Base ten decimal number -0.095 000 000 000 000 01 converted to 64 bit double precision IEEE 754 binary floating point standard

How to convert the decimal number -0.095 000 000 000 000 01(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:

|-0.095 000 000 000 000 01| = 0.095 000 000 000 000 01

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: 0.095 000 000 000 000 01. 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) 0.095 000 000 000 000 01 × 2 = 0 + 0.190 000 000 000 000 02;
  • 2) 0.190 000 000 000 000 02 × 2 = 0 + 0.380 000 000 000 000 04;
  • 3) 0.380 000 000 000 000 04 × 2 = 0 + 0.760 000 000 000 000 08;
  • 4) 0.760 000 000 000 000 08 × 2 = 1 + 0.520 000 000 000 000 16;
  • 5) 0.520 000 000 000 000 16 × 2 = 1 + 0.040 000 000 000 000 32;
  • 6) 0.040 000 000 000 000 32 × 2 = 0 + 0.080 000 000 000 000 64;
  • 7) 0.080 000 000 000 000 64 × 2 = 0 + 0.160 000 000 000 001 28;
  • 8) 0.160 000 000 000 001 28 × 2 = 0 + 0.320 000 000 000 002 56;
  • 9) 0.320 000 000 000 002 56 × 2 = 0 + 0.640 000 000 000 005 12;
  • 10) 0.640 000 000 000 005 12 × 2 = 1 + 0.280 000 000 000 010 24;
  • 11) 0.280 000 000 000 010 24 × 2 = 0 + 0.560 000 000 000 020 48;
  • 12) 0.560 000 000 000 020 48 × 2 = 1 + 0.120 000 000 000 040 96;
  • 13) 0.120 000 000 000 040 96 × 2 = 0 + 0.240 000 000 000 081 92;
  • 14) 0.240 000 000 000 081 92 × 2 = 0 + 0.480 000 000 000 163 84;
  • 15) 0.480 000 000 000 163 84 × 2 = 0 + 0.960 000 000 000 327 68;
  • 16) 0.960 000 000 000 327 68 × 2 = 1 + 0.920 000 000 000 655 36;
  • 17) 0.920 000 000 000 655 36 × 2 = 1 + 0.840 000 000 001 310 72;
  • 18) 0.840 000 000 001 310 72 × 2 = 1 + 0.680 000 000 002 621 44;
  • 19) 0.680 000 000 002 621 44 × 2 = 1 + 0.360 000 000 005 242 88;
  • 20) 0.360 000 000 005 242 88 × 2 = 0 + 0.720 000 000 010 485 76;
  • 21) 0.720 000 000 010 485 76 × 2 = 1 + 0.440 000 000 020 971 52;
  • 22) 0.440 000 000 020 971 52 × 2 = 0 + 0.880 000 000 041 943 04;
  • 23) 0.880 000 000 041 943 04 × 2 = 1 + 0.760 000 000 083 886 08;
  • 24) 0.760 000 000 083 886 08 × 2 = 1 + 0.520 000 000 167 772 16;
  • 25) 0.520 000 000 167 772 16 × 2 = 1 + 0.040 000 000 335 544 32;
  • 26) 0.040 000 000 335 544 32 × 2 = 0 + 0.080 000 000 671 088 64;
  • 27) 0.080 000 000 671 088 64 × 2 = 0 + 0.160 000 001 342 177 28;
  • 28) 0.160 000 001 342 177 28 × 2 = 0 + 0.320 000 002 684 354 56;
  • 29) 0.320 000 002 684 354 56 × 2 = 0 + 0.640 000 005 368 709 12;
  • 30) 0.640 000 005 368 709 12 × 2 = 1 + 0.280 000 010 737 418 24;
  • 31) 0.280 000 010 737 418 24 × 2 = 0 + 0.560 000 021 474 836 48;
  • 32) 0.560 000 021 474 836 48 × 2 = 1 + 0.120 000 042 949 672 96;
  • 33) 0.120 000 042 949 672 96 × 2 = 0 + 0.240 000 085 899 345 92;
  • 34) 0.240 000 085 899 345 92 × 2 = 0 + 0.480 000 171 798 691 84;
  • 35) 0.480 000 171 798 691 84 × 2 = 0 + 0.960 000 343 597 383 68;
  • 36) 0.960 000 343 597 383 68 × 2 = 1 + 0.920 000 687 194 767 36;
  • 37) 0.920 000 687 194 767 36 × 2 = 1 + 0.840 001 374 389 534 72;
  • 38) 0.840 001 374 389 534 72 × 2 = 1 + 0.680 002 748 779 069 44;
  • 39) 0.680 002 748 779 069 44 × 2 = 1 + 0.360 005 497 558 138 88;
  • 40) 0.360 005 497 558 138 88 × 2 = 0 + 0.720 010 995 116 277 76;
  • 41) 0.720 010 995 116 277 76 × 2 = 1 + 0.440 021 990 232 555 52;
  • 42) 0.440 021 990 232 555 52 × 2 = 0 + 0.880 043 980 465 111 04;
  • 43) 0.880 043 980 465 111 04 × 2 = 1 + 0.760 087 960 930 222 08;
  • 44) 0.760 087 960 930 222 08 × 2 = 1 + 0.520 175 921 860 444 16;
  • 45) 0.520 175 921 860 444 16 × 2 = 1 + 0.040 351 843 720 888 32;
  • 46) 0.040 351 843 720 888 32 × 2 = 0 + 0.080 703 687 441 776 64;
  • 47) 0.080 703 687 441 776 64 × 2 = 0 + 0.161 407 374 883 553 28;
  • 48) 0.161 407 374 883 553 28 × 2 = 0 + 0.322 814 749 767 106 56;
  • 49) 0.322 814 749 767 106 56 × 2 = 0 + 0.645 629 499 534 213 12;
  • 50) 0.645 629 499 534 213 12 × 2 = 1 + 0.291 258 999 068 426 24;
  • 51) 0.291 258 999 068 426 24 × 2 = 0 + 0.582 517 998 136 852 48;
  • 52) 0.582 517 998 136 852 48 × 2 = 1 + 0.165 035 996 273 704 96;
  • 53) 0.165 035 996 273 704 96 × 2 = 0 + 0.330 071 992 547 409 92;
  • 54) 0.330 071 992 547 409 92 × 2 = 0 + 0.660 143 985 094 819 84;
  • 55) 0.660 143 985 094 819 84 × 2 = 1 + 0.320 287 970 189 639 68;
  • 56) 0.320 287 970 189 639 68 × 2 = 0 + 0.640 575 940 379 279 36;

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:

0.095 000 000 000 000 01(10) =


0.0001 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010(2)

Positive number before normalization:

0.095 000 000 000 000 01(10) =


0.0001 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010(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:

0.095 000 000 000 000 01(10) =


0.0001 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010(2) =


0.0001 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010(2) × 20 =


1.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010(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): 1.1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010

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. 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010 =


1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010

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) =
1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010

Number -0.095 000 000 000 000 01, a decimal, converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:


1 - 011 1111 1011 - 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010

(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
    • 0

      48
    • 0

      47
    • 1

      46
    • 0

      45
    • 1

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 1

      40
    • 1

      39
    • 1

      38
    • 1

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 1

      32
    • 1

      31
    • 0

      30
    • 0

      29
    • 0

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 1

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 1

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 0

      14
    • 1

      13
    • 1

      12
    • 1

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 0

      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

-0.095 000 000 000 000 01 = 1 - 011 1111 1011 - 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010 Jun 24 16:23 UTC (GMT)
11.25 = 0 - 100 0000 0010 - 0110 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Jun 24 16:22 UTC (GMT)
13.438 = 0 - 100 0000 0010 - 1010 1110 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 Jun 24 16:22 UTC (GMT)
3.14 = 0 - 100 0000 0000 - 1001 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 Jun 24 16:17 UTC (GMT)
2.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 3 = 0 - 100 0000 0000 - 0010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 Jun 24 16:12 UTC (GMT)
-0.000 352 859 497 070 312 5 = 1 - 011 1111 0011 - 0111 0010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Jun 24 16:11 UTC (GMT)
24.75 = 0 - 100 0000 0011 - 1000 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Jun 24 16:10 UTC (GMT)
7 899 554.1 = 0 - 100 0001 0101 - 1110 0010 0010 0110 1000 1000 0110 0110 0110 0110 0110 0110 0110 Jun 24 16:10 UTC (GMT)
-1.298 38 = 1 - 011 1111 1111 - 0100 1100 0110 0010 1010 0001 1011 0101 1100 0111 1100 1101 1000 Jun 24 16:07 UTC (GMT)
11 011 = 0 - 100 0000 1100 - 0101 1000 0001 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Jun 24 16:07 UTC (GMT)
-117 = 1 - 100 0000 0101 - 1101 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Jun 24 16:06 UTC (GMT)
92 274 688 = 0 - 100 0001 1001 - 0110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Jun 24 16:04 UTC (GMT)
87.006 = 0 - 100 0000 0101 - 0101 1100 0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 Jun 24 16:03 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