Base ten decimal number -0.100 000 000 000 000 02 converted to 64 bit double precision IEEE 754 binary floating point standard

How to convert the decimal number -0.100 000 000 000 000 02(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:

|-0.100 000 000 000 000 02| = 0.100 000 000 000 000 02

2. First, convert to binary (base 2) the integer part: 0. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

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:

0(10) =


0(2)

4. Convert to binary (base 2) the fractional part: 0.100 000 000 000 000 02. 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) 0.100 000 000 000 000 02 × 2 = 0 + 0.200 000 000 000 000 04;
  • 2) 0.200 000 000 000 000 04 × 2 = 0 + 0.400 000 000 000 000 08;
  • 3) 0.400 000 000 000 000 08 × 2 = 0 + 0.800 000 000 000 000 16;
  • 4) 0.800 000 000 000 000 16 × 2 = 1 + 0.600 000 000 000 000 32;
  • 5) 0.600 000 000 000 000 32 × 2 = 1 + 0.200 000 000 000 000 64;
  • 6) 0.200 000 000 000 000 64 × 2 = 0 + 0.400 000 000 000 001 28;
  • 7) 0.400 000 000 000 001 28 × 2 = 0 + 0.800 000 000 000 002 56;
  • 8) 0.800 000 000 000 002 56 × 2 = 1 + 0.600 000 000 000 005 12;
  • 9) 0.600 000 000 000 005 12 × 2 = 1 + 0.200 000 000 000 010 24;
  • 10) 0.200 000 000 000 010 24 × 2 = 0 + 0.400 000 000 000 020 48;
  • 11) 0.400 000 000 000 020 48 × 2 = 0 + 0.800 000 000 000 040 96;
  • 12) 0.800 000 000 000 040 96 × 2 = 1 + 0.600 000 000 000 081 92;
  • 13) 0.600 000 000 000 081 92 × 2 = 1 + 0.200 000 000 000 163 84;
  • 14) 0.200 000 000 000 163 84 × 2 = 0 + 0.400 000 000 000 327 68;
  • 15) 0.400 000 000 000 327 68 × 2 = 0 + 0.800 000 000 000 655 36;
  • 16) 0.800 000 000 000 655 36 × 2 = 1 + 0.600 000 000 001 310 72;
  • 17) 0.600 000 000 001 310 72 × 2 = 1 + 0.200 000 000 002 621 44;
  • 18) 0.200 000 000 002 621 44 × 2 = 0 + 0.400 000 000 005 242 88;
  • 19) 0.400 000 000 005 242 88 × 2 = 0 + 0.800 000 000 010 485 76;
  • 20) 0.800 000 000 010 485 76 × 2 = 1 + 0.600 000 000 020 971 52;
  • 21) 0.600 000 000 020 971 52 × 2 = 1 + 0.200 000 000 041 943 04;
  • 22) 0.200 000 000 041 943 04 × 2 = 0 + 0.400 000 000 083 886 08;
  • 23) 0.400 000 000 083 886 08 × 2 = 0 + 0.800 000 000 167 772 16;
  • 24) 0.800 000 000 167 772 16 × 2 = 1 + 0.600 000 000 335 544 32;
  • 25) 0.600 000 000 335 544 32 × 2 = 1 + 0.200 000 000 671 088 64;
  • 26) 0.200 000 000 671 088 64 × 2 = 0 + 0.400 000 001 342 177 28;
  • 27) 0.400 000 001 342 177 28 × 2 = 0 + 0.800 000 002 684 354 56;
  • 28) 0.800 000 002 684 354 56 × 2 = 1 + 0.600 000 005 368 709 12;
  • 29) 0.600 000 005 368 709 12 × 2 = 1 + 0.200 000 010 737 418 24;
  • 30) 0.200 000 010 737 418 24 × 2 = 0 + 0.400 000 021 474 836 48;
  • 31) 0.400 000 021 474 836 48 × 2 = 0 + 0.800 000 042 949 672 96;
  • 32) 0.800 000 042 949 672 96 × 2 = 1 + 0.600 000 085 899 345 92;
  • 33) 0.600 000 085 899 345 92 × 2 = 1 + 0.200 000 171 798 691 84;
  • 34) 0.200 000 171 798 691 84 × 2 = 0 + 0.400 000 343 597 383 68;
  • 35) 0.400 000 343 597 383 68 × 2 = 0 + 0.800 000 687 194 767 36;
  • 36) 0.800 000 687 194 767 36 × 2 = 1 + 0.600 001 374 389 534 72;
  • 37) 0.600 001 374 389 534 72 × 2 = 1 + 0.200 002 748 779 069 44;
  • 38) 0.200 002 748 779 069 44 × 2 = 0 + 0.400 005 497 558 138 88;
  • 39) 0.400 005 497 558 138 88 × 2 = 0 + 0.800 010 995 116 277 76;
  • 40) 0.800 010 995 116 277 76 × 2 = 1 + 0.600 021 990 232 555 52;
  • 41) 0.600 021 990 232 555 52 × 2 = 1 + 0.200 043 980 465 111 04;
  • 42) 0.200 043 980 465 111 04 × 2 = 0 + 0.400 087 960 930 222 08;
  • 43) 0.400 087 960 930 222 08 × 2 = 0 + 0.800 175 921 860 444 16;
  • 44) 0.800 175 921 860 444 16 × 2 = 1 + 0.600 351 843 720 888 32;
  • 45) 0.600 351 843 720 888 32 × 2 = 1 + 0.200 703 687 441 776 64;
  • 46) 0.200 703 687 441 776 64 × 2 = 0 + 0.401 407 374 883 553 28;
  • 47) 0.401 407 374 883 553 28 × 2 = 0 + 0.802 814 749 767 106 56;
  • 48) 0.802 814 749 767 106 56 × 2 = 1 + 0.605 629 499 534 213 12;
  • 49) 0.605 629 499 534 213 12 × 2 = 1 + 0.211 258 999 068 426 24;
  • 50) 0.211 258 999 068 426 24 × 2 = 0 + 0.422 517 998 136 852 48;
  • 51) 0.422 517 998 136 852 48 × 2 = 0 + 0.845 035 996 273 704 96;
  • 52) 0.845 035 996 273 704 96 × 2 = 1 + 0.690 071 992 547 409 92;
  • 53) 0.690 071 992 547 409 92 × 2 = 1 + 0.380 143 985 094 819 84;
  • 54) 0.380 143 985 094 819 84 × 2 = 0 + 0.760 287 970 189 639 68;
  • 55) 0.760 287 970 189 639 68 × 2 = 1 + 0.520 575 940 379 279 36;
  • 56) 0.520 575 940 379 279 36 × 2 = 1 + 0.041 151 880 758 558 72;

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:

0.100 000 000 000 000 02(10) =


0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011(2)

Positive number before normalization:

0.100 000 000 000 000 02(10) =


0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011(2)

6. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the right so that only one non zero digit remains to the left of it:

0.100 000 000 000 000 02(10) =


0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011(2) =


0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011(2) × 20 =


1.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011(2) × 2-4

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): 1.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011

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 =


-4 + 2(11-1) - 1 =


(-4 + 1 023)(10) =


1 019(10)


  • 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;

Exponent (adjusted) =


1019(10) =


011 1111 1011(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, only if necessary (not the case here):

Mantissa (normalized) =


1. 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 =


1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011

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) =
011 1111 1011


Mantissa (52 bits) =
1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011

Number -0.100 000 000 000 000 02, a decimal, converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:


1 - 011 1111 1011 - 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011

(64 bits IEEE 754)
  • Sign (1 bit):

    • 1

      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):

    • 1

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 1

      47
    • 0

      46
    • 0

      45
    • 1

      44
    • 1

      43
    • 0

      42
    • 0

      41
    • 1

      40
    • 1

      39
    • 0

      38
    • 0

      37
    • 1

      36
    • 1

      35
    • 0

      34
    • 0

      33
    • 1

      32
    • 1

      31
    • 0

      30
    • 0

      29
    • 1

      28
    • 1

      27
    • 0

      26
    • 0

      25
    • 1

      24
    • 1

      23
    • 0

      22
    • 0

      21
    • 1

      20
    • 1

      19
    • 0

      18
    • 0

      17
    • 1

      16
    • 1

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 0

      10
    • 0

      9
    • 1

      8
    • 1

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 1

      3
    • 0

      2
    • 1

      1
    • 1

      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

-0.100 000 000 000 000 02 = 1 - 011 1111 1011 - 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 Sep 18 07:16 UTC (GMT)
-160.207 117 = 1 - 100 0000 0110 - 0100 0000 0110 1010 0000 1011 0011 1101 0100 1010 1110 0100 0010 Sep 18 07:15 UTC (GMT)
2 555.8 = 0 - 100 0000 1010 - 0011 1111 0111 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 Sep 18 07:15 UTC (GMT)
0.65 = 0 - 011 1111 1110 - 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 Sep 18 07:14 UTC (GMT)
0.65 = 0 - 011 1111 1110 - 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 Sep 18 07:13 UTC (GMT)
39 = 0 - 100 0000 0100 - 0011 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Sep 18 07:13 UTC (GMT)
5.1 = 0 - 100 0000 0001 - 0100 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 Sep 18 07:11 UTC (GMT)
-0.16 = 1 - 011 1111 1100 - 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 Sep 18 07:11 UTC (GMT)
0.110 100 1 = 0 - 011 1111 1011 - 1100 0010 1111 1000 0101 0010 1000 1100 1001 0100 1100 1110 0101 Sep 18 07:10 UTC (GMT)
1.432 5 = 0 - 011 1111 1111 - 0110 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 Sep 18 07:09 UTC (GMT)
39 = 0 - 100 0000 0100 - 0011 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Sep 18 07:08 UTC (GMT)
4.66 = 0 - 100 0000 0001 - 0010 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1010 0011 Sep 18 07:06 UTC (GMT)
19.005 8 = 0 - 100 0000 0011 - 0011 0000 0001 0111 1100 0001 1011 1101 1010 0101 0001 0001 1001 Sep 18 07:05 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