-0.000 000 000 000 176 556 68 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 68(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
-0.000 000 000 000 176 556 68(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:

|-0.000 000 000 000 176 556 68| = 0.000 000 000 000 176 556 68


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: 0.000 000 000 000 176 556 68.

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) 0.000 000 000 000 176 556 68 × 2 = 0 + 0.000 000 000 000 353 113 36;
  • 2) 0.000 000 000 000 353 113 36 × 2 = 0 + 0.000 000 000 000 706 226 72;
  • 3) 0.000 000 000 000 706 226 72 × 2 = 0 + 0.000 000 000 001 412 453 44;
  • 4) 0.000 000 000 001 412 453 44 × 2 = 0 + 0.000 000 000 002 824 906 88;
  • 5) 0.000 000 000 002 824 906 88 × 2 = 0 + 0.000 000 000 005 649 813 76;
  • 6) 0.000 000 000 005 649 813 76 × 2 = 0 + 0.000 000 000 011 299 627 52;
  • 7) 0.000 000 000 011 299 627 52 × 2 = 0 + 0.000 000 000 022 599 255 04;
  • 8) 0.000 000 000 022 599 255 04 × 2 = 0 + 0.000 000 000 045 198 510 08;
  • 9) 0.000 000 000 045 198 510 08 × 2 = 0 + 0.000 000 000 090 397 020 16;
  • 10) 0.000 000 000 090 397 020 16 × 2 = 0 + 0.000 000 000 180 794 040 32;
  • 11) 0.000 000 000 180 794 040 32 × 2 = 0 + 0.000 000 000 361 588 080 64;
  • 12) 0.000 000 000 361 588 080 64 × 2 = 0 + 0.000 000 000 723 176 161 28;
  • 13) 0.000 000 000 723 176 161 28 × 2 = 0 + 0.000 000 001 446 352 322 56;
  • 14) 0.000 000 001 446 352 322 56 × 2 = 0 + 0.000 000 002 892 704 645 12;
  • 15) 0.000 000 002 892 704 645 12 × 2 = 0 + 0.000 000 005 785 409 290 24;
  • 16) 0.000 000 005 785 409 290 24 × 2 = 0 + 0.000 000 011 570 818 580 48;
  • 17) 0.000 000 011 570 818 580 48 × 2 = 0 + 0.000 000 023 141 637 160 96;
  • 18) 0.000 000 023 141 637 160 96 × 2 = 0 + 0.000 000 046 283 274 321 92;
  • 19) 0.000 000 046 283 274 321 92 × 2 = 0 + 0.000 000 092 566 548 643 84;
  • 20) 0.000 000 092 566 548 643 84 × 2 = 0 + 0.000 000 185 133 097 287 68;
  • 21) 0.000 000 185 133 097 287 68 × 2 = 0 + 0.000 000 370 266 194 575 36;
  • 22) 0.000 000 370 266 194 575 36 × 2 = 0 + 0.000 000 740 532 389 150 72;
  • 23) 0.000 000 740 532 389 150 72 × 2 = 0 + 0.000 001 481 064 778 301 44;
  • 24) 0.000 001 481 064 778 301 44 × 2 = 0 + 0.000 002 962 129 556 602 88;
  • 25) 0.000 002 962 129 556 602 88 × 2 = 0 + 0.000 005 924 259 113 205 76;
  • 26) 0.000 005 924 259 113 205 76 × 2 = 0 + 0.000 011 848 518 226 411 52;
  • 27) 0.000 011 848 518 226 411 52 × 2 = 0 + 0.000 023 697 036 452 823 04;
  • 28) 0.000 023 697 036 452 823 04 × 2 = 0 + 0.000 047 394 072 905 646 08;
  • 29) 0.000 047 394 072 905 646 08 × 2 = 0 + 0.000 094 788 145 811 292 16;
  • 30) 0.000 094 788 145 811 292 16 × 2 = 0 + 0.000 189 576 291 622 584 32;
  • 31) 0.000 189 576 291 622 584 32 × 2 = 0 + 0.000 379 152 583 245 168 64;
  • 32) 0.000 379 152 583 245 168 64 × 2 = 0 + 0.000 758 305 166 490 337 28;
  • 33) 0.000 758 305 166 490 337 28 × 2 = 0 + 0.001 516 610 332 980 674 56;
  • 34) 0.001 516 610 332 980 674 56 × 2 = 0 + 0.003 033 220 665 961 349 12;
  • 35) 0.003 033 220 665 961 349 12 × 2 = 0 + 0.006 066 441 331 922 698 24;
  • 36) 0.006 066 441 331 922 698 24 × 2 = 0 + 0.012 132 882 663 845 396 48;
  • 37) 0.012 132 882 663 845 396 48 × 2 = 0 + 0.024 265 765 327 690 792 96;
  • 38) 0.024 265 765 327 690 792 96 × 2 = 0 + 0.048 531 530 655 381 585 92;
  • 39) 0.048 531 530 655 381 585 92 × 2 = 0 + 0.097 063 061 310 763 171 84;
  • 40) 0.097 063 061 310 763 171 84 × 2 = 0 + 0.194 126 122 621 526 343 68;
  • 41) 0.194 126 122 621 526 343 68 × 2 = 0 + 0.388 252 245 243 052 687 36;
  • 42) 0.388 252 245 243 052 687 36 × 2 = 0 + 0.776 504 490 486 105 374 72;
  • 43) 0.776 504 490 486 105 374 72 × 2 = 1 + 0.553 008 980 972 210 749 44;
  • 44) 0.553 008 980 972 210 749 44 × 2 = 1 + 0.106 017 961 944 421 498 88;
  • 45) 0.106 017 961 944 421 498 88 × 2 = 0 + 0.212 035 923 888 842 997 76;
  • 46) 0.212 035 923 888 842 997 76 × 2 = 0 + 0.424 071 847 777 685 995 52;
  • 47) 0.424 071 847 777 685 995 52 × 2 = 0 + 0.848 143 695 555 371 991 04;
  • 48) 0.848 143 695 555 371 991 04 × 2 = 1 + 0.696 287 391 110 743 982 08;
  • 49) 0.696 287 391 110 743 982 08 × 2 = 1 + 0.392 574 782 221 487 964 16;
  • 50) 0.392 574 782 221 487 964 16 × 2 = 0 + 0.785 149 564 442 975 928 32;
  • 51) 0.785 149 564 442 975 928 32 × 2 = 1 + 0.570 299 128 885 951 856 64;
  • 52) 0.570 299 128 885 951 856 64 × 2 = 1 + 0.140 598 257 771 903 713 28;
  • 53) 0.140 598 257 771 903 713 28 × 2 = 0 + 0.281 196 515 543 807 426 56;
  • 54) 0.281 196 515 543 807 426 56 × 2 = 0 + 0.562 393 031 087 614 853 12;
  • 55) 0.562 393 031 087 614 853 12 × 2 = 1 + 0.124 786 062 175 229 706 24;
  • 56) 0.124 786 062 175 229 706 24 × 2 = 0 + 0.249 572 124 350 459 412 48;
  • 57) 0.249 572 124 350 459 412 48 × 2 = 0 + 0.499 144 248 700 918 824 96;
  • 58) 0.499 144 248 700 918 824 96 × 2 = 0 + 0.998 288 497 401 837 649 92;
  • 59) 0.998 288 497 401 837 649 92 × 2 = 1 + 0.996 576 994 803 675 299 84;
  • 60) 0.996 576 994 803 675 299 84 × 2 = 1 + 0.993 153 989 607 350 599 68;
  • 61) 0.993 153 989 607 350 599 68 × 2 = 1 + 0.986 307 979 214 701 199 36;
  • 62) 0.986 307 979 214 701 199 36 × 2 = 1 + 0.972 615 958 429 402 398 72;
  • 63) 0.972 615 958 429 402 398 72 × 2 = 1 + 0.945 231 916 858 804 797 44;
  • 64) 0.945 231 916 858 804 797 44 × 2 = 1 + 0.890 463 833 717 609 594 88;
  • 65) 0.890 463 833 717 609 594 88 × 2 = 1 + 0.780 927 667 435 219 189 76;
  • 66) 0.780 927 667 435 219 189 76 × 2 = 1 + 0.561 855 334 870 438 379 52;
  • 67) 0.561 855 334 870 438 379 52 × 2 = 1 + 0.123 710 669 740 876 759 04;
  • 68) 0.123 710 669 740 876 759 04 × 2 = 0 + 0.247 421 339 481 753 518 08;
  • 69) 0.247 421 339 481 753 518 08 × 2 = 0 + 0.494 842 678 963 507 036 16;
  • 70) 0.494 842 678 963 507 036 16 × 2 = 0 + 0.989 685 357 927 014 072 32;
  • 71) 0.989 685 357 927 014 072 32 × 2 = 1 + 0.979 370 715 854 028 144 64;
  • 72) 0.979 370 715 854 028 144 64 × 2 = 1 + 0.958 741 431 708 056 289 28;
  • 73) 0.958 741 431 708 056 289 28 × 2 = 1 + 0.917 482 863 416 112 578 56;
  • 74) 0.917 482 863 416 112 578 56 × 2 = 1 + 0.834 965 726 832 225 157 12;
  • 75) 0.834 965 726 832 225 157 12 × 2 = 1 + 0.669 931 453 664 450 314 24;
  • 76) 0.669 931 453 664 450 314 24 × 2 = 1 + 0.339 862 907 328 900 628 48;
  • 77) 0.339 862 907 328 900 628 48 × 2 = 0 + 0.679 725 814 657 801 256 96;
  • 78) 0.679 725 814 657 801 256 96 × 2 = 1 + 0.359 451 629 315 602 513 92;
  • 79) 0.359 451 629 315 602 513 92 × 2 = 0 + 0.718 903 258 631 205 027 84;
  • 80) 0.718 903 258 631 205 027 84 × 2 = 1 + 0.437 806 517 262 410 055 68;
  • 81) 0.437 806 517 262 410 055 68 × 2 = 0 + 0.875 613 034 524 820 111 36;
  • 82) 0.875 613 034 524 820 111 36 × 2 = 1 + 0.751 226 069 049 640 222 72;
  • 83) 0.751 226 069 049 640 222 72 × 2 = 1 + 0.502 452 138 099 280 445 44;
  • 84) 0.502 452 138 099 280 445 44 × 2 = 1 + 0.004 904 276 198 560 890 88;
  • 85) 0.004 904 276 198 560 890 88 × 2 = 0 + 0.009 808 552 397 121 781 76;
  • 86) 0.009 808 552 397 121 781 76 × 2 = 0 + 0.019 617 104 794 243 563 52;
  • 87) 0.019 617 104 794 243 563 52 × 2 = 0 + 0.039 234 209 588 487 127 04;
  • 88) 0.039 234 209 588 487 127 04 × 2 = 0 + 0.078 468 419 176 974 254 08;
  • 89) 0.078 468 419 176 974 254 08 × 2 = 0 + 0.156 936 838 353 948 508 16;
  • 90) 0.156 936 838 353 948 508 16 × 2 = 0 + 0.313 873 676 707 897 016 32;
  • 91) 0.313 873 676 707 897 016 32 × 2 = 0 + 0.627 747 353 415 794 032 64;
  • 92) 0.627 747 353 415 794 032 64 × 2 = 1 + 0.255 494 706 831 588 065 28;
  • 93) 0.255 494 706 831 588 065 28 × 2 = 0 + 0.510 989 413 663 176 130 56;
  • 94) 0.510 989 413 663 176 130 56 × 2 = 1 + 0.021 978 827 326 352 261 12;
  • 95) 0.021 978 827 326 352 261 12 × 2 = 0 + 0.043 957 654 652 704 522 24;

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:


0.000 000 000 000 176 556 68(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1111 1110 0011 1111 0101 0111 0000 0001 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 68(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1111 1110 0011 1111 0101 0111 0000 0001 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 556 68(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1111 1110 0011 1111 0101 0111 0000 0001 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1111 1110 0011 1111 0101 0111 0000 0001 010(2) × 20 =

1.1000 1101 1001 0001 1111 1111 0001 1111 1010 1011 1000 0000 1010(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 1001 0001 1111 1111 0001 1111 1010 1011 1000 0000 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

980(10) =

011 1101 0100(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. 1000 1101 1001 0001 1111 1111 0001 1111 1010 1011 1000 0000 1010 =

1000 1101 1001 0001 1111 1111 0001 1111 1010 1011 1000 0000 1010


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 1101 0100

Mantissa (52 bits) =
1000 1101 1001 0001 1111 1111 0001 1111 1010 1011 1000 0000 1010


Decimal number -0.000 000 000 000 176 556 68 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 1111 1111 0001 1111 1010 1011 1000 0000 1010

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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