × 2 = 0 + 0.999 04;
  • 44) 0.999 04 × 2 = 1 + 0.998 08;
  • 45) 0.998 08 × 2 = 1 + 0.996 16;
  • 46) 0.996 16 × 2 = 1 + 0.992 32;
  • 47) 0.992 32 × 2 = 1 + 0.984 64;
  • 48) 0.984 64 × 2 = 1 + 0.969 28;
  • 49) 0.969 28 × 2 = 1 + 0.938 56;
  • 50) 0.938 56 × 2 = 1 + 0.877 12;
  • 51) 0.877 12 × 2 = 1 + 0.754 24;
  • 52) 0.754 24 × 2 = 1 + 0.508 48;
  • 53) 0.508 48 × 2 = 1 + 0.016 96;
  • 54) 0.016 96 × 2 = 0 + 0.033 92;
  • 55) 0.033 92 × 2 = 0 + 0.067 84;
  • 56) 0.067 84 × 2 = 0 + 0.135 68;

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.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.070 38(10) =


0.0001 0010 0000 0100 0110 1100 0111 0110 0100 1010 1101 1111 1111 1000(2)


5. Positive number before normalization:

0.070 38(10) =


0.0001 0010 0000 0100 0110 1100 0111 0110 0100 1010 1101 1111 1111 1000(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right so that only one non zero digit remains to the left of it:

0.070 38(10) =


0.0001 0010 0000 0100 0110 1100 0111 0110 0100 1010 1101 1111 1111 1000(2) =


0.0001 0010 0000 0100 0110 1100 0111 0110 0100 1010 1101 1111 1111 1000(2) × 20 =


1.0010 0000 0100 0110 1100 0111 0110 0100 1010 1101 1111 1111 1000(2) × 2-4


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):
1.0010 0000 0100 0110 1100 0111 0110 0100 1010 1101 1111 1111 1000


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) =


1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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) =


1019(10) =


011 1111 1011(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 0000 0100 0110 1100 0111 0110 0100 1010 1101 1111 1111 1000 =


0010 0000 0100 0110 1100 0111 0110 0100 1010 1101 1111 1111 1000


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 1011


Mantissa (52 bits) =
0010 0000 0100 0110 1100 0111 0110 0100 1010 1101 1111 1111 1000


Conclusion:

Number 0.070 38 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1011 - 0010 0000 0100 0110 1100 0111 0110 0100 1010 1101 1111 1111 1000

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 0

      54
    • 1

      53
    • 1

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 1

      49
    • 0

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 0

      44
    • 0

      43
    • 1

      42
    • 0

      41
    • 0

      40
    • 0

      39
    • 1

      38
    • 1

      37
    • 0

      36
    • 1

      35
    • 1

      34
    • 0

      33
    • 0

      32
    • 0

      31
    • 1

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 1

      25
    • 0

      24
    • 0

      23
    • 1

      22
    • 0

      21
    • 0

      20
    • 1

      19
    • 0

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 1

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 1

      9
    • 1

      8
    • 1

      7
    • 1

      6
    • 1

      5
    • 1

      4
    • 1

      3
    • 0

      2
    • 0

      1
    • 0

      0

More operations of this kind:

0.070 37 = ? ... 0.070 39 = ?


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)

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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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