-0.000 000 000 000 176 564 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 564 7(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
-0.000 000 000 000 176 564 7(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:

|-0.000 000 000 000 176 564 7| = 0.000 000 000 000 176 564 7


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: 0.000 000 000 000 176 564 7.

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) 0.000 000 000 000 176 564 7 × 2 = 0 + 0.000 000 000 000 353 129 4;
  • 2) 0.000 000 000 000 353 129 4 × 2 = 0 + 0.000 000 000 000 706 258 8;
  • 3) 0.000 000 000 000 706 258 8 × 2 = 0 + 0.000 000 000 001 412 517 6;
  • 4) 0.000 000 000 001 412 517 6 × 2 = 0 + 0.000 000 000 002 825 035 2;
  • 5) 0.000 000 000 002 825 035 2 × 2 = 0 + 0.000 000 000 005 650 070 4;
  • 6) 0.000 000 000 005 650 070 4 × 2 = 0 + 0.000 000 000 011 300 140 8;
  • 7) 0.000 000 000 011 300 140 8 × 2 = 0 + 0.000 000 000 022 600 281 6;
  • 8) 0.000 000 000 022 600 281 6 × 2 = 0 + 0.000 000 000 045 200 563 2;
  • 9) 0.000 000 000 045 200 563 2 × 2 = 0 + 0.000 000 000 090 401 126 4;
  • 10) 0.000 000 000 090 401 126 4 × 2 = 0 + 0.000 000 000 180 802 252 8;
  • 11) 0.000 000 000 180 802 252 8 × 2 = 0 + 0.000 000 000 361 604 505 6;
  • 12) 0.000 000 000 361 604 505 6 × 2 = 0 + 0.000 000 000 723 209 011 2;
  • 13) 0.000 000 000 723 209 011 2 × 2 = 0 + 0.000 000 001 446 418 022 4;
  • 14) 0.000 000 001 446 418 022 4 × 2 = 0 + 0.000 000 002 892 836 044 8;
  • 15) 0.000 000 002 892 836 044 8 × 2 = 0 + 0.000 000 005 785 672 089 6;
  • 16) 0.000 000 005 785 672 089 6 × 2 = 0 + 0.000 000 011 571 344 179 2;
  • 17) 0.000 000 011 571 344 179 2 × 2 = 0 + 0.000 000 023 142 688 358 4;
  • 18) 0.000 000 023 142 688 358 4 × 2 = 0 + 0.000 000 046 285 376 716 8;
  • 19) 0.000 000 046 285 376 716 8 × 2 = 0 + 0.000 000 092 570 753 433 6;
  • 20) 0.000 000 092 570 753 433 6 × 2 = 0 + 0.000 000 185 141 506 867 2;
  • 21) 0.000 000 185 141 506 867 2 × 2 = 0 + 0.000 000 370 283 013 734 4;
  • 22) 0.000 000 370 283 013 734 4 × 2 = 0 + 0.000 000 740 566 027 468 8;
  • 23) 0.000 000 740 566 027 468 8 × 2 = 0 + 0.000 001 481 132 054 937 6;
  • 24) 0.000 001 481 132 054 937 6 × 2 = 0 + 0.000 002 962 264 109 875 2;
  • 25) 0.000 002 962 264 109 875 2 × 2 = 0 + 0.000 005 924 528 219 750 4;
  • 26) 0.000 005 924 528 219 750 4 × 2 = 0 + 0.000 011 849 056 439 500 8;
  • 27) 0.000 011 849 056 439 500 8 × 2 = 0 + 0.000 023 698 112 879 001 6;
  • 28) 0.000 023 698 112 879 001 6 × 2 = 0 + 0.000 047 396 225 758 003 2;
  • 29) 0.000 047 396 225 758 003 2 × 2 = 0 + 0.000 094 792 451 516 006 4;
  • 30) 0.000 094 792 451 516 006 4 × 2 = 0 + 0.000 189 584 903 032 012 8;
  • 31) 0.000 189 584 903 032 012 8 × 2 = 0 + 0.000 379 169 806 064 025 6;
  • 32) 0.000 379 169 806 064 025 6 × 2 = 0 + 0.000 758 339 612 128 051 2;
  • 33) 0.000 758 339 612 128 051 2 × 2 = 0 + 0.001 516 679 224 256 102 4;
  • 34) 0.001 516 679 224 256 102 4 × 2 = 0 + 0.003 033 358 448 512 204 8;
  • 35) 0.003 033 358 448 512 204 8 × 2 = 0 + 0.006 066 716 897 024 409 6;
  • 36) 0.006 066 716 897 024 409 6 × 2 = 0 + 0.012 133 433 794 048 819 2;
  • 37) 0.012 133 433 794 048 819 2 × 2 = 0 + 0.024 266 867 588 097 638 4;
  • 38) 0.024 266 867 588 097 638 4 × 2 = 0 + 0.048 533 735 176 195 276 8;
  • 39) 0.048 533 735 176 195 276 8 × 2 = 0 + 0.097 067 470 352 390 553 6;
  • 40) 0.097 067 470 352 390 553 6 × 2 = 0 + 0.194 134 940 704 781 107 2;
  • 41) 0.194 134 940 704 781 107 2 × 2 = 0 + 0.388 269 881 409 562 214 4;
  • 42) 0.388 269 881 409 562 214 4 × 2 = 0 + 0.776 539 762 819 124 428 8;
  • 43) 0.776 539 762 819 124 428 8 × 2 = 1 + 0.553 079 525 638 248 857 6;
  • 44) 0.553 079 525 638 248 857 6 × 2 = 1 + 0.106 159 051 276 497 715 2;
  • 45) 0.106 159 051 276 497 715 2 × 2 = 0 + 0.212 318 102 552 995 430 4;
  • 46) 0.212 318 102 552 995 430 4 × 2 = 0 + 0.424 636 205 105 990 860 8;
  • 47) 0.424 636 205 105 990 860 8 × 2 = 0 + 0.849 272 410 211 981 721 6;
  • 48) 0.849 272 410 211 981 721 6 × 2 = 1 + 0.698 544 820 423 963 443 2;
  • 49) 0.698 544 820 423 963 443 2 × 2 = 1 + 0.397 089 640 847 926 886 4;
  • 50) 0.397 089 640 847 926 886 4 × 2 = 0 + 0.794 179 281 695 853 772 8;
  • 51) 0.794 179 281 695 853 772 8 × 2 = 1 + 0.588 358 563 391 707 545 6;
  • 52) 0.588 358 563 391 707 545 6 × 2 = 1 + 0.176 717 126 783 415 091 2;
  • 53) 0.176 717 126 783 415 091 2 × 2 = 0 + 0.353 434 253 566 830 182 4;
  • 54) 0.353 434 253 566 830 182 4 × 2 = 0 + 0.706 868 507 133 660 364 8;
  • 55) 0.706 868 507 133 660 364 8 × 2 = 1 + 0.413 737 014 267 320 729 6;
  • 56) 0.413 737 014 267 320 729 6 × 2 = 0 + 0.827 474 028 534 641 459 2;
  • 57) 0.827 474 028 534 641 459 2 × 2 = 1 + 0.654 948 057 069 282 918 4;
  • 58) 0.654 948 057 069 282 918 4 × 2 = 1 + 0.309 896 114 138 565 836 8;
  • 59) 0.309 896 114 138 565 836 8 × 2 = 0 + 0.619 792 228 277 131 673 6;
  • 60) 0.619 792 228 277 131 673 6 × 2 = 1 + 0.239 584 456 554 263 347 2;
  • 61) 0.239 584 456 554 263 347 2 × 2 = 0 + 0.479 168 913 108 526 694 4;
  • 62) 0.479 168 913 108 526 694 4 × 2 = 0 + 0.958 337 826 217 053 388 8;
  • 63) 0.958 337 826 217 053 388 8 × 2 = 1 + 0.916 675 652 434 106 777 6;
  • 64) 0.916 675 652 434 106 777 6 × 2 = 1 + 0.833 351 304 868 213 555 2;
  • 65) 0.833 351 304 868 213 555 2 × 2 = 1 + 0.666 702 609 736 427 110 4;
  • 66) 0.666 702 609 736 427 110 4 × 2 = 1 + 0.333 405 219 472 854 220 8;
  • 67) 0.333 405 219 472 854 220 8 × 2 = 0 + 0.666 810 438 945 708 441 6;
  • 68) 0.666 810 438 945 708 441 6 × 2 = 1 + 0.333 620 877 891 416 883 2;
  • 69) 0.333 620 877 891 416 883 2 × 2 = 0 + 0.667 241 755 782 833 766 4;
  • 70) 0.667 241 755 782 833 766 4 × 2 = 1 + 0.334 483 511 565 667 532 8;
  • 71) 0.334 483 511 565 667 532 8 × 2 = 0 + 0.668 967 023 131 335 065 6;
  • 72) 0.668 967 023 131 335 065 6 × 2 = 1 + 0.337 934 046 262 670 131 2;
  • 73) 0.337 934 046 262 670 131 2 × 2 = 0 + 0.675 868 092 525 340 262 4;
  • 74) 0.675 868 092 525 340 262 4 × 2 = 1 + 0.351 736 185 050 680 524 8;
  • 75) 0.351 736 185 050 680 524 8 × 2 = 0 + 0.703 472 370 101 361 049 6;
  • 76) 0.703 472 370 101 361 049 6 × 2 = 1 + 0.406 944 740 202 722 099 2;
  • 77) 0.406 944 740 202 722 099 2 × 2 = 0 + 0.813 889 480 405 444 198 4;
  • 78) 0.813 889 480 405 444 198 4 × 2 = 1 + 0.627 778 960 810 888 396 8;
  • 79) 0.627 778 960 810 888 396 8 × 2 = 1 + 0.255 557 921 621 776 793 6;
  • 80) 0.255 557 921 621 776 793 6 × 2 = 0 + 0.511 115 843 243 553 587 2;
  • 81) 0.511 115 843 243 553 587 2 × 2 = 1 + 0.022 231 686 487 107 174 4;
  • 82) 0.022 231 686 487 107 174 4 × 2 = 0 + 0.044 463 372 974 214 348 8;
  • 83) 0.044 463 372 974 214 348 8 × 2 = 0 + 0.088 926 745 948 428 697 6;
  • 84) 0.088 926 745 948 428 697 6 × 2 = 0 + 0.177 853 491 896 857 395 2;
  • 85) 0.177 853 491 896 857 395 2 × 2 = 0 + 0.355 706 983 793 714 790 4;
  • 86) 0.355 706 983 793 714 790 4 × 2 = 0 + 0.711 413 967 587 429 580 8;
  • 87) 0.711 413 967 587 429 580 8 × 2 = 1 + 0.422 827 935 174 859 161 6;
  • 88) 0.422 827 935 174 859 161 6 × 2 = 0 + 0.845 655 870 349 718 323 2;
  • 89) 0.845 655 870 349 718 323 2 × 2 = 1 + 0.691 311 740 699 436 646 4;
  • 90) 0.691 311 740 699 436 646 4 × 2 = 1 + 0.382 623 481 398 873 292 8;
  • 91) 0.382 623 481 398 873 292 8 × 2 = 0 + 0.765 246 962 797 746 585 6;
  • 92) 0.765 246 962 797 746 585 6 × 2 = 1 + 0.530 493 925 595 493 171 2;
  • 93) 0.530 493 925 595 493 171 2 × 2 = 1 + 0.060 987 851 190 986 342 4;
  • 94) 0.060 987 851 190 986 342 4 × 2 = 0 + 0.121 975 702 381 972 684 8;
  • 95) 0.121 975 702 381 972 684 8 × 2 = 0 + 0.243 951 404 763 945 369 6;

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:


0.000 000 000 000 176 564 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1101 0011 1101 0101 0101 0110 1000 0010 1101 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 564 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1101 0011 1101 0101 0101 0110 1000 0010 1101 100(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:


0.000 000 000 000 176 564 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1101 0011 1101 0101 0101 0110 1000 0010 1101 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1101 0011 1101 0101 0101 0110 1000 0010 1101 100(2) × 20 =

1.1000 1101 1001 0110 1001 1110 1010 1010 1011 0100 0001 0110 1100(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 0110 1001 1110 1010 1010 1011 0100 0001 0110 1100


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 0110 1001 1110 1010 1010 1011 0100 0001 0110 1100 =

1000 1101 1001 0110 1001 1110 1010 1010 1011 0100 0001 0110 1100


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 0110 1001 1110 1010 1010 1011 0100 0001 0110 1100


Decimal number -0.000 000 000 000 176 564 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0110 1001 1110 1010 1010 1011 0100 0001 0110 1100

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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