-0.016 738 891 601 562 531 086 244 689 504 383 130 43 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 43(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 43(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 43| = 0.016 738 891 601 562 531 086 244 689 504 383 130 43


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 43.

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 43 × 2 = 0 + 0.033 477 783 203 125 062 172 489 379 008 766 260 86;
  • 2) 0.033 477 783 203 125 062 172 489 379 008 766 260 86 × 2 = 0 + 0.066 955 566 406 250 124 344 978 758 017 532 521 72;
  • 3) 0.066 955 566 406 250 124 344 978 758 017 532 521 72 × 2 = 0 + 0.133 911 132 812 500 248 689 957 516 035 065 043 44;
  • 4) 0.133 911 132 812 500 248 689 957 516 035 065 043 44 × 2 = 0 + 0.267 822 265 625 000 497 379 915 032 070 130 086 88;
  • 5) 0.267 822 265 625 000 497 379 915 032 070 130 086 88 × 2 = 0 + 0.535 644 531 250 000 994 759 830 064 140 260 173 76;
  • 6) 0.535 644 531 250 000 994 759 830 064 140 260 173 76 × 2 = 1 + 0.071 289 062 500 001 989 519 660 128 280 520 347 52;
  • 7) 0.071 289 062 500 001 989 519 660 128 280 520 347 52 × 2 = 0 + 0.142 578 125 000 003 979 039 320 256 561 040 695 04;
  • 8) 0.142 578 125 000 003 979 039 320 256 561 040 695 04 × 2 = 0 + 0.285 156 250 000 007 958 078 640 513 122 081 390 08;
  • 9) 0.285 156 250 000 007 958 078 640 513 122 081 390 08 × 2 = 0 + 0.570 312 500 000 015 916 157 281 026 244 162 780 16;
  • 10) 0.570 312 500 000 015 916 157 281 026 244 162 780 16 × 2 = 1 + 0.140 625 000 000 031 832 314 562 052 488 325 560 32;
  • 11) 0.140 625 000 000 031 832 314 562 052 488 325 560 32 × 2 = 0 + 0.281 250 000 000 063 664 629 124 104 976 651 120 64;
  • 12) 0.281 250 000 000 063 664 629 124 104 976 651 120 64 × 2 = 0 + 0.562 500 000 000 127 329 258 248 209 953 302 241 28;
  • 13) 0.562 500 000 000 127 329 258 248 209 953 302 241 28 × 2 = 1 + 0.125 000 000 000 254 658 516 496 419 906 604 482 56;
  • 14) 0.125 000 000 000 254 658 516 496 419 906 604 482 56 × 2 = 0 + 0.250 000 000 000 509 317 032 992 839 813 208 965 12;
  • 15) 0.250 000 000 000 509 317 032 992 839 813 208 965 12 × 2 = 0 + 0.500 000 000 001 018 634 065 985 679 626 417 930 24;
  • 16) 0.500 000 000 001 018 634 065 985 679 626 417 930 24 × 2 = 1 + 0.000 000 000 002 037 268 131 971 359 252 835 860 48;
  • 17) 0.000 000 000 002 037 268 131 971 359 252 835 860 48 × 2 = 0 + 0.000 000 000 004 074 536 263 942 718 505 671 720 96;
  • 18) 0.000 000 000 004 074 536 263 942 718 505 671 720 96 × 2 = 0 + 0.000 000 000 008 149 072 527 885 437 011 343 441 92;
  • 19) 0.000 000 000 008 149 072 527 885 437 011 343 441 92 × 2 = 0 + 0.000 000 000 016 298 145 055 770 874 022 686 883 84;
  • 20) 0.000 000 000 016 298 145 055 770 874 022 686 883 84 × 2 = 0 + 0.000 000 000 032 596 290 111 541 748 045 373 767 68;
  • 21) 0.000 000 000 032 596 290 111 541 748 045 373 767 68 × 2 = 0 + 0.000 000 000 065 192 580 223 083 496 090 747 535 36;
  • 22) 0.000 000 000 065 192 580 223 083 496 090 747 535 36 × 2 = 0 + 0.000 000 000 130 385 160 446 166 992 181 495 070 72;
  • 23) 0.000 000 000 130 385 160 446 166 992 181 495 070 72 × 2 = 0 + 0.000 000 000 260 770 320 892 333 984 362 990 141 44;
  • 24) 0.000 000 000 260 770 320 892 333 984 362 990 141 44 × 2 = 0 + 0.000 000 000 521 540 641 784 667 968 725 980 282 88;
  • 25) 0.000 000 000 521 540 641 784 667 968 725 980 282 88 × 2 = 0 + 0.000 000 001 043 081 283 569 335 937 451 960 565 76;
  • 26) 0.000 000 001 043 081 283 569 335 937 451 960 565 76 × 2 = 0 + 0.000 000 002 086 162 567 138 671 874 903 921 131 52;
  • 27) 0.000 000 002 086 162 567 138 671 874 903 921 131 52 × 2 = 0 + 0.000 000 004 172 325 134 277 343 749 807 842 263 04;
  • 28) 0.000 000 004 172 325 134 277 343 749 807 842 263 04 × 2 = 0 + 0.000 000 008 344 650 268 554 687 499 615 684 526 08;
  • 29) 0.000 000 008 344 650 268 554 687 499 615 684 526 08 × 2 = 0 + 0.000 000 016 689 300 537 109 374 999 231 369 052 16;
  • 30) 0.000 000 016 689 300 537 109 374 999 231 369 052 16 × 2 = 0 + 0.000 000 033 378 601 074 218 749 998 462 738 104 32;
  • 31) 0.000 000 033 378 601 074 218 749 998 462 738 104 32 × 2 = 0 + 0.000 000 066 757 202 148 437 499 996 925 476 208 64;
  • 32) 0.000 000 066 757 202 148 437 499 996 925 476 208 64 × 2 = 0 + 0.000 000 133 514 404 296 874 999 993 850 952 417 28;
  • 33) 0.000 000 133 514 404 296 874 999 993 850 952 417 28 × 2 = 0 + 0.000 000 267 028 808 593 749 999 987 701 904 834 56;
  • 34) 0.000 000 267 028 808 593 749 999 987 701 904 834 56 × 2 = 0 + 0.000 000 534 057 617 187 499 999 975 403 809 669 12;
  • 35) 0.000 000 534 057 617 187 499 999 975 403 809 669 12 × 2 = 0 + 0.000 001 068 115 234 374 999 999 950 807 619 338 24;
  • 36) 0.000 001 068 115 234 374 999 999 950 807 619 338 24 × 2 = 0 + 0.000 002 136 230 468 749 999 999 901 615 238 676 48;
  • 37) 0.000 002 136 230 468 749 999 999 901 615 238 676 48 × 2 = 0 + 0.000 004 272 460 937 499 999 999 803 230 477 352 96;
  • 38) 0.000 004 272 460 937 499 999 999 803 230 477 352 96 × 2 = 0 + 0.000 008 544 921 874 999 999 999 606 460 954 705 92;
  • 39) 0.000 008 544 921 874 999 999 999 606 460 954 705 92 × 2 = 0 + 0.000 017 089 843 749 999 999 999 212 921 909 411 84;
  • 40) 0.000 017 089 843 749 999 999 999 212 921 909 411 84 × 2 = 0 + 0.000 034 179 687 499 999 999 998 425 843 818 823 68;
  • 41) 0.000 034 179 687 499 999 999 998 425 843 818 823 68 × 2 = 0 + 0.000 068 359 374 999 999 999 996 851 687 637 647 36;
  • 42) 0.000 068 359 374 999 999 999 996 851 687 637 647 36 × 2 = 0 + 0.000 136 718 749 999 999 999 993 703 375 275 294 72;
  • 43) 0.000 136 718 749 999 999 999 993 703 375 275 294 72 × 2 = 0 + 0.000 273 437 499 999 999 999 987 406 750 550 589 44;
  • 44) 0.000 273 437 499 999 999 999 987 406 750 550 589 44 × 2 = 0 + 0.000 546 874 999 999 999 999 974 813 501 101 178 88;
  • 45) 0.000 546 874 999 999 999 999 974 813 501 101 178 88 × 2 = 0 + 0.001 093 749 999 999 999 999 949 627 002 202 357 76;
  • 46) 0.001 093 749 999 999 999 999 949 627 002 202 357 76 × 2 = 0 + 0.002 187 499 999 999 999 999 899 254 004 404 715 52;
  • 47) 0.002 187 499 999 999 999 999 899 254 004 404 715 52 × 2 = 0 + 0.004 374 999 999 999 999 999 798 508 008 809 431 04;
  • 48) 0.004 374 999 999 999 999 999 798 508 008 809 431 04 × 2 = 0 + 0.008 749 999 999 999 999 999 597 016 017 618 862 08;
  • 49) 0.008 749 999 999 999 999 999 597 016 017 618 862 08 × 2 = 0 + 0.017 499 999 999 999 999 999 194 032 035 237 724 16;
  • 50) 0.017 499 999 999 999 999 999 194 032 035 237 724 16 × 2 = 0 + 0.034 999 999 999 999 999 998 388 064 070 475 448 32;
  • 51) 0.034 999 999 999 999 999 998 388 064 070 475 448 32 × 2 = 0 + 0.069 999 999 999 999 999 996 776 128 140 950 896 64;
  • 52) 0.069 999 999 999 999 999 996 776 128 140 950 896 64 × 2 = 0 + 0.139 999 999 999 999 999 993 552 256 281 901 793 28;
  • 53) 0.139 999 999 999 999 999 993 552 256 281 901 793 28 × 2 = 0 + 0.279 999 999 999 999 999 987 104 512 563 803 586 56;
  • 54) 0.279 999 999 999 999 999 987 104 512 563 803 586 56 × 2 = 0 + 0.559 999 999 999 999 999 974 209 025 127 607 173 12;
  • 55) 0.559 999 999 999 999 999 974 209 025 127 607 173 12 × 2 = 1 + 0.119 999 999 999 999 999 948 418 050 255 214 346 24;
  • 56) 0.119 999 999 999 999 999 948 418 050 255 214 346 24 × 2 = 0 + 0.239 999 999 999 999 999 896 836 100 510 428 692 48;
  • 57) 0.239 999 999 999 999 999 896 836 100 510 428 692 48 × 2 = 0 + 0.479 999 999 999 999 999 793 672 201 020 857 384 96;
  • 58) 0.479 999 999 999 999 999 793 672 201 020 857 384 96 × 2 = 0 + 0.959 999 999 999 999 999 587 344 402 041 714 769 92;

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 43(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 43(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 43(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 43 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 -0.016 738 891 601 562 531 086 244 689 504 383 130 35 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