64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 12.333 333 333 333 333 333 333 333 333 333 333 333 333 336 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 12.333 333 333 333 333 333 333 333 333 333 333 333 333 336(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 12.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when 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.

Take 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: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 336.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 336 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 672;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 672 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 344;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 344 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 688;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 688 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 376;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 376 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 752;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 752 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 504;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 504 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 667 008;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 667 008 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 334 016;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 334 016 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 668 032;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 668 032 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 336 064;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 336 064 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 672 128;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 672 128 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 344 256;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 344 256 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 688 512;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 688 512 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 377 024;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 377 024 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 754 048;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 754 048 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 508 096;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 508 096 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 667 016 192;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 667 016 192 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 334 032 384;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 334 032 384 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 668 064 768;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 668 064 768 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 336 129 536;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 336 129 536 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 672 259 072;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 672 259 072 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 344 518 144;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 344 518 144 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 689 036 288;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 689 036 288 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 378 072 576;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 378 072 576 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 756 145 152;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 756 145 152 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 512 290 304;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 512 290 304 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 667 024 580 608;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 667 024 580 608 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 334 049 161 216;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 334 049 161 216 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 668 098 322 432;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 668 098 322 432 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 336 196 644 864;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 336 196 644 864 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 672 393 289 728;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 672 393 289 728 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 344 786 579 456;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 344 786 579 456 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 689 573 158 912;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 689 573 158 912 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 379 146 317 824;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 379 146 317 824 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 758 292 635 648;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 758 292 635 648 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 516 585 271 296;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 516 585 271 296 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 667 033 170 542 592;
  • 38) 0.666 666 666 666 666 666 666 666 666 667 033 170 542 592 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 334 066 341 085 184;
  • 39) 0.333 333 333 333 333 333 333 333 333 334 066 341 085 184 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 668 132 682 170 368;
  • 40) 0.666 666 666 666 666 666 666 666 666 668 132 682 170 368 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 336 265 364 340 736;
  • 41) 0.333 333 333 333 333 333 333 333 333 336 265 364 340 736 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 672 530 728 681 472;
  • 42) 0.666 666 666 666 666 666 666 666 666 672 530 728 681 472 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 345 061 457 362 944;
  • 43) 0.333 333 333 333 333 333 333 333 333 345 061 457 362 944 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 690 122 914 725 888;
  • 44) 0.666 666 666 666 666 666 666 666 666 690 122 914 725 888 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 380 245 829 451 776;
  • 45) 0.333 333 333 333 333 333 333 333 333 380 245 829 451 776 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 760 491 658 903 552;
  • 46) 0.666 666 666 666 666 666 666 666 666 760 491 658 903 552 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 520 983 317 807 104;
  • 47) 0.333 333 333 333 333 333 333 333 333 520 983 317 807 104 × 2 = 0 + 0.666 666 666 666 666 666 666 666 667 041 966 635 614 208;
  • 48) 0.666 666 666 666 666 666 666 666 667 041 966 635 614 208 × 2 = 1 + 0.333 333 333 333 333 333 333 333 334 083 933 271 228 416;
  • 49) 0.333 333 333 333 333 333 333 333 334 083 933 271 228 416 × 2 = 0 + 0.666 666 666 666 666 666 666 666 668 167 866 542 456 832;
  • 50) 0.666 666 666 666 666 666 666 666 668 167 866 542 456 832 × 2 = 1 + 0.333 333 333 333 333 333 333 333 336 335 733 084 913 664;
  • 51) 0.333 333 333 333 333 333 333 333 336 335 733 084 913 664 × 2 = 0 + 0.666 666 666 666 666 666 666 666 672 671 466 169 827 328;
  • 52) 0.666 666 666 666 666 666 666 666 672 671 466 169 827 328 × 2 = 1 + 0.333 333 333 333 333 333 333 333 345 342 932 339 654 656;
  • 53) 0.333 333 333 333 333 333 333 333 345 342 932 339 654 656 × 2 = 0 + 0.666 666 666 666 666 666 666 666 690 685 864 679 309 312;

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.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 336(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2)


5. Positive number before normalization:

12.333 333 333 333 333 333 333 333 333 333 333 333 333 336(10) =

1100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


12.333 333 333 333 333 333 333 333 333 333 333 333 333 336(10) =

1100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) =

1100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) × 20 =

1.1000 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010(2) × 23


7. 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 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

3 + 2(11-1) - 1 =

(3 + 1 023)(10) =

1 026(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1026(10) =

100 0000 0010(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) 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 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 =

1000 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


12. 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 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


The base ten decimal number 12.333 333 333 333 333 333 333 333 333 333 333 333 333 336 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0010 - 1000 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 12.333 333 333 333 333 333 333 333 333 333 333 333 333 335 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 12.333 333 333 333 333 333 333 333 333 333 333 333 333 337 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

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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