1.999 999 999 999 999 888 977 697 537 484 345 957 636 832 636 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.999 999 999 999 999 888 977 697 537 484 345 957 636 832 636(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
1.999 999 999 999 999 888 977 697 537 484 345 957 636 832 636(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. First, convert to binary (in base 2) the integer part: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.999 999 999 999 999 888 977 697 537 484 345 957 636 832 636.

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) 0.999 999 999 999 999 888 977 697 537 484 345 957 636 832 636 × 2 = 1 + 0.999 999 999 999 999 777 955 395 074 968 691 915 273 665 272;
  • 2) 0.999 999 999 999 999 777 955 395 074 968 691 915 273 665 272 × 2 = 1 + 0.999 999 999 999 999 555 910 790 149 937 383 830 547 330 544;
  • 3) 0.999 999 999 999 999 555 910 790 149 937 383 830 547 330 544 × 2 = 1 + 0.999 999 999 999 999 111 821 580 299 874 767 661 094 661 088;
  • 4) 0.999 999 999 999 999 111 821 580 299 874 767 661 094 661 088 × 2 = 1 + 0.999 999 999 999 998 223 643 160 599 749 535 322 189 322 176;
  • 5) 0.999 999 999 999 998 223 643 160 599 749 535 322 189 322 176 × 2 = 1 + 0.999 999 999 999 996 447 286 321 199 499 070 644 378 644 352;
  • 6) 0.999 999 999 999 996 447 286 321 199 499 070 644 378 644 352 × 2 = 1 + 0.999 999 999 999 992 894 572 642 398 998 141 288 757 288 704;
  • 7) 0.999 999 999 999 992 894 572 642 398 998 141 288 757 288 704 × 2 = 1 + 0.999 999 999 999 985 789 145 284 797 996 282 577 514 577 408;
  • 8) 0.999 999 999 999 985 789 145 284 797 996 282 577 514 577 408 × 2 = 1 + 0.999 999 999 999 971 578 290 569 595 992 565 155 029 154 816;
  • 9) 0.999 999 999 999 971 578 290 569 595 992 565 155 029 154 816 × 2 = 1 + 0.999 999 999 999 943 156 581 139 191 985 130 310 058 309 632;
  • 10) 0.999 999 999 999 943 156 581 139 191 985 130 310 058 309 632 × 2 = 1 + 0.999 999 999 999 886 313 162 278 383 970 260 620 116 619 264;
  • 11) 0.999 999 999 999 886 313 162 278 383 970 260 620 116 619 264 × 2 = 1 + 0.999 999 999 999 772 626 324 556 767 940 521 240 233 238 528;
  • 12) 0.999 999 999 999 772 626 324 556 767 940 521 240 233 238 528 × 2 = 1 + 0.999 999 999 999 545 252 649 113 535 881 042 480 466 477 056;
  • 13) 0.999 999 999 999 545 252 649 113 535 881 042 480 466 477 056 × 2 = 1 + 0.999 999 999 999 090 505 298 227 071 762 084 960 932 954 112;
  • 14) 0.999 999 999 999 090 505 298 227 071 762 084 960 932 954 112 × 2 = 1 + 0.999 999 999 998 181 010 596 454 143 524 169 921 865 908 224;
  • 15) 0.999 999 999 998 181 010 596 454 143 524 169 921 865 908 224 × 2 = 1 + 0.999 999 999 996 362 021 192 908 287 048 339 843 731 816 448;
  • 16) 0.999 999 999 996 362 021 192 908 287 048 339 843 731 816 448 × 2 = 1 + 0.999 999 999 992 724 042 385 816 574 096 679 687 463 632 896;
  • 17) 0.999 999 999 992 724 042 385 816 574 096 679 687 463 632 896 × 2 = 1 + 0.999 999 999 985 448 084 771 633 148 193 359 374 927 265 792;
  • 18) 0.999 999 999 985 448 084 771 633 148 193 359 374 927 265 792 × 2 = 1 + 0.999 999 999 970 896 169 543 266 296 386 718 749 854 531 584;
  • 19) 0.999 999 999 970 896 169 543 266 296 386 718 749 854 531 584 × 2 = 1 + 0.999 999 999 941 792 339 086 532 592 773 437 499 709 063 168;
  • 20) 0.999 999 999 941 792 339 086 532 592 773 437 499 709 063 168 × 2 = 1 + 0.999 999 999 883 584 678 173 065 185 546 874 999 418 126 336;
  • 21) 0.999 999 999 883 584 678 173 065 185 546 874 999 418 126 336 × 2 = 1 + 0.999 999 999 767 169 356 346 130 371 093 749 998 836 252 672;
  • 22) 0.999 999 999 767 169 356 346 130 371 093 749 998 836 252 672 × 2 = 1 + 0.999 999 999 534 338 712 692 260 742 187 499 997 672 505 344;
  • 23) 0.999 999 999 534 338 712 692 260 742 187 499 997 672 505 344 × 2 = 1 + 0.999 999 999 068 677 425 384 521 484 374 999 995 345 010 688;
  • 24) 0.999 999 999 068 677 425 384 521 484 374 999 995 345 010 688 × 2 = 1 + 0.999 999 998 137 354 850 769 042 968 749 999 990 690 021 376;
  • 25) 0.999 999 998 137 354 850 769 042 968 749 999 990 690 021 376 × 2 = 1 + 0.999 999 996 274 709 701 538 085 937 499 999 981 380 042 752;
  • 26) 0.999 999 996 274 709 701 538 085 937 499 999 981 380 042 752 × 2 = 1 + 0.999 999 992 549 419 403 076 171 874 999 999 962 760 085 504;
  • 27) 0.999 999 992 549 419 403 076 171 874 999 999 962 760 085 504 × 2 = 1 + 0.999 999 985 098 838 806 152 343 749 999 999 925 520 171 008;
  • 28) 0.999 999 985 098 838 806 152 343 749 999 999 925 520 171 008 × 2 = 1 + 0.999 999 970 197 677 612 304 687 499 999 999 851 040 342 016;
  • 29) 0.999 999 970 197 677 612 304 687 499 999 999 851 040 342 016 × 2 = 1 + 0.999 999 940 395 355 224 609 374 999 999 999 702 080 684 032;
  • 30) 0.999 999 940 395 355 224 609 374 999 999 999 702 080 684 032 × 2 = 1 + 0.999 999 880 790 710 449 218 749 999 999 999 404 161 368 064;
  • 31) 0.999 999 880 790 710 449 218 749 999 999 999 404 161 368 064 × 2 = 1 + 0.999 999 761 581 420 898 437 499 999 999 998 808 322 736 128;
  • 32) 0.999 999 761 581 420 898 437 499 999 999 998 808 322 736 128 × 2 = 1 + 0.999 999 523 162 841 796 874 999 999 999 997 616 645 472 256;
  • 33) 0.999 999 523 162 841 796 874 999 999 999 997 616 645 472 256 × 2 = 1 + 0.999 999 046 325 683 593 749 999 999 999 995 233 290 944 512;
  • 34) 0.999 999 046 325 683 593 749 999 999 999 995 233 290 944 512 × 2 = 1 + 0.999 998 092 651 367 187 499 999 999 999 990 466 581 889 024;
  • 35) 0.999 998 092 651 367 187 499 999 999 999 990 466 581 889 024 × 2 = 1 + 0.999 996 185 302 734 374 999 999 999 999 980 933 163 778 048;
  • 36) 0.999 996 185 302 734 374 999 999 999 999 980 933 163 778 048 × 2 = 1 + 0.999 992 370 605 468 749 999 999 999 999 961 866 327 556 096;
  • 37) 0.999 992 370 605 468 749 999 999 999 999 961 866 327 556 096 × 2 = 1 + 0.999 984 741 210 937 499 999 999 999 999 923 732 655 112 192;
  • 38) 0.999 984 741 210 937 499 999 999 999 999 923 732 655 112 192 × 2 = 1 + 0.999 969 482 421 874 999 999 999 999 999 847 465 310 224 384;
  • 39) 0.999 969 482 421 874 999 999 999 999 999 847 465 310 224 384 × 2 = 1 + 0.999 938 964 843 749 999 999 999 999 999 694 930 620 448 768;
  • 40) 0.999 938 964 843 749 999 999 999 999 999 694 930 620 448 768 × 2 = 1 + 0.999 877 929 687 499 999 999 999 999 999 389 861 240 897 536;
  • 41) 0.999 877 929 687 499 999 999 999 999 999 389 861 240 897 536 × 2 = 1 + 0.999 755 859 374 999 999 999 999 999 998 779 722 481 795 072;
  • 42) 0.999 755 859 374 999 999 999 999 999 998 779 722 481 795 072 × 2 = 1 + 0.999 511 718 749 999 999 999 999 999 997 559 444 963 590 144;
  • 43) 0.999 511 718 749 999 999 999 999 999 997 559 444 963 590 144 × 2 = 1 + 0.999 023 437 499 999 999 999 999 999 995 118 889 927 180 288;
  • 44) 0.999 023 437 499 999 999 999 999 999 995 118 889 927 180 288 × 2 = 1 + 0.998 046 874 999 999 999 999 999 999 990 237 779 854 360 576;
  • 45) 0.998 046 874 999 999 999 999 999 999 990 237 779 854 360 576 × 2 = 1 + 0.996 093 749 999 999 999 999 999 999 980 475 559 708 721 152;
  • 46) 0.996 093 749 999 999 999 999 999 999 980 475 559 708 721 152 × 2 = 1 + 0.992 187 499 999 999 999 999 999 999 960 951 119 417 442 304;
  • 47) 0.992 187 499 999 999 999 999 999 999 960 951 119 417 442 304 × 2 = 1 + 0.984 374 999 999 999 999 999 999 999 921 902 238 834 884 608;
  • 48) 0.984 374 999 999 999 999 999 999 999 921 902 238 834 884 608 × 2 = 1 + 0.968 749 999 999 999 999 999 999 999 843 804 477 669 769 216;
  • 49) 0.968 749 999 999 999 999 999 999 999 843 804 477 669 769 216 × 2 = 1 + 0.937 499 999 999 999 999 999 999 999 687 608 955 339 538 432;
  • 50) 0.937 499 999 999 999 999 999 999 999 687 608 955 339 538 432 × 2 = 1 + 0.874 999 999 999 999 999 999 999 999 375 217 910 679 076 864;
  • 51) 0.874 999 999 999 999 999 999 999 999 375 217 910 679 076 864 × 2 = 1 + 0.749 999 999 999 999 999 999 999 998 750 435 821 358 153 728;
  • 52) 0.749 999 999 999 999 999 999 999 998 750 435 821 358 153 728 × 2 = 1 + 0.499 999 999 999 999 999 999 999 997 500 871 642 716 307 456;
  • 53) 0.499 999 999 999 999 999 999 999 997 500 871 642 716 307 456 × 2 = 0 + 0.999 999 999 999 999 999 999 999 995 001 743 285 432 614 912;

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).


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:


0.999 999 999 999 999 888 977 697 537 484 345 957 636 832 636(10) =

0.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0(2)

5. Positive number before normalization:

1.999 999 999 999 999 888 977 697 537 484 345 957 636 832 636(10) =

1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.999 999 999 999 999 888 977 697 537 484 345 957 636 832 636(10) =

1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0(2) =

1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0(2) × 20


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): 0

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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;

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) =

1023(10) =

011 1111 1111(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. 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0 =

1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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) =
011 1111 1111

Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


Decimal number 1.999 999 999 999 999 888 977 697 537 484 345 957 636 832 636 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

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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