-0.000 000 000 000 176 570 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 570 6(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
-0.000 000 000 000 176 570 6(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:

|-0.000 000 000 000 176 570 6| = 0.000 000 000 000 176 570 6


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: 0.000 000 000 000 176 570 6.

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) 0.000 000 000 000 176 570 6 × 2 = 0 + 0.000 000 000 000 353 141 2;
  • 2) 0.000 000 000 000 353 141 2 × 2 = 0 + 0.000 000 000 000 706 282 4;
  • 3) 0.000 000 000 000 706 282 4 × 2 = 0 + 0.000 000 000 001 412 564 8;
  • 4) 0.000 000 000 001 412 564 8 × 2 = 0 + 0.000 000 000 002 825 129 6;
  • 5) 0.000 000 000 002 825 129 6 × 2 = 0 + 0.000 000 000 005 650 259 2;
  • 6) 0.000 000 000 005 650 259 2 × 2 = 0 + 0.000 000 000 011 300 518 4;
  • 7) 0.000 000 000 011 300 518 4 × 2 = 0 + 0.000 000 000 022 601 036 8;
  • 8) 0.000 000 000 022 601 036 8 × 2 = 0 + 0.000 000 000 045 202 073 6;
  • 9) 0.000 000 000 045 202 073 6 × 2 = 0 + 0.000 000 000 090 404 147 2;
  • 10) 0.000 000 000 090 404 147 2 × 2 = 0 + 0.000 000 000 180 808 294 4;
  • 11) 0.000 000 000 180 808 294 4 × 2 = 0 + 0.000 000 000 361 616 588 8;
  • 12) 0.000 000 000 361 616 588 8 × 2 = 0 + 0.000 000 000 723 233 177 6;
  • 13) 0.000 000 000 723 233 177 6 × 2 = 0 + 0.000 000 001 446 466 355 2;
  • 14) 0.000 000 001 446 466 355 2 × 2 = 0 + 0.000 000 002 892 932 710 4;
  • 15) 0.000 000 002 892 932 710 4 × 2 = 0 + 0.000 000 005 785 865 420 8;
  • 16) 0.000 000 005 785 865 420 8 × 2 = 0 + 0.000 000 011 571 730 841 6;
  • 17) 0.000 000 011 571 730 841 6 × 2 = 0 + 0.000 000 023 143 461 683 2;
  • 18) 0.000 000 023 143 461 683 2 × 2 = 0 + 0.000 000 046 286 923 366 4;
  • 19) 0.000 000 046 286 923 366 4 × 2 = 0 + 0.000 000 092 573 846 732 8;
  • 20) 0.000 000 092 573 846 732 8 × 2 = 0 + 0.000 000 185 147 693 465 6;
  • 21) 0.000 000 185 147 693 465 6 × 2 = 0 + 0.000 000 370 295 386 931 2;
  • 22) 0.000 000 370 295 386 931 2 × 2 = 0 + 0.000 000 740 590 773 862 4;
  • 23) 0.000 000 740 590 773 862 4 × 2 = 0 + 0.000 001 481 181 547 724 8;
  • 24) 0.000 001 481 181 547 724 8 × 2 = 0 + 0.000 002 962 363 095 449 6;
  • 25) 0.000 002 962 363 095 449 6 × 2 = 0 + 0.000 005 924 726 190 899 2;
  • 26) 0.000 005 924 726 190 899 2 × 2 = 0 + 0.000 011 849 452 381 798 4;
  • 27) 0.000 011 849 452 381 798 4 × 2 = 0 + 0.000 023 698 904 763 596 8;
  • 28) 0.000 023 698 904 763 596 8 × 2 = 0 + 0.000 047 397 809 527 193 6;
  • 29) 0.000 047 397 809 527 193 6 × 2 = 0 + 0.000 094 795 619 054 387 2;
  • 30) 0.000 094 795 619 054 387 2 × 2 = 0 + 0.000 189 591 238 108 774 4;
  • 31) 0.000 189 591 238 108 774 4 × 2 = 0 + 0.000 379 182 476 217 548 8;
  • 32) 0.000 379 182 476 217 548 8 × 2 = 0 + 0.000 758 364 952 435 097 6;
  • 33) 0.000 758 364 952 435 097 6 × 2 = 0 + 0.001 516 729 904 870 195 2;
  • 34) 0.001 516 729 904 870 195 2 × 2 = 0 + 0.003 033 459 809 740 390 4;
  • 35) 0.003 033 459 809 740 390 4 × 2 = 0 + 0.006 066 919 619 480 780 8;
  • 36) 0.006 066 919 619 480 780 8 × 2 = 0 + 0.012 133 839 238 961 561 6;
  • 37) 0.012 133 839 238 961 561 6 × 2 = 0 + 0.024 267 678 477 923 123 2;
  • 38) 0.024 267 678 477 923 123 2 × 2 = 0 + 0.048 535 356 955 846 246 4;
  • 39) 0.048 535 356 955 846 246 4 × 2 = 0 + 0.097 070 713 911 692 492 8;
  • 40) 0.097 070 713 911 692 492 8 × 2 = 0 + 0.194 141 427 823 384 985 6;
  • 41) 0.194 141 427 823 384 985 6 × 2 = 0 + 0.388 282 855 646 769 971 2;
  • 42) 0.388 282 855 646 769 971 2 × 2 = 0 + 0.776 565 711 293 539 942 4;
  • 43) 0.776 565 711 293 539 942 4 × 2 = 1 + 0.553 131 422 587 079 884 8;
  • 44) 0.553 131 422 587 079 884 8 × 2 = 1 + 0.106 262 845 174 159 769 6;
  • 45) 0.106 262 845 174 159 769 6 × 2 = 0 + 0.212 525 690 348 319 539 2;
  • 46) 0.212 525 690 348 319 539 2 × 2 = 0 + 0.425 051 380 696 639 078 4;
  • 47) 0.425 051 380 696 639 078 4 × 2 = 0 + 0.850 102 761 393 278 156 8;
  • 48) 0.850 102 761 393 278 156 8 × 2 = 1 + 0.700 205 522 786 556 313 6;
  • 49) 0.700 205 522 786 556 313 6 × 2 = 1 + 0.400 411 045 573 112 627 2;
  • 50) 0.400 411 045 573 112 627 2 × 2 = 0 + 0.800 822 091 146 225 254 4;
  • 51) 0.800 822 091 146 225 254 4 × 2 = 1 + 0.601 644 182 292 450 508 8;
  • 52) 0.601 644 182 292 450 508 8 × 2 = 1 + 0.203 288 364 584 901 017 6;
  • 53) 0.203 288 364 584 901 017 6 × 2 = 0 + 0.406 576 729 169 802 035 2;
  • 54) 0.406 576 729 169 802 035 2 × 2 = 0 + 0.813 153 458 339 604 070 4;
  • 55) 0.813 153 458 339 604 070 4 × 2 = 1 + 0.626 306 916 679 208 140 8;
  • 56) 0.626 306 916 679 208 140 8 × 2 = 1 + 0.252 613 833 358 416 281 6;
  • 57) 0.252 613 833 358 416 281 6 × 2 = 0 + 0.505 227 666 716 832 563 2;
  • 58) 0.505 227 666 716 832 563 2 × 2 = 1 + 0.010 455 333 433 665 126 4;
  • 59) 0.010 455 333 433 665 126 4 × 2 = 0 + 0.020 910 666 867 330 252 8;
  • 60) 0.020 910 666 867 330 252 8 × 2 = 0 + 0.041 821 333 734 660 505 6;
  • 61) 0.041 821 333 734 660 505 6 × 2 = 0 + 0.083 642 667 469 321 011 2;
  • 62) 0.083 642 667 469 321 011 2 × 2 = 0 + 0.167 285 334 938 642 022 4;
  • 63) 0.167 285 334 938 642 022 4 × 2 = 0 + 0.334 570 669 877 284 044 8;
  • 64) 0.334 570 669 877 284 044 8 × 2 = 0 + 0.669 141 339 754 568 089 6;
  • 65) 0.669 141 339 754 568 089 6 × 2 = 1 + 0.338 282 679 509 136 179 2;
  • 66) 0.338 282 679 509 136 179 2 × 2 = 0 + 0.676 565 359 018 272 358 4;
  • 67) 0.676 565 359 018 272 358 4 × 2 = 1 + 0.353 130 718 036 544 716 8;
  • 68) 0.353 130 718 036 544 716 8 × 2 = 0 + 0.706 261 436 073 089 433 6;
  • 69) 0.706 261 436 073 089 433 6 × 2 = 1 + 0.412 522 872 146 178 867 2;
  • 70) 0.412 522 872 146 178 867 2 × 2 = 0 + 0.825 045 744 292 357 734 4;
  • 71) 0.825 045 744 292 357 734 4 × 2 = 1 + 0.650 091 488 584 715 468 8;
  • 72) 0.650 091 488 584 715 468 8 × 2 = 1 + 0.300 182 977 169 430 937 6;
  • 73) 0.300 182 977 169 430 937 6 × 2 = 0 + 0.600 365 954 338 861 875 2;
  • 74) 0.600 365 954 338 861 875 2 × 2 = 1 + 0.200 731 908 677 723 750 4;
  • 75) 0.200 731 908 677 723 750 4 × 2 = 0 + 0.401 463 817 355 447 500 8;
  • 76) 0.401 463 817 355 447 500 8 × 2 = 0 + 0.802 927 634 710 895 001 6;
  • 77) 0.802 927 634 710 895 001 6 × 2 = 1 + 0.605 855 269 421 790 003 2;
  • 78) 0.605 855 269 421 790 003 2 × 2 = 1 + 0.211 710 538 843 580 006 4;
  • 79) 0.211 710 538 843 580 006 4 × 2 = 0 + 0.423 421 077 687 160 012 8;
  • 80) 0.423 421 077 687 160 012 8 × 2 = 0 + 0.846 842 155 374 320 025 6;
  • 81) 0.846 842 155 374 320 025 6 × 2 = 1 + 0.693 684 310 748 640 051 2;
  • 82) 0.693 684 310 748 640 051 2 × 2 = 1 + 0.387 368 621 497 280 102 4;
  • 83) 0.387 368 621 497 280 102 4 × 2 = 0 + 0.774 737 242 994 560 204 8;
  • 84) 0.774 737 242 994 560 204 8 × 2 = 1 + 0.549 474 485 989 120 409 6;
  • 85) 0.549 474 485 989 120 409 6 × 2 = 1 + 0.098 948 971 978 240 819 2;
  • 86) 0.098 948 971 978 240 819 2 × 2 = 0 + 0.197 897 943 956 481 638 4;
  • 87) 0.197 897 943 956 481 638 4 × 2 = 0 + 0.395 795 887 912 963 276 8;
  • 88) 0.395 795 887 912 963 276 8 × 2 = 0 + 0.791 591 775 825 926 553 6;
  • 89) 0.791 591 775 825 926 553 6 × 2 = 1 + 0.583 183 551 651 853 107 2;
  • 90) 0.583 183 551 651 853 107 2 × 2 = 1 + 0.166 367 103 303 706 214 4;
  • 91) 0.166 367 103 303 706 214 4 × 2 = 0 + 0.332 734 206 607 412 428 8;
  • 92) 0.332 734 206 607 412 428 8 × 2 = 0 + 0.665 468 413 214 824 857 6;
  • 93) 0.665 468 413 214 824 857 6 × 2 = 1 + 0.330 936 826 429 649 715 2;
  • 94) 0.330 936 826 429 649 715 2 × 2 = 0 + 0.661 873 652 859 299 430 4;
  • 95) 0.661 873 652 859 299 430 4 × 2 = 1 + 0.323 747 305 718 598 860 8;

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:


0.000 000 000 000 176 570 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0100 0000 1010 1011 0100 1100 1101 1000 1100 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 570 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0100 0000 1010 1011 0100 1100 1101 1000 1100 101(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:


0.000 000 000 000 176 570 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0100 0000 1010 1011 0100 1100 1101 1000 1100 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0100 0000 1010 1011 0100 1100 1101 1000 1100 101(2) × 20 =

1.1000 1101 1001 1010 0000 0101 0101 1010 0110 0110 1100 0110 0101(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 1010 0000 0101 0101 1010 0110 0110 1100 0110 0101


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 1010 0000 0101 0101 1010 0110 0110 1100 0110 0101 =

1000 1101 1001 1010 0000 0101 0101 1010 0110 0110 1100 0110 0101


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 1010 0000 0101 0101 1010 0110 0110 1100 0110 0101


Decimal number -0.000 000 000 000 176 570 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 1010 0000 0101 0101 1010 0110 0110 1100 0110 0101

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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