Base ten decimal number -9 876.543 21 converted to 64 bit double precision IEEE 754 binary floating point standard

How to convert the decimal number -9 876.543 21(10)
to
64 bit double precision IEEE 754 binary floating point
(1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. We start with the positive version of the number:

|-9 876.543 21| = 9 876.543 21

2. First, convert to binary (base 2) the integer part: 9 876. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 9 876 ÷ 2 = 4 938 + 0;
  • 4 938 ÷ 2 = 2 469 + 0;
  • 2 469 ÷ 2 = 1 234 + 1;
  • 1 234 ÷ 2 = 617 + 0;
  • 617 ÷ 2 = 308 + 1;
  • 308 ÷ 2 = 154 + 0;
  • 154 ÷ 2 = 77 + 0;
  • 77 ÷ 2 = 38 + 1;
  • 38 ÷ 2 = 19 + 0;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. 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:

9 876(10) =


10 0110 1001 0100(2)

4. Convert to binary (base 2) the fractional part: 0.543 21. 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) 0.543 21 × 2 = 1 + 0.086 42;
  • 2) 0.086 42 × 2 = 0 + 0.172 84;
  • 3) 0.172 84 × 2 = 0 + 0.345 68;
  • 4) 0.345 68 × 2 = 0 + 0.691 36;
  • 5) 0.691 36 × 2 = 1 + 0.382 72;
  • 6) 0.382 72 × 2 = 0 + 0.765 44;
  • 7) 0.765 44 × 2 = 1 + 0.530 88;
  • 8) 0.530 88 × 2 = 1 + 0.061 76;
  • 9) 0.061 76 × 2 = 0 + 0.123 52;
  • 10) 0.123 52 × 2 = 0 + 0.247 04;
  • 11) 0.247 04 × 2 = 0 + 0.494 08;
  • 12) 0.494 08 × 2 = 0 + 0.988 16;
  • 13) 0.988 16 × 2 = 1 + 0.976 32;
  • 14) 0.976 32 × 2 = 1 + 0.952 64;
  • 15) 0.952 64 × 2 = 1 + 0.905 28;
  • 16) 0.905 28 × 2 = 1 + 0.810 56;
  • 17) 0.810 56 × 2 = 1 + 0.621 12;
  • 18) 0.621 12 × 2 = 1 + 0.242 24;
  • 19) 0.242 24 × 2 = 0 + 0.484 48;
  • 20) 0.484 48 × 2 = 0 + 0.968 96;
  • 21) 0.968 96 × 2 = 1 + 0.937 92;
  • 22) 0.937 92 × 2 = 1 + 0.875 84;
  • 23) 0.875 84 × 2 = 1 + 0.751 68;
  • 24) 0.751 68 × 2 = 1 + 0.503 36;
  • 25) 0.503 36 × 2 = 1 + 0.006 72;
  • 26) 0.006 72 × 2 = 0 + 0.013 44;
  • 27) 0.013 44 × 2 = 0 + 0.026 88;
  • 28) 0.026 88 × 2 = 0 + 0.053 76;
  • 29) 0.053 76 × 2 = 0 + 0.107 52;
  • 30) 0.107 52 × 2 = 0 + 0.215 04;
  • 31) 0.215 04 × 2 = 0 + 0.430 08;
  • 32) 0.430 08 × 2 = 0 + 0.860 16;
  • 33) 0.860 16 × 2 = 1 + 0.720 32;
  • 34) 0.720 32 × 2 = 1 + 0.440 64;
  • 35) 0.440 64 × 2 = 0 + 0.881 28;
  • 36) 0.881 28 × 2 = 1 + 0.762 56;
  • 37) 0.762 56 × 2 = 1 + 0.525 12;
  • 38) 0.525 12 × 2 = 1 + 0.050 24;
  • 39) 0.050 24 × 2 = 0 + 0.100 48;
  • 40) 0.100 48 × 2 = 0 + 0.200 96;
  • 41) 0.200 96 × 2 = 0 + 0.401 92;
  • 42) 0.401 92 × 2 = 0 + 0.803 84;
  • 43) 0.803 84 × 2 = 1 + 0.607 68;
  • 44) 0.607 68 × 2 = 1 + 0.215 36;
  • 45) 0.215 36 × 2 = 0 + 0.430 72;
  • 46) 0.430 72 × 2 = 0 + 0.861 44;
  • 47) 0.861 44 × 2 = 1 + 0.722 88;
  • 48) 0.722 88 × 2 = 1 + 0.445 76;
  • 49) 0.445 76 × 2 = 0 + 0.891 52;
  • 50) 0.891 52 × 2 = 1 + 0.783 04;
  • 51) 0.783 04 × 2 = 1 + 0.566 08;
  • 52) 0.566 08 × 2 = 1 + 0.132 16;
  • 53) 0.132 16 × 2 = 0 + 0.264 32;

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...)

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 constructed list above:

0.543 21(10) =


0.1000 1011 0000 1111 1100 1111 1000 0000 1101 1100 0011 0011 0111 0(2)

Positive number before normalization:

9 876.543 21(10) =


10 0110 1001 0100.1000 1011 0000 1111 1100 1111 1000 0000 1101 1100 0011 0011 0111 0(2)

6. Normalize the binary representation of the number, shifting the decimal mark 13 positions to the left so that only one non zero digit remains to the left of it:

9 876.543 21(10) =


10 0110 1001 0100.1000 1011 0000 1111 1100 1111 1000 0000 1101 1100 0011 0011 0111 0(2) =


10 0110 1001 0100.1000 1011 0000 1111 1100 1111 1000 0000 1101 1100 0011 0011 0111 0(2) × 20 =


1.0011 0100 1010 0100 0101 1000 0111 1110 0111 1100 0000 0110 1110 0001 1001 1011 10(2) × 213

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): 13


Mantissa (not normalized): 1.0011 0100 1010 0100 0101 1000 0111 1110 0111 1100 0000 0110 1110 0001 1001 1011 10

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:

Exponent (adjusted) =


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


13 + 2(11-1) - 1 =


(13 + 1 023)(10) =


1 036(10)


  • division = quotient + remainder;
  • 1 036 ÷ 2 = 518 + 0;
  • 518 ÷ 2 = 259 + 0;
  • 259 ÷ 2 = 129 + 1;
  • 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) =


1036(10) =


100 0000 1100(2)

8. 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. 0011 0100 1010 0100 0101 1000 0111 1110 0111 1100 0000 0110 1110 00 0110 0110 1110 =


0011 0100 1010 0100 0101 1000 0111 1110 0111 1100 0000 0110 1110

Conclusion:

The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)


Exponent (11 bits) =
100 0000 1100


Mantissa (52 bits) =
0011 0100 1010 0100 0101 1000 0111 1110 0111 1100 0000 0110 1110

Number -9 876.543 21, a decimal, converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:


1 - 100 0000 1100 - 0011 0100 1010 0100 0101 1000 0111 1110 0111 1100 0000 0110 1110

(64 bits IEEE 754)
  • Sign (1 bit):

    • 1

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 1

      55
    • 1

      54
    • 0

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 1

      49
    • 1

      48
    • 0

      47
    • 1

      46
    • 0

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 1

      41
    • 0

      40
    • 0

      39
    • 1

      38
    • 0

      37
    • 0

      36
    • 0

      35
    • 1

      34
    • 0

      33
    • 1

      32
    • 1

      31
    • 0

      30
    • 0

      29
    • 0

      28
    • 0

      27
    • 1

      26
    • 1

      25
    • 1

      24
    • 1

      23
    • 1

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 1

      15
    • 1

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 1

      6
    • 1

      5
    • 0

      4
    • 1

      3
    • 1

      2
    • 1

      1
    • 0

      0

Convert decimal numbers from base ten 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 1 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

-9 876.543 21 = 1 - 100 0000 1100 - 0011 0100 1010 0100 0101 1000 0111 1110 0111 1100 0000 0110 1110 May 20 05:01 UTC (GMT)
44 = 0 - 100 0000 0100 - 0110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 05:01 UTC (GMT)
18 446 744 073 709 550 593 = 0 - 100 0011 1110 - 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 May 20 05:00 UTC (GMT)
-5.625 = 1 - 100 0000 0001 - 0110 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 05:00 UTC (GMT)
2.76 = 0 - 100 0000 0000 - 0110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 May 20 04:59 UTC (GMT)
404 = 0 - 100 0000 0111 - 1001 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 04:58 UTC (GMT)
6.3 = 0 - 100 0000 0001 - 1001 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 May 20 04:57 UTC (GMT)
78.65 = 0 - 100 0000 0101 - 0011 1010 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 May 20 04:56 UTC (GMT)
0.136 = 0 - 011 1111 1100 - 0001 0110 1000 0111 0010 1011 0000 0010 0000 1100 0100 1001 1011 May 20 04:55 UTC (GMT)
255 = 0 - 100 0000 0110 - 1111 1110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 04:54 UTC (GMT)
736 914.953 035 463 = 0 - 100 0001 0010 - 0110 0111 1101 0010 0101 1110 0111 1111 0100 0100 0011 1010 0011 May 20 04:51 UTC (GMT)
-1.625 = 1 - 011 1111 1111 - 1010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 04:49 UTC (GMT)
1 000 321 323.000 022 12 = 0 - 100 0001 1100 - 1101 1100 1111 1101 1000 1001 0101 1000 0000 0000 0000 1011 1001 May 20 04:47 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