-0.016 738 891 601 562 531 086 244 689 504 383 130 07 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 531 086 244 689 504 383 130 07(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
-0.016 738 891 601 562 531 086 244 689 504 383 130 07(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:

|-0.016 738 891 601 562 531 086 244 689 504 383 130 07| = 0.016 738 891 601 562 531 086 244 689 504 383 130 07


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

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 562 531 086 244 689 504 383 130 07.

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) 0.016 738 891 601 562 531 086 244 689 504 383 130 07 × 2 = 0 + 0.033 477 783 203 125 062 172 489 379 008 766 260 14;
  • 2) 0.033 477 783 203 125 062 172 489 379 008 766 260 14 × 2 = 0 + 0.066 955 566 406 250 124 344 978 758 017 532 520 28;
  • 3) 0.066 955 566 406 250 124 344 978 758 017 532 520 28 × 2 = 0 + 0.133 911 132 812 500 248 689 957 516 035 065 040 56;
  • 4) 0.133 911 132 812 500 248 689 957 516 035 065 040 56 × 2 = 0 + 0.267 822 265 625 000 497 379 915 032 070 130 081 12;
  • 5) 0.267 822 265 625 000 497 379 915 032 070 130 081 12 × 2 = 0 + 0.535 644 531 250 000 994 759 830 064 140 260 162 24;
  • 6) 0.535 644 531 250 000 994 759 830 064 140 260 162 24 × 2 = 1 + 0.071 289 062 500 001 989 519 660 128 280 520 324 48;
  • 7) 0.071 289 062 500 001 989 519 660 128 280 520 324 48 × 2 = 0 + 0.142 578 125 000 003 979 039 320 256 561 040 648 96;
  • 8) 0.142 578 125 000 003 979 039 320 256 561 040 648 96 × 2 = 0 + 0.285 156 250 000 007 958 078 640 513 122 081 297 92;
  • 9) 0.285 156 250 000 007 958 078 640 513 122 081 297 92 × 2 = 0 + 0.570 312 500 000 015 916 157 281 026 244 162 595 84;
  • 10) 0.570 312 500 000 015 916 157 281 026 244 162 595 84 × 2 = 1 + 0.140 625 000 000 031 832 314 562 052 488 325 191 68;
  • 11) 0.140 625 000 000 031 832 314 562 052 488 325 191 68 × 2 = 0 + 0.281 250 000 000 063 664 629 124 104 976 650 383 36;
  • 12) 0.281 250 000 000 063 664 629 124 104 976 650 383 36 × 2 = 0 + 0.562 500 000 000 127 329 258 248 209 953 300 766 72;
  • 13) 0.562 500 000 000 127 329 258 248 209 953 300 766 72 × 2 = 1 + 0.125 000 000 000 254 658 516 496 419 906 601 533 44;
  • 14) 0.125 000 000 000 254 658 516 496 419 906 601 533 44 × 2 = 0 + 0.250 000 000 000 509 317 032 992 839 813 203 066 88;
  • 15) 0.250 000 000 000 509 317 032 992 839 813 203 066 88 × 2 = 0 + 0.500 000 000 001 018 634 065 985 679 626 406 133 76;
  • 16) 0.500 000 000 001 018 634 065 985 679 626 406 133 76 × 2 = 1 + 0.000 000 000 002 037 268 131 971 359 252 812 267 52;
  • 17) 0.000 000 000 002 037 268 131 971 359 252 812 267 52 × 2 = 0 + 0.000 000 000 004 074 536 263 942 718 505 624 535 04;
  • 18) 0.000 000 000 004 074 536 263 942 718 505 624 535 04 × 2 = 0 + 0.000 000 000 008 149 072 527 885 437 011 249 070 08;
  • 19) 0.000 000 000 008 149 072 527 885 437 011 249 070 08 × 2 = 0 + 0.000 000 000 016 298 145 055 770 874 022 498 140 16;
  • 20) 0.000 000 000 016 298 145 055 770 874 022 498 140 16 × 2 = 0 + 0.000 000 000 032 596 290 111 541 748 044 996 280 32;
  • 21) 0.000 000 000 032 596 290 111 541 748 044 996 280 32 × 2 = 0 + 0.000 000 000 065 192 580 223 083 496 089 992 560 64;
  • 22) 0.000 000 000 065 192 580 223 083 496 089 992 560 64 × 2 = 0 + 0.000 000 000 130 385 160 446 166 992 179 985 121 28;
  • 23) 0.000 000 000 130 385 160 446 166 992 179 985 121 28 × 2 = 0 + 0.000 000 000 260 770 320 892 333 984 359 970 242 56;
  • 24) 0.000 000 000 260 770 320 892 333 984 359 970 242 56 × 2 = 0 + 0.000 000 000 521 540 641 784 667 968 719 940 485 12;
  • 25) 0.000 000 000 521 540 641 784 667 968 719 940 485 12 × 2 = 0 + 0.000 000 001 043 081 283 569 335 937 439 880 970 24;
  • 26) 0.000 000 001 043 081 283 569 335 937 439 880 970 24 × 2 = 0 + 0.000 000 002 086 162 567 138 671 874 879 761 940 48;
  • 27) 0.000 000 002 086 162 567 138 671 874 879 761 940 48 × 2 = 0 + 0.000 000 004 172 325 134 277 343 749 759 523 880 96;
  • 28) 0.000 000 004 172 325 134 277 343 749 759 523 880 96 × 2 = 0 + 0.000 000 008 344 650 268 554 687 499 519 047 761 92;
  • 29) 0.000 000 008 344 650 268 554 687 499 519 047 761 92 × 2 = 0 + 0.000 000 016 689 300 537 109 374 999 038 095 523 84;
  • 30) 0.000 000 016 689 300 537 109 374 999 038 095 523 84 × 2 = 0 + 0.000 000 033 378 601 074 218 749 998 076 191 047 68;
  • 31) 0.000 000 033 378 601 074 218 749 998 076 191 047 68 × 2 = 0 + 0.000 000 066 757 202 148 437 499 996 152 382 095 36;
  • 32) 0.000 000 066 757 202 148 437 499 996 152 382 095 36 × 2 = 0 + 0.000 000 133 514 404 296 874 999 992 304 764 190 72;
  • 33) 0.000 000 133 514 404 296 874 999 992 304 764 190 72 × 2 = 0 + 0.000 000 267 028 808 593 749 999 984 609 528 381 44;
  • 34) 0.000 000 267 028 808 593 749 999 984 609 528 381 44 × 2 = 0 + 0.000 000 534 057 617 187 499 999 969 219 056 762 88;
  • 35) 0.000 000 534 057 617 187 499 999 969 219 056 762 88 × 2 = 0 + 0.000 001 068 115 234 374 999 999 938 438 113 525 76;
  • 36) 0.000 001 068 115 234 374 999 999 938 438 113 525 76 × 2 = 0 + 0.000 002 136 230 468 749 999 999 876 876 227 051 52;
  • 37) 0.000 002 136 230 468 749 999 999 876 876 227 051 52 × 2 = 0 + 0.000 004 272 460 937 499 999 999 753 752 454 103 04;
  • 38) 0.000 004 272 460 937 499 999 999 753 752 454 103 04 × 2 = 0 + 0.000 008 544 921 874 999 999 999 507 504 908 206 08;
  • 39) 0.000 008 544 921 874 999 999 999 507 504 908 206 08 × 2 = 0 + 0.000 017 089 843 749 999 999 999 015 009 816 412 16;
  • 40) 0.000 017 089 843 749 999 999 999 015 009 816 412 16 × 2 = 0 + 0.000 034 179 687 499 999 999 998 030 019 632 824 32;
  • 41) 0.000 034 179 687 499 999 999 998 030 019 632 824 32 × 2 = 0 + 0.000 068 359 374 999 999 999 996 060 039 265 648 64;
  • 42) 0.000 068 359 374 999 999 999 996 060 039 265 648 64 × 2 = 0 + 0.000 136 718 749 999 999 999 992 120 078 531 297 28;
  • 43) 0.000 136 718 749 999 999 999 992 120 078 531 297 28 × 2 = 0 + 0.000 273 437 499 999 999 999 984 240 157 062 594 56;
  • 44) 0.000 273 437 499 999 999 999 984 240 157 062 594 56 × 2 = 0 + 0.000 546 874 999 999 999 999 968 480 314 125 189 12;
  • 45) 0.000 546 874 999 999 999 999 968 480 314 125 189 12 × 2 = 0 + 0.001 093 749 999 999 999 999 936 960 628 250 378 24;
  • 46) 0.001 093 749 999 999 999 999 936 960 628 250 378 24 × 2 = 0 + 0.002 187 499 999 999 999 999 873 921 256 500 756 48;
  • 47) 0.002 187 499 999 999 999 999 873 921 256 500 756 48 × 2 = 0 + 0.004 374 999 999 999 999 999 747 842 513 001 512 96;
  • 48) 0.004 374 999 999 999 999 999 747 842 513 001 512 96 × 2 = 0 + 0.008 749 999 999 999 999 999 495 685 026 003 025 92;
  • 49) 0.008 749 999 999 999 999 999 495 685 026 003 025 92 × 2 = 0 + 0.017 499 999 999 999 999 998 991 370 052 006 051 84;
  • 50) 0.017 499 999 999 999 999 998 991 370 052 006 051 84 × 2 = 0 + 0.034 999 999 999 999 999 997 982 740 104 012 103 68;
  • 51) 0.034 999 999 999 999 999 997 982 740 104 012 103 68 × 2 = 0 + 0.069 999 999 999 999 999 995 965 480 208 024 207 36;
  • 52) 0.069 999 999 999 999 999 995 965 480 208 024 207 36 × 2 = 0 + 0.139 999 999 999 999 999 991 930 960 416 048 414 72;
  • 53) 0.139 999 999 999 999 999 991 930 960 416 048 414 72 × 2 = 0 + 0.279 999 999 999 999 999 983 861 920 832 096 829 44;
  • 54) 0.279 999 999 999 999 999 983 861 920 832 096 829 44 × 2 = 0 + 0.559 999 999 999 999 999 967 723 841 664 193 658 88;
  • 55) 0.559 999 999 999 999 999 967 723 841 664 193 658 88 × 2 = 1 + 0.119 999 999 999 999 999 935 447 683 328 387 317 76;
  • 56) 0.119 999 999 999 999 999 935 447 683 328 387 317 76 × 2 = 0 + 0.239 999 999 999 999 999 870 895 366 656 774 635 52;
  • 57) 0.239 999 999 999 999 999 870 895 366 656 774 635 52 × 2 = 0 + 0.479 999 999 999 999 999 741 790 733 313 549 271 04;
  • 58) 0.479 999 999 999 999 999 741 790 733 313 549 271 04 × 2 = 0 + 0.959 999 999 999 999 999 483 581 466 627 098 542 08;

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:


0.016 738 891 601 562 531 086 244 689 504 383 130 07(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2)

6. Positive number before normalization:

0.016 738 891 601 562 531 086 244 689 504 383 130 07(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 531 086 244 689 504 383 130 07(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2) × 20 =

1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 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;

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) =

1017(10) =

011 1111 1001(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 =

0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000


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) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000


Decimal number -0.016 738 891 601 562 531 086 244 689 504 383 130 07 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000

» Convert decimal -0.016 738 891 601 562 531 086 244 689 504 383 129 8 to 64 bit double precision IEEE 754 binary floating point representation standard

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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