-0.000 000 000 000 176 551 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 551 9(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
-0.000 000 000 000 176 551 9(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:

|-0.000 000 000 000 176 551 9| = 0.000 000 000 000 176 551 9


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: 0.000 000 000 000 176 551 9.

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) 0.000 000 000 000 176 551 9 × 2 = 0 + 0.000 000 000 000 353 103 8;
  • 2) 0.000 000 000 000 353 103 8 × 2 = 0 + 0.000 000 000 000 706 207 6;
  • 3) 0.000 000 000 000 706 207 6 × 2 = 0 + 0.000 000 000 001 412 415 2;
  • 4) 0.000 000 000 001 412 415 2 × 2 = 0 + 0.000 000 000 002 824 830 4;
  • 5) 0.000 000 000 002 824 830 4 × 2 = 0 + 0.000 000 000 005 649 660 8;
  • 6) 0.000 000 000 005 649 660 8 × 2 = 0 + 0.000 000 000 011 299 321 6;
  • 7) 0.000 000 000 011 299 321 6 × 2 = 0 + 0.000 000 000 022 598 643 2;
  • 8) 0.000 000 000 022 598 643 2 × 2 = 0 + 0.000 000 000 045 197 286 4;
  • 9) 0.000 000 000 045 197 286 4 × 2 = 0 + 0.000 000 000 090 394 572 8;
  • 10) 0.000 000 000 090 394 572 8 × 2 = 0 + 0.000 000 000 180 789 145 6;
  • 11) 0.000 000 000 180 789 145 6 × 2 = 0 + 0.000 000 000 361 578 291 2;
  • 12) 0.000 000 000 361 578 291 2 × 2 = 0 + 0.000 000 000 723 156 582 4;
  • 13) 0.000 000 000 723 156 582 4 × 2 = 0 + 0.000 000 001 446 313 164 8;
  • 14) 0.000 000 001 446 313 164 8 × 2 = 0 + 0.000 000 002 892 626 329 6;
  • 15) 0.000 000 002 892 626 329 6 × 2 = 0 + 0.000 000 005 785 252 659 2;
  • 16) 0.000 000 005 785 252 659 2 × 2 = 0 + 0.000 000 011 570 505 318 4;
  • 17) 0.000 000 011 570 505 318 4 × 2 = 0 + 0.000 000 023 141 010 636 8;
  • 18) 0.000 000 023 141 010 636 8 × 2 = 0 + 0.000 000 046 282 021 273 6;
  • 19) 0.000 000 046 282 021 273 6 × 2 = 0 + 0.000 000 092 564 042 547 2;
  • 20) 0.000 000 092 564 042 547 2 × 2 = 0 + 0.000 000 185 128 085 094 4;
  • 21) 0.000 000 185 128 085 094 4 × 2 = 0 + 0.000 000 370 256 170 188 8;
  • 22) 0.000 000 370 256 170 188 8 × 2 = 0 + 0.000 000 740 512 340 377 6;
  • 23) 0.000 000 740 512 340 377 6 × 2 = 0 + 0.000 001 481 024 680 755 2;
  • 24) 0.000 001 481 024 680 755 2 × 2 = 0 + 0.000 002 962 049 361 510 4;
  • 25) 0.000 002 962 049 361 510 4 × 2 = 0 + 0.000 005 924 098 723 020 8;
  • 26) 0.000 005 924 098 723 020 8 × 2 = 0 + 0.000 011 848 197 446 041 6;
  • 27) 0.000 011 848 197 446 041 6 × 2 = 0 + 0.000 023 696 394 892 083 2;
  • 28) 0.000 023 696 394 892 083 2 × 2 = 0 + 0.000 047 392 789 784 166 4;
  • 29) 0.000 047 392 789 784 166 4 × 2 = 0 + 0.000 094 785 579 568 332 8;
  • 30) 0.000 094 785 579 568 332 8 × 2 = 0 + 0.000 189 571 159 136 665 6;
  • 31) 0.000 189 571 159 136 665 6 × 2 = 0 + 0.000 379 142 318 273 331 2;
  • 32) 0.000 379 142 318 273 331 2 × 2 = 0 + 0.000 758 284 636 546 662 4;
  • 33) 0.000 758 284 636 546 662 4 × 2 = 0 + 0.001 516 569 273 093 324 8;
  • 34) 0.001 516 569 273 093 324 8 × 2 = 0 + 0.003 033 138 546 186 649 6;
  • 35) 0.003 033 138 546 186 649 6 × 2 = 0 + 0.006 066 277 092 373 299 2;
  • 36) 0.006 066 277 092 373 299 2 × 2 = 0 + 0.012 132 554 184 746 598 4;
  • 37) 0.012 132 554 184 746 598 4 × 2 = 0 + 0.024 265 108 369 493 196 8;
  • 38) 0.024 265 108 369 493 196 8 × 2 = 0 + 0.048 530 216 738 986 393 6;
  • 39) 0.048 530 216 738 986 393 6 × 2 = 0 + 0.097 060 433 477 972 787 2;
  • 40) 0.097 060 433 477 972 787 2 × 2 = 0 + 0.194 120 866 955 945 574 4;
  • 41) 0.194 120 866 955 945 574 4 × 2 = 0 + 0.388 241 733 911 891 148 8;
  • 42) 0.388 241 733 911 891 148 8 × 2 = 0 + 0.776 483 467 823 782 297 6;
  • 43) 0.776 483 467 823 782 297 6 × 2 = 1 + 0.552 966 935 647 564 595 2;
  • 44) 0.552 966 935 647 564 595 2 × 2 = 1 + 0.105 933 871 295 129 190 4;
  • 45) 0.105 933 871 295 129 190 4 × 2 = 0 + 0.211 867 742 590 258 380 8;
  • 46) 0.211 867 742 590 258 380 8 × 2 = 0 + 0.423 735 485 180 516 761 6;
  • 47) 0.423 735 485 180 516 761 6 × 2 = 0 + 0.847 470 970 361 033 523 2;
  • 48) 0.847 470 970 361 033 523 2 × 2 = 1 + 0.694 941 940 722 067 046 4;
  • 49) 0.694 941 940 722 067 046 4 × 2 = 1 + 0.389 883 881 444 134 092 8;
  • 50) 0.389 883 881 444 134 092 8 × 2 = 0 + 0.779 767 762 888 268 185 6;
  • 51) 0.779 767 762 888 268 185 6 × 2 = 1 + 0.559 535 525 776 536 371 2;
  • 52) 0.559 535 525 776 536 371 2 × 2 = 1 + 0.119 071 051 553 072 742 4;
  • 53) 0.119 071 051 553 072 742 4 × 2 = 0 + 0.238 142 103 106 145 484 8;
  • 54) 0.238 142 103 106 145 484 8 × 2 = 0 + 0.476 284 206 212 290 969 6;
  • 55) 0.476 284 206 212 290 969 6 × 2 = 0 + 0.952 568 412 424 581 939 2;
  • 56) 0.952 568 412 424 581 939 2 × 2 = 1 + 0.905 136 824 849 163 878 4;
  • 57) 0.905 136 824 849 163 878 4 × 2 = 1 + 0.810 273 649 698 327 756 8;
  • 58) 0.810 273 649 698 327 756 8 × 2 = 1 + 0.620 547 299 396 655 513 6;
  • 59) 0.620 547 299 396 655 513 6 × 2 = 1 + 0.241 094 598 793 311 027 2;
  • 60) 0.241 094 598 793 311 027 2 × 2 = 0 + 0.482 189 197 586 622 054 4;
  • 61) 0.482 189 197 586 622 054 4 × 2 = 0 + 0.964 378 395 173 244 108 8;
  • 62) 0.964 378 395 173 244 108 8 × 2 = 1 + 0.928 756 790 346 488 217 6;
  • 63) 0.928 756 790 346 488 217 6 × 2 = 1 + 0.857 513 580 692 976 435 2;
  • 64) 0.857 513 580 692 976 435 2 × 2 = 1 + 0.715 027 161 385 952 870 4;
  • 65) 0.715 027 161 385 952 870 4 × 2 = 1 + 0.430 054 322 771 905 740 8;
  • 66) 0.430 054 322 771 905 740 8 × 2 = 0 + 0.860 108 645 543 811 481 6;
  • 67) 0.860 108 645 543 811 481 6 × 2 = 1 + 0.720 217 291 087 622 963 2;
  • 68) 0.720 217 291 087 622 963 2 × 2 = 1 + 0.440 434 582 175 245 926 4;
  • 69) 0.440 434 582 175 245 926 4 × 2 = 0 + 0.880 869 164 350 491 852 8;
  • 70) 0.880 869 164 350 491 852 8 × 2 = 1 + 0.761 738 328 700 983 705 6;
  • 71) 0.761 738 328 700 983 705 6 × 2 = 1 + 0.523 476 657 401 967 411 2;
  • 72) 0.523 476 657 401 967 411 2 × 2 = 1 + 0.046 953 314 803 934 822 4;
  • 73) 0.046 953 314 803 934 822 4 × 2 = 0 + 0.093 906 629 607 869 644 8;
  • 74) 0.093 906 629 607 869 644 8 × 2 = 0 + 0.187 813 259 215 739 289 6;
  • 75) 0.187 813 259 215 739 289 6 × 2 = 0 + 0.375 626 518 431 478 579 2;
  • 76) 0.375 626 518 431 478 579 2 × 2 = 0 + 0.751 253 036 862 957 158 4;
  • 77) 0.751 253 036 862 957 158 4 × 2 = 1 + 0.502 506 073 725 914 316 8;
  • 78) 0.502 506 073 725 914 316 8 × 2 = 1 + 0.005 012 147 451 828 633 6;
  • 79) 0.005 012 147 451 828 633 6 × 2 = 0 + 0.010 024 294 903 657 267 2;
  • 80) 0.010 024 294 903 657 267 2 × 2 = 0 + 0.020 048 589 807 314 534 4;
  • 81) 0.020 048 589 807 314 534 4 × 2 = 0 + 0.040 097 179 614 629 068 8;
  • 82) 0.040 097 179 614 629 068 8 × 2 = 0 + 0.080 194 359 229 258 137 6;
  • 83) 0.080 194 359 229 258 137 6 × 2 = 0 + 0.160 388 718 458 516 275 2;
  • 84) 0.160 388 718 458 516 275 2 × 2 = 0 + 0.320 777 436 917 032 550 4;
  • 85) 0.320 777 436 917 032 550 4 × 2 = 0 + 0.641 554 873 834 065 100 8;
  • 86) 0.641 554 873 834 065 100 8 × 2 = 1 + 0.283 109 747 668 130 201 6;
  • 87) 0.283 109 747 668 130 201 6 × 2 = 0 + 0.566 219 495 336 260 403 2;
  • 88) 0.566 219 495 336 260 403 2 × 2 = 1 + 0.132 438 990 672 520 806 4;
  • 89) 0.132 438 990 672 520 806 4 × 2 = 0 + 0.264 877 981 345 041 612 8;
  • 90) 0.264 877 981 345 041 612 8 × 2 = 0 + 0.529 755 962 690 083 225 6;
  • 91) 0.529 755 962 690 083 225 6 × 2 = 1 + 0.059 511 925 380 166 451 2;
  • 92) 0.059 511 925 380 166 451 2 × 2 = 0 + 0.119 023 850 760 332 902 4;
  • 93) 0.119 023 850 760 332 902 4 × 2 = 0 + 0.238 047 701 520 665 804 8;
  • 94) 0.238 047 701 520 665 804 8 × 2 = 0 + 0.476 095 403 041 331 609 6;
  • 95) 0.476 095 403 041 331 609 6 × 2 = 0 + 0.952 190 806 082 663 219 2;

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:


0.000 000 000 000 176 551 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1110 0111 1011 0111 0000 1100 0000 0101 0010 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 551 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1110 0111 1011 0111 0000 1100 0000 0101 0010 000(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:


0.000 000 000 000 176 551 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1110 0111 1011 0111 0000 1100 0000 0101 0010 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1110 0111 1011 0111 0000 1100 0000 0101 0010 000(2) × 20 =

1.1000 1101 1000 1111 0011 1101 1011 1000 0110 0000 0010 1001 0000(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 1111 0011 1101 1011 1000 0110 0000 0010 1001 0000


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 1111 0011 1101 1011 1000 0110 0000 0010 1001 0000 =

1000 1101 1000 1111 0011 1101 1011 1000 0110 0000 0010 1001 0000


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 1111 0011 1101 1011 1000 0110 0000 0010 1001 0000


Decimal number -0.000 000 000 000 176 551 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1111 0011 1101 1011 1000 0110 0000 0010 1001 0000

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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