Base ten decimal number 100.093 610 8 converted to 64 bit double precision IEEE 754 binary floating point standard

How to convert the decimal number 100.093 610 8(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: 100. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 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:

100(10) =


110 0100(2)

3. Convert to binary (base 2) the fractional part: 0.093 610 8. 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) 0.093 610 8 × 2 = 0 + 0.187 221 6;
  • 2) 0.187 221 6 × 2 = 0 + 0.374 443 2;
  • 3) 0.374 443 2 × 2 = 0 + 0.748 886 4;
  • 4) 0.748 886 4 × 2 = 1 + 0.497 772 8;
  • 5) 0.497 772 8 × 2 = 0 + 0.995 545 6;
  • 6) 0.995 545 6 × 2 = 1 + 0.991 091 2;
  • 7) 0.991 091 2 × 2 = 1 + 0.982 182 4;
  • 8) 0.982 182 4 × 2 = 1 + 0.964 364 8;
  • 9) 0.964 364 8 × 2 = 1 + 0.928 729 6;
  • 10) 0.928 729 6 × 2 = 1 + 0.857 459 2;
  • 11) 0.857 459 2 × 2 = 1 + 0.714 918 4;
  • 12) 0.714 918 4 × 2 = 1 + 0.429 836 8;
  • 13) 0.429 836 8 × 2 = 0 + 0.859 673 6;
  • 14) 0.859 673 6 × 2 = 1 + 0.719 347 2;
  • 15) 0.719 347 2 × 2 = 1 + 0.438 694 4;
  • 16) 0.438 694 4 × 2 = 0 + 0.877 388 8;
  • 17) 0.877 388 8 × 2 = 1 + 0.754 777 6;
  • 18) 0.754 777 6 × 2 = 1 + 0.509 555 2;
  • 19) 0.509 555 2 × 2 = 1 + 0.019 110 4;
  • 20) 0.019 110 4 × 2 = 0 + 0.038 220 8;
  • 21) 0.038 220 8 × 2 = 0 + 0.076 441 6;
  • 22) 0.076 441 6 × 2 = 0 + 0.152 883 2;
  • 23) 0.152 883 2 × 2 = 0 + 0.305 766 4;
  • 24) 0.305 766 4 × 2 = 0 + 0.611 532 8;
  • 25) 0.611 532 8 × 2 = 1 + 0.223 065 6;
  • 26) 0.223 065 6 × 2 = 0 + 0.446 131 2;
  • 27) 0.446 131 2 × 2 = 0 + 0.892 262 4;
  • 28) 0.892 262 4 × 2 = 1 + 0.784 524 8;
  • 29) 0.784 524 8 × 2 = 1 + 0.569 049 6;
  • 30) 0.569 049 6 × 2 = 1 + 0.138 099 2;
  • 31) 0.138 099 2 × 2 = 0 + 0.276 198 4;
  • 32) 0.276 198 4 × 2 = 0 + 0.552 396 8;
  • 33) 0.552 396 8 × 2 = 1 + 0.104 793 6;
  • 34) 0.104 793 6 × 2 = 0 + 0.209 587 2;
  • 35) 0.209 587 2 × 2 = 0 + 0.419 174 4;
  • 36) 0.419 174 4 × 2 = 0 + 0.838 348 8;
  • 37) 0.838 348 8 × 2 = 1 + 0.676 697 6;
  • 38) 0.676 697 6 × 2 = 1 + 0.353 395 2;
  • 39) 0.353 395 2 × 2 = 0 + 0.706 790 4;
  • 40) 0.706 790 4 × 2 = 1 + 0.413 580 8;
  • 41) 0.413 580 8 × 2 = 0 + 0.827 161 6;
  • 42) 0.827 161 6 × 2 = 1 + 0.654 323 2;
  • 43) 0.654 323 2 × 2 = 1 + 0.308 646 4;
  • 44) 0.308 646 4 × 2 = 0 + 0.617 292 8;
  • 45) 0.617 292 8 × 2 = 1 + 0.234 585 6;
  • 46) 0.234 585 6 × 2 = 0 + 0.469 171 2;
  • 47) 0.469 171 2 × 2 = 0 + 0.938 342 4;
  • 48) 0.938 342 4 × 2 = 1 + 0.876 684 8;
  • 49) 0.876 684 8 × 2 = 1 + 0.753 369 6;
  • 50) 0.753 369 6 × 2 = 1 + 0.506 739 2;
  • 51) 0.506 739 2 × 2 = 1 + 0.013 478 4;
  • 52) 0.013 478 4 × 2 = 0 + 0.026 956 8;
  • 53) 0.026 956 8 × 2 = 0 + 0.053 913 6;

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:

0.093 610 8(10) =


0.0001 0111 1111 0110 1110 0000 1001 1100 1000 1101 0110 1001 1110 0(2)

Positive number before normalization:

100.093 610 8(10) =


110 0100.0001 0111 1111 0110 1110 0000 1001 1100 1000 1101 0110 1001 1110 0(2)

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

100.093 610 8(10) =


110 0100.0001 0111 1111 0110 1110 0000 1001 1100 1000 1101 0110 1001 1110 0(2) =


110 0100.0001 0111 1111 0110 1110 0000 1001 1100 1000 1101 0110 1001 1110 0(2) × 20 =


1.1001 0000 0101 1111 1101 1011 1000 0010 0111 0010 0011 0101 1010 0111 100(2) × 26

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): 6


Mantissa (not normalized): 1.1001 0000 0101 1111 1101 1011 1000 0010 0111 0010 0011 0101 1010 0111 100

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 =


6 + 2(11-1) - 1 =


(6 + 1 023)(10) =


1 029(10)


  • division = quotient + remainder;
  • 1 029 ÷ 2 = 514 + 1;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 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) =


1029(10) =


100 0000 0101(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. 1001 0000 0101 1111 1101 1011 1000 0010 0111 0010 0011 0101 1010 011 1100 =


1001 0000 0101 1111 1101 1011 1000 0010 0111 0010 0011 0101 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 0101


Mantissa (52 bits) =
1001 0000 0101 1111 1101 1011 1000 0010 0111 0010 0011 0101 1010

Number 100.093 610 8, a decimal, converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:


0 - 100 0000 0101 - 1001 0000 0101 1111 1101 1011 1000 0010 0111 0010 0011 0101 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
    • 1

      54
    • 0

      53
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 0

      44
    • 0

      43
    • 1

      42
    • 0

      41
    • 1

      40
    • 1

      39
    • 1

      38
    • 1

      37
    • 1

      36
    • 1

      35
    • 1

      34
    • 0

      33
    • 1

      32
    • 1

      31
    • 0

      30
    • 1

      29
    • 1

      28
    • 1

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 0

      15
    • 0

      14
    • 1

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 1

      9
    • 1

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 1

      3
    • 0

      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

100.093 610 8 = 0 - 100 0000 0101 - 1001 0000 0101 1111 1101 1011 1000 0010 0111 0010 0011 0101 1010 Dec 13 07:27 UTC (GMT)
56.3 = 0 - 100 0000 0100 - 1100 0010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 Dec 13 07:26 UTC (GMT)
-0.016 738 891 601 562 531 086 244 689 504 383 131 861 686 706 542 968 75 = 1 - 011 1111 1001 - 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 Dec 13 07:22 UTC (GMT)
-10.1 = 1 - 100 0000 0010 - 0100 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 Dec 13 07:20 UTC (GMT)
-0.120 000 000 000 000 01 = 1 - 011 1111 1011 - 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1001 Dec 13 07:20 UTC (GMT)
192.5 = 0 - 100 0000 0110 - 1000 0001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Dec 13 07:15 UTC (GMT)
420 = 0 - 100 0000 0111 - 1010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Dec 13 07:15 UTC (GMT)
-1.21 = 1 - 011 1111 1111 - 0011 0101 1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 Dec 13 07:14 UTC (GMT)
-1.21 = 1 - 011 1111 1111 - 0011 0101 1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 Dec 13 07:14 UTC (GMT)
-0.235 = 1 - 011 1111 1100 - 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 Dec 13 07:13 UTC (GMT)
-1 234.75 = 1 - 100 0000 1001 - 0011 0100 1011 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Dec 13 07:12 UTC (GMT)
4 160 750 272 = 0 - 100 0001 1110 - 1111 0000 0000 0000 0000 0101 1000 0000 0000 0000 0000 0000 0000 Dec 13 07:12 UTC (GMT)
1.456 = 0 - 011 1111 1111 - 0111 0100 1011 1100 0110 1010 0111 1110 1111 1001 1101 1011 0010 Dec 13 07:09 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