-24.901 899 999 999 997 703 525 878 023 356 199 264 526 19 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -24.901 899 999 999 997 703 525 878 023 356 199 264 526 19(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-24.901 899 999 999 997 703 525 878 023 356 199 264 526 19(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-24.901 899 999 999 997 703 525 878 023 356 199 264 526 19| = 24.901 899 999 999 997 703 525 878 023 356 199 264 526 19


2. First, convert to binary (in base 2) the integer part: 24.
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;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

3. 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.

24(10) =

1 1000(2)


4. Convert to binary (base 2) the fractional part: 0.901 899 999 999 997 703 525 878 023 356 199 264 526 19.

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) 0.901 899 999 999 997 703 525 878 023 356 199 264 526 19 × 2 = 1 + 0.803 799 999 999 995 407 051 756 046 712 398 529 052 38;
  • 2) 0.803 799 999 999 995 407 051 756 046 712 398 529 052 38 × 2 = 1 + 0.607 599 999 999 990 814 103 512 093 424 797 058 104 76;
  • 3) 0.607 599 999 999 990 814 103 512 093 424 797 058 104 76 × 2 = 1 + 0.215 199 999 999 981 628 207 024 186 849 594 116 209 52;
  • 4) 0.215 199 999 999 981 628 207 024 186 849 594 116 209 52 × 2 = 0 + 0.430 399 999 999 963 256 414 048 373 699 188 232 419 04;
  • 5) 0.430 399 999 999 963 256 414 048 373 699 188 232 419 04 × 2 = 0 + 0.860 799 999 999 926 512 828 096 747 398 376 464 838 08;
  • 6) 0.860 799 999 999 926 512 828 096 747 398 376 464 838 08 × 2 = 1 + 0.721 599 999 999 853 025 656 193 494 796 752 929 676 16;
  • 7) 0.721 599 999 999 853 025 656 193 494 796 752 929 676 16 × 2 = 1 + 0.443 199 999 999 706 051 312 386 989 593 505 859 352 32;
  • 8) 0.443 199 999 999 706 051 312 386 989 593 505 859 352 32 × 2 = 0 + 0.886 399 999 999 412 102 624 773 979 187 011 718 704 64;
  • 9) 0.886 399 999 999 412 102 624 773 979 187 011 718 704 64 × 2 = 1 + 0.772 799 999 998 824 205 249 547 958 374 023 437 409 28;
  • 10) 0.772 799 999 998 824 205 249 547 958 374 023 437 409 28 × 2 = 1 + 0.545 599 999 997 648 410 499 095 916 748 046 874 818 56;
  • 11) 0.545 599 999 997 648 410 499 095 916 748 046 874 818 56 × 2 = 1 + 0.091 199 999 995 296 820 998 191 833 496 093 749 637 12;
  • 12) 0.091 199 999 995 296 820 998 191 833 496 093 749 637 12 × 2 = 0 + 0.182 399 999 990 593 641 996 383 666 992 187 499 274 24;
  • 13) 0.182 399 999 990 593 641 996 383 666 992 187 499 274 24 × 2 = 0 + 0.364 799 999 981 187 283 992 767 333 984 374 998 548 48;
  • 14) 0.364 799 999 981 187 283 992 767 333 984 374 998 548 48 × 2 = 0 + 0.729 599 999 962 374 567 985 534 667 968 749 997 096 96;
  • 15) 0.729 599 999 962 374 567 985 534 667 968 749 997 096 96 × 2 = 1 + 0.459 199 999 924 749 135 971 069 335 937 499 994 193 92;
  • 16) 0.459 199 999 924 749 135 971 069 335 937 499 994 193 92 × 2 = 0 + 0.918 399 999 849 498 271 942 138 671 874 999 988 387 84;
  • 17) 0.918 399 999 849 498 271 942 138 671 874 999 988 387 84 × 2 = 1 + 0.836 799 999 698 996 543 884 277 343 749 999 976 775 68;
  • 18) 0.836 799 999 698 996 543 884 277 343 749 999 976 775 68 × 2 = 1 + 0.673 599 999 397 993 087 768 554 687 499 999 953 551 36;
  • 19) 0.673 599 999 397 993 087 768 554 687 499 999 953 551 36 × 2 = 1 + 0.347 199 998 795 986 175 537 109 374 999 999 907 102 72;
  • 20) 0.347 199 998 795 986 175 537 109 374 999 999 907 102 72 × 2 = 0 + 0.694 399 997 591 972 351 074 218 749 999 999 814 205 44;
  • 21) 0.694 399 997 591 972 351 074 218 749 999 999 814 205 44 × 2 = 1 + 0.388 799 995 183 944 702 148 437 499 999 999 628 410 88;
  • 22) 0.388 799 995 183 944 702 148 437 499 999 999 628 410 88 × 2 = 0 + 0.777 599 990 367 889 404 296 874 999 999 999 256 821 76;
  • 23) 0.777 599 990 367 889 404 296 874 999 999 999 256 821 76 × 2 = 1 + 0.555 199 980 735 778 808 593 749 999 999 998 513 643 52;
  • 24) 0.555 199 980 735 778 808 593 749 999 999 998 513 643 52 × 2 = 1 + 0.110 399 961 471 557 617 187 499 999 999 997 027 287 04;
  • 25) 0.110 399 961 471 557 617 187 499 999 999 997 027 287 04 × 2 = 0 + 0.220 799 922 943 115 234 374 999 999 999 994 054 574 08;
  • 26) 0.220 799 922 943 115 234 374 999 999 999 994 054 574 08 × 2 = 0 + 0.441 599 845 886 230 468 749 999 999 999 988 109 148 16;
  • 27) 0.441 599 845 886 230 468 749 999 999 999 988 109 148 16 × 2 = 0 + 0.883 199 691 772 460 937 499 999 999 999 976 218 296 32;
  • 28) 0.883 199 691 772 460 937 499 999 999 999 976 218 296 32 × 2 = 1 + 0.766 399 383 544 921 874 999 999 999 999 952 436 592 64;
  • 29) 0.766 399 383 544 921 874 999 999 999 999 952 436 592 64 × 2 = 1 + 0.532 798 767 089 843 749 999 999 999 999 904 873 185 28;
  • 30) 0.532 798 767 089 843 749 999 999 999 999 904 873 185 28 × 2 = 1 + 0.065 597 534 179 687 499 999 999 999 999 809 746 370 56;
  • 31) 0.065 597 534 179 687 499 999 999 999 999 809 746 370 56 × 2 = 0 + 0.131 195 068 359 374 999 999 999 999 999 619 492 741 12;
  • 32) 0.131 195 068 359 374 999 999 999 999 999 619 492 741 12 × 2 = 0 + 0.262 390 136 718 749 999 999 999 999 999 238 985 482 24;
  • 33) 0.262 390 136 718 749 999 999 999 999 999 238 985 482 24 × 2 = 0 + 0.524 780 273 437 499 999 999 999 999 998 477 970 964 48;
  • 34) 0.524 780 273 437 499 999 999 999 999 998 477 970 964 48 × 2 = 1 + 0.049 560 546 874 999 999 999 999 999 996 955 941 928 96;
  • 35) 0.049 560 546 874 999 999 999 999 999 996 955 941 928 96 × 2 = 0 + 0.099 121 093 749 999 999 999 999 999 993 911 883 857 92;
  • 36) 0.099 121 093 749 999 999 999 999 999 993 911 883 857 92 × 2 = 0 + 0.198 242 187 499 999 999 999 999 999 987 823 767 715 84;
  • 37) 0.198 242 187 499 999 999 999 999 999 987 823 767 715 84 × 2 = 0 + 0.396 484 374 999 999 999 999 999 999 975 647 535 431 68;
  • 38) 0.396 484 374 999 999 999 999 999 999 975 647 535 431 68 × 2 = 0 + 0.792 968 749 999 999 999 999 999 999 951 295 070 863 36;
  • 39) 0.792 968 749 999 999 999 999 999 999 951 295 070 863 36 × 2 = 1 + 0.585 937 499 999 999 999 999 999 999 902 590 141 726 72;
  • 40) 0.585 937 499 999 999 999 999 999 999 902 590 141 726 72 × 2 = 1 + 0.171 874 999 999 999 999 999 999 999 805 180 283 453 44;
  • 41) 0.171 874 999 999 999 999 999 999 999 805 180 283 453 44 × 2 = 0 + 0.343 749 999 999 999 999 999 999 999 610 360 566 906 88;
  • 42) 0.343 749 999 999 999 999 999 999 999 610 360 566 906 88 × 2 = 0 + 0.687 499 999 999 999 999 999 999 999 220 721 133 813 76;
  • 43) 0.687 499 999 999 999 999 999 999 999 220 721 133 813 76 × 2 = 1 + 0.374 999 999 999 999 999 999 999 998 441 442 267 627 52;
  • 44) 0.374 999 999 999 999 999 999 999 998 441 442 267 627 52 × 2 = 0 + 0.749 999 999 999 999 999 999 999 996 882 884 535 255 04;
  • 45) 0.749 999 999 999 999 999 999 999 996 882 884 535 255 04 × 2 = 1 + 0.499 999 999 999 999 999 999 999 993 765 769 070 510 08;
  • 46) 0.499 999 999 999 999 999 999 999 993 765 769 070 510 08 × 2 = 0 + 0.999 999 999 999 999 999 999 999 987 531 538 141 020 16;
  • 47) 0.999 999 999 999 999 999 999 999 987 531 538 141 020 16 × 2 = 1 + 0.999 999 999 999 999 999 999 999 975 063 076 282 040 32;
  • 48) 0.999 999 999 999 999 999 999 999 975 063 076 282 040 32 × 2 = 1 + 0.999 999 999 999 999 999 999 999 950 126 152 564 080 64;
  • 49) 0.999 999 999 999 999 999 999 999 950 126 152 564 080 64 × 2 = 1 + 0.999 999 999 999 999 999 999 999 900 252 305 128 161 28;
  • 50) 0.999 999 999 999 999 999 999 999 900 252 305 128 161 28 × 2 = 1 + 0.999 999 999 999 999 999 999 999 800 504 610 256 322 56;
  • 51) 0.999 999 999 999 999 999 999 999 800 504 610 256 322 56 × 2 = 1 + 0.999 999 999 999 999 999 999 999 601 009 220 512 645 12;
  • 52) 0.999 999 999 999 999 999 999 999 601 009 220 512 645 12 × 2 = 1 + 0.999 999 999 999 999 999 999 999 202 018 441 025 290 24;
  • 53) 0.999 999 999 999 999 999 999 999 202 018 441 025 290 24 × 2 = 1 + 0.999 999 999 999 999 999 999 998 404 036 882 050 580 48;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


5. 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:


0.901 899 999 999 997 703 525 878 023 356 199 264 526 19(10) =

0.1110 0110 1110 0010 1110 1011 0001 1100 0100 0011 0010 1011 1111 1(2)

6. Positive number before normalization:

24.901 899 999 999 997 703 525 878 023 356 199 264 526 19(10) =

1 1000.1110 0110 1110 0010 1110 1011 0001 1100 0100 0011 0010 1011 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


24.901 899 999 999 997 703 525 878 023 356 199 264 526 19(10) =

1 1000.1110 0110 1110 0010 1110 1011 0001 1100 0100 0011 0010 1011 1111 1(2) =

1 1000.1110 0110 1110 0010 1110 1011 0001 1100 0100 0011 0010 1011 1111 1(2) × 20 =

1.1000 1110 0110 1110 0010 1110 1011 0001 1100 0100 0011 0010 1011 1111 1(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):
1.1000 1110 0110 1110 0010 1110 1011 0001 1100 0100 0011 0010 1011 1111 1


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

4 + 2(11-1) - 1 =

(4 + 1 023)(10) =

1 027(10)


10. 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;
  • 1 027 ÷ 2 = 513 + 1;
  • 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;

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

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


Exponent (adjusted) =

1027(10) =

100 0000 0011(2)


12. 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 1110 0110 1110 0010 1110 1011 0001 1100 0100 0011 0010 1011 1 1111 =

1000 1110 0110 1110 0010 1110 1011 0001 1100 0100 0011 0010 1011


13. 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 0011

Mantissa (52 bits) =
1000 1110 0110 1110 0010 1110 1011 0001 1100 0100 0011 0010 1011


Decimal number -24.901 899 999 999 997 703 525 878 023 356 199 264 526 19 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0011 - 1000 1110 0110 1110 0010 1110 1011 0001 1100 0100 0011 0010 1011

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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