-0.000 000 000 000 176 548 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 548 8(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
-0.000 000 000 000 176 548 8(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:

|-0.000 000 000 000 176 548 8| = 0.000 000 000 000 176 548 8


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: 0.000 000 000 000 176 548 8.

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) 0.000 000 000 000 176 548 8 × 2 = 0 + 0.000 000 000 000 353 097 6;
  • 2) 0.000 000 000 000 353 097 6 × 2 = 0 + 0.000 000 000 000 706 195 2;
  • 3) 0.000 000 000 000 706 195 2 × 2 = 0 + 0.000 000 000 001 412 390 4;
  • 4) 0.000 000 000 001 412 390 4 × 2 = 0 + 0.000 000 000 002 824 780 8;
  • 5) 0.000 000 000 002 824 780 8 × 2 = 0 + 0.000 000 000 005 649 561 6;
  • 6) 0.000 000 000 005 649 561 6 × 2 = 0 + 0.000 000 000 011 299 123 2;
  • 7) 0.000 000 000 011 299 123 2 × 2 = 0 + 0.000 000 000 022 598 246 4;
  • 8) 0.000 000 000 022 598 246 4 × 2 = 0 + 0.000 000 000 045 196 492 8;
  • 9) 0.000 000 000 045 196 492 8 × 2 = 0 + 0.000 000 000 090 392 985 6;
  • 10) 0.000 000 000 090 392 985 6 × 2 = 0 + 0.000 000 000 180 785 971 2;
  • 11) 0.000 000 000 180 785 971 2 × 2 = 0 + 0.000 000 000 361 571 942 4;
  • 12) 0.000 000 000 361 571 942 4 × 2 = 0 + 0.000 000 000 723 143 884 8;
  • 13) 0.000 000 000 723 143 884 8 × 2 = 0 + 0.000 000 001 446 287 769 6;
  • 14) 0.000 000 001 446 287 769 6 × 2 = 0 + 0.000 000 002 892 575 539 2;
  • 15) 0.000 000 002 892 575 539 2 × 2 = 0 + 0.000 000 005 785 151 078 4;
  • 16) 0.000 000 005 785 151 078 4 × 2 = 0 + 0.000 000 011 570 302 156 8;
  • 17) 0.000 000 011 570 302 156 8 × 2 = 0 + 0.000 000 023 140 604 313 6;
  • 18) 0.000 000 023 140 604 313 6 × 2 = 0 + 0.000 000 046 281 208 627 2;
  • 19) 0.000 000 046 281 208 627 2 × 2 = 0 + 0.000 000 092 562 417 254 4;
  • 20) 0.000 000 092 562 417 254 4 × 2 = 0 + 0.000 000 185 124 834 508 8;
  • 21) 0.000 000 185 124 834 508 8 × 2 = 0 + 0.000 000 370 249 669 017 6;
  • 22) 0.000 000 370 249 669 017 6 × 2 = 0 + 0.000 000 740 499 338 035 2;
  • 23) 0.000 000 740 499 338 035 2 × 2 = 0 + 0.000 001 480 998 676 070 4;
  • 24) 0.000 001 480 998 676 070 4 × 2 = 0 + 0.000 002 961 997 352 140 8;
  • 25) 0.000 002 961 997 352 140 8 × 2 = 0 + 0.000 005 923 994 704 281 6;
  • 26) 0.000 005 923 994 704 281 6 × 2 = 0 + 0.000 011 847 989 408 563 2;
  • 27) 0.000 011 847 989 408 563 2 × 2 = 0 + 0.000 023 695 978 817 126 4;
  • 28) 0.000 023 695 978 817 126 4 × 2 = 0 + 0.000 047 391 957 634 252 8;
  • 29) 0.000 047 391 957 634 252 8 × 2 = 0 + 0.000 094 783 915 268 505 6;
  • 30) 0.000 094 783 915 268 505 6 × 2 = 0 + 0.000 189 567 830 537 011 2;
  • 31) 0.000 189 567 830 537 011 2 × 2 = 0 + 0.000 379 135 661 074 022 4;
  • 32) 0.000 379 135 661 074 022 4 × 2 = 0 + 0.000 758 271 322 148 044 8;
  • 33) 0.000 758 271 322 148 044 8 × 2 = 0 + 0.001 516 542 644 296 089 6;
  • 34) 0.001 516 542 644 296 089 6 × 2 = 0 + 0.003 033 085 288 592 179 2;
  • 35) 0.003 033 085 288 592 179 2 × 2 = 0 + 0.006 066 170 577 184 358 4;
  • 36) 0.006 066 170 577 184 358 4 × 2 = 0 + 0.012 132 341 154 368 716 8;
  • 37) 0.012 132 341 154 368 716 8 × 2 = 0 + 0.024 264 682 308 737 433 6;
  • 38) 0.024 264 682 308 737 433 6 × 2 = 0 + 0.048 529 364 617 474 867 2;
  • 39) 0.048 529 364 617 474 867 2 × 2 = 0 + 0.097 058 729 234 949 734 4;
  • 40) 0.097 058 729 234 949 734 4 × 2 = 0 + 0.194 117 458 469 899 468 8;
  • 41) 0.194 117 458 469 899 468 8 × 2 = 0 + 0.388 234 916 939 798 937 6;
  • 42) 0.388 234 916 939 798 937 6 × 2 = 0 + 0.776 469 833 879 597 875 2;
  • 43) 0.776 469 833 879 597 875 2 × 2 = 1 + 0.552 939 667 759 195 750 4;
  • 44) 0.552 939 667 759 195 750 4 × 2 = 1 + 0.105 879 335 518 391 500 8;
  • 45) 0.105 879 335 518 391 500 8 × 2 = 0 + 0.211 758 671 036 783 001 6;
  • 46) 0.211 758 671 036 783 001 6 × 2 = 0 + 0.423 517 342 073 566 003 2;
  • 47) 0.423 517 342 073 566 003 2 × 2 = 0 + 0.847 034 684 147 132 006 4;
  • 48) 0.847 034 684 147 132 006 4 × 2 = 1 + 0.694 069 368 294 264 012 8;
  • 49) 0.694 069 368 294 264 012 8 × 2 = 1 + 0.388 138 736 588 528 025 6;
  • 50) 0.388 138 736 588 528 025 6 × 2 = 0 + 0.776 277 473 177 056 051 2;
  • 51) 0.776 277 473 177 056 051 2 × 2 = 1 + 0.552 554 946 354 112 102 4;
  • 52) 0.552 554 946 354 112 102 4 × 2 = 1 + 0.105 109 892 708 224 204 8;
  • 53) 0.105 109 892 708 224 204 8 × 2 = 0 + 0.210 219 785 416 448 409 6;
  • 54) 0.210 219 785 416 448 409 6 × 2 = 0 + 0.420 439 570 832 896 819 2;
  • 55) 0.420 439 570 832 896 819 2 × 2 = 0 + 0.840 879 141 665 793 638 4;
  • 56) 0.840 879 141 665 793 638 4 × 2 = 1 + 0.681 758 283 331 587 276 8;
  • 57) 0.681 758 283 331 587 276 8 × 2 = 1 + 0.363 516 566 663 174 553 6;
  • 58) 0.363 516 566 663 174 553 6 × 2 = 0 + 0.727 033 133 326 349 107 2;
  • 59) 0.727 033 133 326 349 107 2 × 2 = 1 + 0.454 066 266 652 698 214 4;
  • 60) 0.454 066 266 652 698 214 4 × 2 = 0 + 0.908 132 533 305 396 428 8;
  • 61) 0.908 132 533 305 396 428 8 × 2 = 1 + 0.816 265 066 610 792 857 6;
  • 62) 0.816 265 066 610 792 857 6 × 2 = 1 + 0.632 530 133 221 585 715 2;
  • 63) 0.632 530 133 221 585 715 2 × 2 = 1 + 0.265 060 266 443 171 430 4;
  • 64) 0.265 060 266 443 171 430 4 × 2 = 0 + 0.530 120 532 886 342 860 8;
  • 65) 0.530 120 532 886 342 860 8 × 2 = 1 + 0.060 241 065 772 685 721 6;
  • 66) 0.060 241 065 772 685 721 6 × 2 = 0 + 0.120 482 131 545 371 443 2;
  • 67) 0.120 482 131 545 371 443 2 × 2 = 0 + 0.240 964 263 090 742 886 4;
  • 68) 0.240 964 263 090 742 886 4 × 2 = 0 + 0.481 928 526 181 485 772 8;
  • 69) 0.481 928 526 181 485 772 8 × 2 = 0 + 0.963 857 052 362 971 545 6;
  • 70) 0.963 857 052 362 971 545 6 × 2 = 1 + 0.927 714 104 725 943 091 2;
  • 71) 0.927 714 104 725 943 091 2 × 2 = 1 + 0.855 428 209 451 886 182 4;
  • 72) 0.855 428 209 451 886 182 4 × 2 = 1 + 0.710 856 418 903 772 364 8;
  • 73) 0.710 856 418 903 772 364 8 × 2 = 1 + 0.421 712 837 807 544 729 6;
  • 74) 0.421 712 837 807 544 729 6 × 2 = 0 + 0.843 425 675 615 089 459 2;
  • 75) 0.843 425 675 615 089 459 2 × 2 = 1 + 0.686 851 351 230 178 918 4;
  • 76) 0.686 851 351 230 178 918 4 × 2 = 1 + 0.373 702 702 460 357 836 8;
  • 77) 0.373 702 702 460 357 836 8 × 2 = 0 + 0.747 405 404 920 715 673 6;
  • 78) 0.747 405 404 920 715 673 6 × 2 = 1 + 0.494 810 809 841 431 347 2;
  • 79) 0.494 810 809 841 431 347 2 × 2 = 0 + 0.989 621 619 682 862 694 4;
  • 80) 0.989 621 619 682 862 694 4 × 2 = 1 + 0.979 243 239 365 725 388 8;
  • 81) 0.979 243 239 365 725 388 8 × 2 = 1 + 0.958 486 478 731 450 777 6;
  • 82) 0.958 486 478 731 450 777 6 × 2 = 1 + 0.916 972 957 462 901 555 2;
  • 83) 0.916 972 957 462 901 555 2 × 2 = 1 + 0.833 945 914 925 803 110 4;
  • 84) 0.833 945 914 925 803 110 4 × 2 = 1 + 0.667 891 829 851 606 220 8;
  • 85) 0.667 891 829 851 606 220 8 × 2 = 1 + 0.335 783 659 703 212 441 6;
  • 86) 0.335 783 659 703 212 441 6 × 2 = 0 + 0.671 567 319 406 424 883 2;
  • 87) 0.671 567 319 406 424 883 2 × 2 = 1 + 0.343 134 638 812 849 766 4;
  • 88) 0.343 134 638 812 849 766 4 × 2 = 0 + 0.686 269 277 625 699 532 8;
  • 89) 0.686 269 277 625 699 532 8 × 2 = 1 + 0.372 538 555 251 399 065 6;
  • 90) 0.372 538 555 251 399 065 6 × 2 = 0 + 0.745 077 110 502 798 131 2;
  • 91) 0.745 077 110 502 798 131 2 × 2 = 1 + 0.490 154 221 005 596 262 4;
  • 92) 0.490 154 221 005 596 262 4 × 2 = 0 + 0.980 308 442 011 192 524 8;
  • 93) 0.980 308 442 011 192 524 8 × 2 = 1 + 0.960 616 884 022 385 049 6;
  • 94) 0.960 616 884 022 385 049 6 × 2 = 1 + 0.921 233 768 044 770 099 2;
  • 95) 0.921 233 768 044 770 099 2 × 2 = 1 + 0.842 467 536 089 540 198 4;

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:


0.000 000 000 000 176 548 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1010 1110 1000 0111 1011 0101 1111 1010 1010 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 548 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1010 1110 1000 0111 1011 0101 1111 1010 1010 111(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:


0.000 000 000 000 176 548 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1010 1110 1000 0111 1011 0101 1111 1010 1010 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1010 1110 1000 0111 1011 0101 1111 1010 1010 111(2) × 20 =

1.1000 1101 1000 1101 0111 0100 0011 1101 1010 1111 1101 0101 0111(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 1000 1101 0111 0100 0011 1101 1010 1111 1101 0101 0111


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 1000 1101 0111 0100 0011 1101 1010 1111 1101 0101 0111 =

1000 1101 1000 1101 0111 0100 0011 1101 1010 1111 1101 0101 0111


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 1000 1101 0111 0100 0011 1101 1010 1111 1101 0101 0111


Decimal number -0.000 000 000 000 176 548 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1101 0111 0100 0011 1101 1010 1111 1101 0101 0111

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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