Convert 12.499 999 999 999 99 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 12.499 999 999 999 99(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: 12. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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:

12(10) =


1100(2)

3. Convert to binary (base 2) the fractional part: 0.499 999 999 999 99. 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) 0.499 999 999 999 99 × 2 = 0 + 0.999 999 999 999 98;
  • 2) 0.999 999 999 999 98 × 2 = 1 + 0.999 999 999 999 96;
  • 3) 0.999 999 999 999 96 × 2 = 1 + 0.999 999 999 999 92;
  • 4) 0.999 999 999 999 92 × 2 = 1 + 0.999 999 999 999 84;
  • 5) 0.999 999 999 999 84 × 2 = 1 + 0.999 999 999 999 68;
  • 6) 0.999 999 999 999 68 × 2 = 1 + 0.999 999 999 999 36;
  • 7) 0.999 999 999 999 36 × 2 = 1 + 0.999 999 999 998 72;
  • 8) 0.999 999 999 998 72 × 2 = 1 + 0.999 999 999 997 44;
  • 9) 0.999 999 999 997 44 × 2 = 1 + 0.999 999 999 994 88;
  • 10) 0.999 999 999 994 88 × 2 = 1 + 0.999 999 999 989 76;
  • 11) 0.999 999 999 989 76 × 2 = 1 + 0.999 999 999 979 52;
  • 12) 0.999 999 999 979 52 × 2 = 1 + 0.999 999 999 959 04;
  • 13) 0.999 999 999 959 04 × 2 = 1 + 0.999 999 999 918 08;
  • 14) 0.999 999 999 918 08 × 2 = 1 + 0.999 999 999 836 16;
  • 15) 0.999 999 999 836 16 × 2 = 1 + 0.999 999 999 672 32;
  • 16) 0.999 999 999 672 32 × 2 = 1 + 0.999 999 999 344 64;
  • 17) 0.999 999 999 344 64 × 2 = 1 + 0.999 999 998 689 28;
  • 18) 0.999 999 998 689 28 × 2 = 1 + 0.999 999 997 378 56;
  • 19) 0.999 999 997 378 56 × 2 = 1 + 0.999 999 994 757 12;
  • 20) 0.999 999 994 757 12 × 2 = 1 + 0.999 999 989 514 24;
  • 21) 0.999 999 989 514 24 × 2 = 1 + 0.999 999 979 028 48;
  • 22) 0.999 999 979 028 48 × 2 = 1 + 0.999 999 958 056 96;
  • 23) 0.999 999 958 056 96 × 2 = 1 + 0.999 999 916 113 92;
  • 24) 0.999 999 916 113 92 × 2 = 1 + 0.999 999 832 227 84;
  • 25) 0.999 999 832 227 84 × 2 = 1 + 0.999 999 664 455 68;
  • 26) 0.999 999 664 455 68 × 2 = 1 + 0.999 999 328 911 36;
  • 27) 0.999 999 328 911 36 × 2 = 1 + 0.999 998 657 822 72;
  • 28) 0.999 998 657 822 72 × 2 = 1 + 0.999 997 315 645 44;
  • 29) 0.999 997 315 645 44 × 2 = 1 + 0.999 994 631 290 88;
  • 30) 0.999 994 631 290 88 × 2 = 1 + 0.999 989 262 581 76;
  • 31) 0.999 989 262 581 76 × 2 = 1 + 0.999 978 525 163 52;
  • 32) 0.999 978 525 163 52 × 2 = 1 + 0.999 957 050 327 04;
  • 33) 0.999 957 050 327 04 × 2 = 1 + 0.999 914 100 654 08;
  • 34) 0.999 914 100 654 08 × 2 = 1 + 0.999 828 201 308 16;
  • 35) 0.999 828 201 308 16 × 2 = 1 + 0.999 656 402 616 32;
  • 36) 0.999 656 402 616 32 × 2 = 1 + 0.999 312 805 232 64;
  • 37) 0.999 312 805 232 64 × 2 = 1 + 0.998 625 610 465 28;
  • 38) 0.998 625 610 465 28 × 2 = 1 + 0.997 251 220 930 56;
  • 39) 0.997 251 220 930 56 × 2 = 1 + 0.994 502 441 861 12;
  • 40) 0.994 502 441 861 12 × 2 = 1 + 0.989 004 883 722 24;
  • 41) 0.989 004 883 722 24 × 2 = 1 + 0.978 009 767 444 48;
  • 42) 0.978 009 767 444 48 × 2 = 1 + 0.956 019 534 888 96;
  • 43) 0.956 019 534 888 96 × 2 = 1 + 0.912 039 069 777 92;
  • 44) 0.912 039 069 777 92 × 2 = 1 + 0.824 078 139 555 84;
  • 45) 0.824 078 139 555 84 × 2 = 1 + 0.648 156 279 111 68;
  • 46) 0.648 156 279 111 68 × 2 = 1 + 0.296 312 558 223 36;
  • 47) 0.296 312 558 223 36 × 2 = 0 + 0.592 625 116 446 72;
  • 48) 0.592 625 116 446 72 × 2 = 1 + 0.185 250 232 893 44;
  • 49) 0.185 250 232 893 44 × 2 = 0 + 0.370 500 465 786 88;
  • 50) 0.370 500 465 786 88 × 2 = 0 + 0.741 000 931 573 76;
  • 51) 0.741 000 931 573 76 × 2 = 1 + 0.482 001 863 147 52;
  • 52) 0.482 001 863 147 52 × 2 = 0 + 0.964 003 726 295 04;
  • 53) 0.964 003 726 295 04 × 2 = 1 + 0.928 007 452 590 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...)

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:

0.499 999 999 999 99(10) =


0.0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 0010 1(2)

Positive number before normalization:

12.499 999 999 999 99(10) =


1100.0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 0010 1(2)

5. Normalize the binary representation of the number, shifting the decimal mark 3 positions to the left so that only one non zero digit remains to the left of it:

12.499 999 999 999 99(10) =


1100.0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 0010 1(2) =


1100.0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 0010 1(2) × 20 =


1.1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010 0101(2) × 23

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): 1.1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010 0101

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 =


3 + 2(11-1) - 1 =


(3 + 1 023)(10) =


1 026(10)


  • division = quotient + remainder;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 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;

Exponent (adjusted) =


1026(10) =


100 0000 0010(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. 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010 0101 =


1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010

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 0010


Mantissa (52 bits) =
1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010

Number 12.499 999 999 999 99 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0010 - 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010

(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
    • 0

      55
    • 0

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 0

      49
    • 0

      48
    • 1

      47
    • 1

      46
    • 1

      45
    • 1

      44
    • 1

      43
    • 1

      42
    • 1

      41
    • 1

      40
    • 1

      39
    • 1

      38
    • 1

      37
    • 1

      36
    • 1

      35
    • 1

      34
    • 1

      33
    • 1

      32
    • 1

      31
    • 1

      30
    • 1

      29
    • 1

      28
    • 1

      27
    • 1

      26
    • 1

      25
    • 1

      24
    • 1

      23
    • 1

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 1

      15
    • 1

      14
    • 1

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 1

      9
    • 1

      8
    • 1

      7
    • 1

      6
    • 1

      5
    • 1

      4
    • 1

      3
    • 0

      2
    • 1

      1
    • 0

      0

12.499 999 999 999 98 = ? ... 12.5 = ?


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

12.499 999 999 999 99 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:25 UTC (GMT)
9.99 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:22 UTC (GMT)
1 536 738 840 000 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:22 UTC (GMT)
201 061 061 514 801 000 910 116 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:22 UTC (GMT)
22.5 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:21 UTC (GMT)
165.106 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:21 UTC (GMT)
7.74 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:20 UTC (GMT)
121 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:20 UTC (GMT)
99 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999 999.999 9 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:20 UTC (GMT)
-0.833 333 333 333 333 333 333 33 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:19 UTC (GMT)
14 695 981 039 346 655 232 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:15 UTC (GMT)
135.187 5 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:15 UTC (GMT)
1.456 to 64 bit double precision IEEE 754 binary floating point = ? Apr 07 10:14 UTC (GMT)
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