-0.000 000 000 000 028 420 84 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 028 420 84(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
-0.000 000 000 000 028 420 84(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:

|-0.000 000 000 000 028 420 84| = 0.000 000 000 000 028 420 84


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: 0.000 000 000 000 028 420 84.

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) 0.000 000 000 000 028 420 84 × 2 = 0 + 0.000 000 000 000 056 841 68;
  • 2) 0.000 000 000 000 056 841 68 × 2 = 0 + 0.000 000 000 000 113 683 36;
  • 3) 0.000 000 000 000 113 683 36 × 2 = 0 + 0.000 000 000 000 227 366 72;
  • 4) 0.000 000 000 000 227 366 72 × 2 = 0 + 0.000 000 000 000 454 733 44;
  • 5) 0.000 000 000 000 454 733 44 × 2 = 0 + 0.000 000 000 000 909 466 88;
  • 6) 0.000 000 000 000 909 466 88 × 2 = 0 + 0.000 000 000 001 818 933 76;
  • 7) 0.000 000 000 001 818 933 76 × 2 = 0 + 0.000 000 000 003 637 867 52;
  • 8) 0.000 000 000 003 637 867 52 × 2 = 0 + 0.000 000 000 007 275 735 04;
  • 9) 0.000 000 000 007 275 735 04 × 2 = 0 + 0.000 000 000 014 551 470 08;
  • 10) 0.000 000 000 014 551 470 08 × 2 = 0 + 0.000 000 000 029 102 940 16;
  • 11) 0.000 000 000 029 102 940 16 × 2 = 0 + 0.000 000 000 058 205 880 32;
  • 12) 0.000 000 000 058 205 880 32 × 2 = 0 + 0.000 000 000 116 411 760 64;
  • 13) 0.000 000 000 116 411 760 64 × 2 = 0 + 0.000 000 000 232 823 521 28;
  • 14) 0.000 000 000 232 823 521 28 × 2 = 0 + 0.000 000 000 465 647 042 56;
  • 15) 0.000 000 000 465 647 042 56 × 2 = 0 + 0.000 000 000 931 294 085 12;
  • 16) 0.000 000 000 931 294 085 12 × 2 = 0 + 0.000 000 001 862 588 170 24;
  • 17) 0.000 000 001 862 588 170 24 × 2 = 0 + 0.000 000 003 725 176 340 48;
  • 18) 0.000 000 003 725 176 340 48 × 2 = 0 + 0.000 000 007 450 352 680 96;
  • 19) 0.000 000 007 450 352 680 96 × 2 = 0 + 0.000 000 014 900 705 361 92;
  • 20) 0.000 000 014 900 705 361 92 × 2 = 0 + 0.000 000 029 801 410 723 84;
  • 21) 0.000 000 029 801 410 723 84 × 2 = 0 + 0.000 000 059 602 821 447 68;
  • 22) 0.000 000 059 602 821 447 68 × 2 = 0 + 0.000 000 119 205 642 895 36;
  • 23) 0.000 000 119 205 642 895 36 × 2 = 0 + 0.000 000 238 411 285 790 72;
  • 24) 0.000 000 238 411 285 790 72 × 2 = 0 + 0.000 000 476 822 571 581 44;
  • 25) 0.000 000 476 822 571 581 44 × 2 = 0 + 0.000 000 953 645 143 162 88;
  • 26) 0.000 000 953 645 143 162 88 × 2 = 0 + 0.000 001 907 290 286 325 76;
  • 27) 0.000 001 907 290 286 325 76 × 2 = 0 + 0.000 003 814 580 572 651 52;
  • 28) 0.000 003 814 580 572 651 52 × 2 = 0 + 0.000 007 629 161 145 303 04;
  • 29) 0.000 007 629 161 145 303 04 × 2 = 0 + 0.000 015 258 322 290 606 08;
  • 30) 0.000 015 258 322 290 606 08 × 2 = 0 + 0.000 030 516 644 581 212 16;
  • 31) 0.000 030 516 644 581 212 16 × 2 = 0 + 0.000 061 033 289 162 424 32;
  • 32) 0.000 061 033 289 162 424 32 × 2 = 0 + 0.000 122 066 578 324 848 64;
  • 33) 0.000 122 066 578 324 848 64 × 2 = 0 + 0.000 244 133 156 649 697 28;
  • 34) 0.000 244 133 156 649 697 28 × 2 = 0 + 0.000 488 266 313 299 394 56;
  • 35) 0.000 488 266 313 299 394 56 × 2 = 0 + 0.000 976 532 626 598 789 12;
  • 36) 0.000 976 532 626 598 789 12 × 2 = 0 + 0.001 953 065 253 197 578 24;
  • 37) 0.001 953 065 253 197 578 24 × 2 = 0 + 0.003 906 130 506 395 156 48;
  • 38) 0.003 906 130 506 395 156 48 × 2 = 0 + 0.007 812 261 012 790 312 96;
  • 39) 0.007 812 261 012 790 312 96 × 2 = 0 + 0.015 624 522 025 580 625 92;
  • 40) 0.015 624 522 025 580 625 92 × 2 = 0 + 0.031 249 044 051 161 251 84;
  • 41) 0.031 249 044 051 161 251 84 × 2 = 0 + 0.062 498 088 102 322 503 68;
  • 42) 0.062 498 088 102 322 503 68 × 2 = 0 + 0.124 996 176 204 645 007 36;
  • 43) 0.124 996 176 204 645 007 36 × 2 = 0 + 0.249 992 352 409 290 014 72;
  • 44) 0.249 992 352 409 290 014 72 × 2 = 0 + 0.499 984 704 818 580 029 44;
  • 45) 0.499 984 704 818 580 029 44 × 2 = 0 + 0.999 969 409 637 160 058 88;
  • 46) 0.999 969 409 637 160 058 88 × 2 = 1 + 0.999 938 819 274 320 117 76;
  • 47) 0.999 938 819 274 320 117 76 × 2 = 1 + 0.999 877 638 548 640 235 52;
  • 48) 0.999 877 638 548 640 235 52 × 2 = 1 + 0.999 755 277 097 280 471 04;
  • 49) 0.999 755 277 097 280 471 04 × 2 = 1 + 0.999 510 554 194 560 942 08;
  • 50) 0.999 510 554 194 560 942 08 × 2 = 1 + 0.999 021 108 389 121 884 16;
  • 51) 0.999 021 108 389 121 884 16 × 2 = 1 + 0.998 042 216 778 243 768 32;
  • 52) 0.998 042 216 778 243 768 32 × 2 = 1 + 0.996 084 433 556 487 536 64;
  • 53) 0.996 084 433 556 487 536 64 × 2 = 1 + 0.992 168 867 112 975 073 28;
  • 54) 0.992 168 867 112 975 073 28 × 2 = 1 + 0.984 337 734 225 950 146 56;
  • 55) 0.984 337 734 225 950 146 56 × 2 = 1 + 0.968 675 468 451 900 293 12;
  • 56) 0.968 675 468 451 900 293 12 × 2 = 1 + 0.937 350 936 903 800 586 24;
  • 57) 0.937 350 936 903 800 586 24 × 2 = 1 + 0.874 701 873 807 601 172 48;
  • 58) 0.874 701 873 807 601 172 48 × 2 = 1 + 0.749 403 747 615 202 344 96;
  • 59) 0.749 403 747 615 202 344 96 × 2 = 1 + 0.498 807 495 230 404 689 92;
  • 60) 0.498 807 495 230 404 689 92 × 2 = 0 + 0.997 614 990 460 809 379 84;
  • 61) 0.997 614 990 460 809 379 84 × 2 = 1 + 0.995 229 980 921 618 759 68;
  • 62) 0.995 229 980 921 618 759 68 × 2 = 1 + 0.990 459 961 843 237 519 36;
  • 63) 0.990 459 961 843 237 519 36 × 2 = 1 + 0.980 919 923 686 475 038 72;
  • 64) 0.980 919 923 686 475 038 72 × 2 = 1 + 0.961 839 847 372 950 077 44;
  • 65) 0.961 839 847 372 950 077 44 × 2 = 1 + 0.923 679 694 745 900 154 88;
  • 66) 0.923 679 694 745 900 154 88 × 2 = 1 + 0.847 359 389 491 800 309 76;
  • 67) 0.847 359 389 491 800 309 76 × 2 = 1 + 0.694 718 778 983 600 619 52;
  • 68) 0.694 718 778 983 600 619 52 × 2 = 1 + 0.389 437 557 967 201 239 04;
  • 69) 0.389 437 557 967 201 239 04 × 2 = 0 + 0.778 875 115 934 402 478 08;
  • 70) 0.778 875 115 934 402 478 08 × 2 = 1 + 0.557 750 231 868 804 956 16;
  • 71) 0.557 750 231 868 804 956 16 × 2 = 1 + 0.115 500 463 737 609 912 32;
  • 72) 0.115 500 463 737 609 912 32 × 2 = 0 + 0.231 000 927 475 219 824 64;
  • 73) 0.231 000 927 475 219 824 64 × 2 = 0 + 0.462 001 854 950 439 649 28;
  • 74) 0.462 001 854 950 439 649 28 × 2 = 0 + 0.924 003 709 900 879 298 56;
  • 75) 0.924 003 709 900 879 298 56 × 2 = 1 + 0.848 007 419 801 758 597 12;
  • 76) 0.848 007 419 801 758 597 12 × 2 = 1 + 0.696 014 839 603 517 194 24;
  • 77) 0.696 014 839 603 517 194 24 × 2 = 1 + 0.392 029 679 207 034 388 48;
  • 78) 0.392 029 679 207 034 388 48 × 2 = 0 + 0.784 059 358 414 068 776 96;
  • 79) 0.784 059 358 414 068 776 96 × 2 = 1 + 0.568 118 716 828 137 553 92;
  • 80) 0.568 118 716 828 137 553 92 × 2 = 1 + 0.136 237 433 656 275 107 84;
  • 81) 0.136 237 433 656 275 107 84 × 2 = 0 + 0.272 474 867 312 550 215 68;
  • 82) 0.272 474 867 312 550 215 68 × 2 = 0 + 0.544 949 734 625 100 431 36;
  • 83) 0.544 949 734 625 100 431 36 × 2 = 1 + 0.089 899 469 250 200 862 72;
  • 84) 0.089 899 469 250 200 862 72 × 2 = 0 + 0.179 798 938 500 401 725 44;
  • 85) 0.179 798 938 500 401 725 44 × 2 = 0 + 0.359 597 877 000 803 450 88;
  • 86) 0.359 597 877 000 803 450 88 × 2 = 0 + 0.719 195 754 001 606 901 76;
  • 87) 0.719 195 754 001 606 901 76 × 2 = 1 + 0.438 391 508 003 213 803 52;
  • 88) 0.438 391 508 003 213 803 52 × 2 = 0 + 0.876 783 016 006 427 607 04;
  • 89) 0.876 783 016 006 427 607 04 × 2 = 1 + 0.753 566 032 012 855 214 08;
  • 90) 0.753 566 032 012 855 214 08 × 2 = 1 + 0.507 132 064 025 710 428 16;
  • 91) 0.507 132 064 025 710 428 16 × 2 = 1 + 0.014 264 128 051 420 856 32;
  • 92) 0.014 264 128 051 420 856 32 × 2 = 0 + 0.028 528 256 102 841 712 64;
  • 93) 0.028 528 256 102 841 712 64 × 2 = 0 + 0.057 056 512 205 683 425 28;
  • 94) 0.057 056 512 205 683 425 28 × 2 = 0 + 0.114 113 024 411 366 850 56;
  • 95) 0.114 113 024 411 366 850 56 × 2 = 0 + 0.228 226 048 822 733 701 12;
  • 96) 0.228 226 048 822 733 701 12 × 2 = 0 + 0.456 452 097 645 467 402 24;
  • 97) 0.456 452 097 645 467 402 24 × 2 = 0 + 0.912 904 195 290 934 804 48;
  • 98) 0.912 904 195 290 934 804 48 × 2 = 1 + 0.825 808 390 581 869 608 96;

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:


0.000 000 000 000 028 420 84(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1110 1111 1111 0110 0011 1011 0010 0010 1110 0000 01(2)

6. Positive number before normalization:

0.000 000 000 000 028 420 84(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1110 1111 1111 0110 0011 1011 0010 0010 1110 0000 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 028 420 84(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1110 1111 1111 0110 0011 1011 0010 0010 1110 0000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1110 1111 1111 0110 0011 1011 0010 0010 1110 0000 01(2) × 20 =

1.1111 1111 1111 1011 1111 1101 1000 1110 1100 1000 1011 1000 0001(2) × 2-46


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): -46

Mantissa (not normalized):
1.1111 1111 1111 1011 1111 1101 1000 1110 1100 1000 1011 1000 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 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) =

977(10) =

011 1101 0001(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. 1111 1111 1111 1011 1111 1101 1000 1110 1100 1000 1011 1000 0001 =

1111 1111 1111 1011 1111 1101 1000 1110 1100 1000 1011 1000 0001


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 0001

Mantissa (52 bits) =
1111 1111 1111 1011 1111 1101 1000 1110 1100 1000 1011 1000 0001


Decimal number -0.000 000 000 000 028 420 84 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 1111 1111 1111 1011 1111 1101 1000 1110 1100 1000 1011 1000 0001

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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