-0.000 000 000 000 176 550 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 550 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
-0.000 000 000 000 176 550 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:

|-0.000 000 000 000 176 550 9| = 0.000 000 000 000 176 550 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: 0.000 000 000 000 176 550 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) 0.000 000 000 000 176 550 9 × 2 = 0 + 0.000 000 000 000 353 101 8;
  • 2) 0.000 000 000 000 353 101 8 × 2 = 0 + 0.000 000 000 000 706 203 6;
  • 3) 0.000 000 000 000 706 203 6 × 2 = 0 + 0.000 000 000 001 412 407 2;
  • 4) 0.000 000 000 001 412 407 2 × 2 = 0 + 0.000 000 000 002 824 814 4;
  • 5) 0.000 000 000 002 824 814 4 × 2 = 0 + 0.000 000 000 005 649 628 8;
  • 6) 0.000 000 000 005 649 628 8 × 2 = 0 + 0.000 000 000 011 299 257 6;
  • 7) 0.000 000 000 011 299 257 6 × 2 = 0 + 0.000 000 000 022 598 515 2;
  • 8) 0.000 000 000 022 598 515 2 × 2 = 0 + 0.000 000 000 045 197 030 4;
  • 9) 0.000 000 000 045 197 030 4 × 2 = 0 + 0.000 000 000 090 394 060 8;
  • 10) 0.000 000 000 090 394 060 8 × 2 = 0 + 0.000 000 000 180 788 121 6;
  • 11) 0.000 000 000 180 788 121 6 × 2 = 0 + 0.000 000 000 361 576 243 2;
  • 12) 0.000 000 000 361 576 243 2 × 2 = 0 + 0.000 000 000 723 152 486 4;
  • 13) 0.000 000 000 723 152 486 4 × 2 = 0 + 0.000 000 001 446 304 972 8;
  • 14) 0.000 000 001 446 304 972 8 × 2 = 0 + 0.000 000 002 892 609 945 6;
  • 15) 0.000 000 002 892 609 945 6 × 2 = 0 + 0.000 000 005 785 219 891 2;
  • 16) 0.000 000 005 785 219 891 2 × 2 = 0 + 0.000 000 011 570 439 782 4;
  • 17) 0.000 000 011 570 439 782 4 × 2 = 0 + 0.000 000 023 140 879 564 8;
  • 18) 0.000 000 023 140 879 564 8 × 2 = 0 + 0.000 000 046 281 759 129 6;
  • 19) 0.000 000 046 281 759 129 6 × 2 = 0 + 0.000 000 092 563 518 259 2;
  • 20) 0.000 000 092 563 518 259 2 × 2 = 0 + 0.000 000 185 127 036 518 4;
  • 21) 0.000 000 185 127 036 518 4 × 2 = 0 + 0.000 000 370 254 073 036 8;
  • 22) 0.000 000 370 254 073 036 8 × 2 = 0 + 0.000 000 740 508 146 073 6;
  • 23) 0.000 000 740 508 146 073 6 × 2 = 0 + 0.000 001 481 016 292 147 2;
  • 24) 0.000 001 481 016 292 147 2 × 2 = 0 + 0.000 002 962 032 584 294 4;
  • 25) 0.000 002 962 032 584 294 4 × 2 = 0 + 0.000 005 924 065 168 588 8;
  • 26) 0.000 005 924 065 168 588 8 × 2 = 0 + 0.000 011 848 130 337 177 6;
  • 27) 0.000 011 848 130 337 177 6 × 2 = 0 + 0.000 023 696 260 674 355 2;
  • 28) 0.000 023 696 260 674 355 2 × 2 = 0 + 0.000 047 392 521 348 710 4;
  • 29) 0.000 047 392 521 348 710 4 × 2 = 0 + 0.000 094 785 042 697 420 8;
  • 30) 0.000 094 785 042 697 420 8 × 2 = 0 + 0.000 189 570 085 394 841 6;
  • 31) 0.000 189 570 085 394 841 6 × 2 = 0 + 0.000 379 140 170 789 683 2;
  • 32) 0.000 379 140 170 789 683 2 × 2 = 0 + 0.000 758 280 341 579 366 4;
  • 33) 0.000 758 280 341 579 366 4 × 2 = 0 + 0.001 516 560 683 158 732 8;
  • 34) 0.001 516 560 683 158 732 8 × 2 = 0 + 0.003 033 121 366 317 465 6;
  • 35) 0.003 033 121 366 317 465 6 × 2 = 0 + 0.006 066 242 732 634 931 2;
  • 36) 0.006 066 242 732 634 931 2 × 2 = 0 + 0.012 132 485 465 269 862 4;
  • 37) 0.012 132 485 465 269 862 4 × 2 = 0 + 0.024 264 970 930 539 724 8;
  • 38) 0.024 264 970 930 539 724 8 × 2 = 0 + 0.048 529 941 861 079 449 6;
  • 39) 0.048 529 941 861 079 449 6 × 2 = 0 + 0.097 059 883 722 158 899 2;
  • 40) 0.097 059 883 722 158 899 2 × 2 = 0 + 0.194 119 767 444 317 798 4;
  • 41) 0.194 119 767 444 317 798 4 × 2 = 0 + 0.388 239 534 888 635 596 8;
  • 42) 0.388 239 534 888 635 596 8 × 2 = 0 + 0.776 479 069 777 271 193 6;
  • 43) 0.776 479 069 777 271 193 6 × 2 = 1 + 0.552 958 139 554 542 387 2;
  • 44) 0.552 958 139 554 542 387 2 × 2 = 1 + 0.105 916 279 109 084 774 4;
  • 45) 0.105 916 279 109 084 774 4 × 2 = 0 + 0.211 832 558 218 169 548 8;
  • 46) 0.211 832 558 218 169 548 8 × 2 = 0 + 0.423 665 116 436 339 097 6;
  • 47) 0.423 665 116 436 339 097 6 × 2 = 0 + 0.847 330 232 872 678 195 2;
  • 48) 0.847 330 232 872 678 195 2 × 2 = 1 + 0.694 660 465 745 356 390 4;
  • 49) 0.694 660 465 745 356 390 4 × 2 = 1 + 0.389 320 931 490 712 780 8;
  • 50) 0.389 320 931 490 712 780 8 × 2 = 0 + 0.778 641 862 981 425 561 6;
  • 51) 0.778 641 862 981 425 561 6 × 2 = 1 + 0.557 283 725 962 851 123 2;
  • 52) 0.557 283 725 962 851 123 2 × 2 = 1 + 0.114 567 451 925 702 246 4;
  • 53) 0.114 567 451 925 702 246 4 × 2 = 0 + 0.229 134 903 851 404 492 8;
  • 54) 0.229 134 903 851 404 492 8 × 2 = 0 + 0.458 269 807 702 808 985 6;
  • 55) 0.458 269 807 702 808 985 6 × 2 = 0 + 0.916 539 615 405 617 971 2;
  • 56) 0.916 539 615 405 617 971 2 × 2 = 1 + 0.833 079 230 811 235 942 4;
  • 57) 0.833 079 230 811 235 942 4 × 2 = 1 + 0.666 158 461 622 471 884 8;
  • 58) 0.666 158 461 622 471 884 8 × 2 = 1 + 0.332 316 923 244 943 769 6;
  • 59) 0.332 316 923 244 943 769 6 × 2 = 0 + 0.664 633 846 489 887 539 2;
  • 60) 0.664 633 846 489 887 539 2 × 2 = 1 + 0.329 267 692 979 775 078 4;
  • 61) 0.329 267 692 979 775 078 4 × 2 = 0 + 0.658 535 385 959 550 156 8;
  • 62) 0.658 535 385 959 550 156 8 × 2 = 1 + 0.317 070 771 919 100 313 6;
  • 63) 0.317 070 771 919 100 313 6 × 2 = 0 + 0.634 141 543 838 200 627 2;
  • 64) 0.634 141 543 838 200 627 2 × 2 = 1 + 0.268 283 087 676 401 254 4;
  • 65) 0.268 283 087 676 401 254 4 × 2 = 0 + 0.536 566 175 352 802 508 8;
  • 66) 0.536 566 175 352 802 508 8 × 2 = 1 + 0.073 132 350 705 605 017 6;
  • 67) 0.073 132 350 705 605 017 6 × 2 = 0 + 0.146 264 701 411 210 035 2;
  • 68) 0.146 264 701 411 210 035 2 × 2 = 0 + 0.292 529 402 822 420 070 4;
  • 69) 0.292 529 402 822 420 070 4 × 2 = 0 + 0.585 058 805 644 840 140 8;
  • 70) 0.585 058 805 644 840 140 8 × 2 = 1 + 0.170 117 611 289 680 281 6;
  • 71) 0.170 117 611 289 680 281 6 × 2 = 0 + 0.340 235 222 579 360 563 2;
  • 72) 0.340 235 222 579 360 563 2 × 2 = 0 + 0.680 470 445 158 721 126 4;
  • 73) 0.680 470 445 158 721 126 4 × 2 = 1 + 0.360 940 890 317 442 252 8;
  • 74) 0.360 940 890 317 442 252 8 × 2 = 0 + 0.721 881 780 634 884 505 6;
  • 75) 0.721 881 780 634 884 505 6 × 2 = 1 + 0.443 763 561 269 769 011 2;
  • 76) 0.443 763 561 269 769 011 2 × 2 = 0 + 0.887 527 122 539 538 022 4;
  • 77) 0.887 527 122 539 538 022 4 × 2 = 1 + 0.775 054 245 079 076 044 8;
  • 78) 0.775 054 245 079 076 044 8 × 2 = 1 + 0.550 108 490 158 152 089 6;
  • 79) 0.550 108 490 158 152 089 6 × 2 = 1 + 0.100 216 980 316 304 179 2;
  • 80) 0.100 216 980 316 304 179 2 × 2 = 0 + 0.200 433 960 632 608 358 4;
  • 81) 0.200 433 960 632 608 358 4 × 2 = 0 + 0.400 867 921 265 216 716 8;
  • 82) 0.400 867 921 265 216 716 8 × 2 = 0 + 0.801 735 842 530 433 433 6;
  • 83) 0.801 735 842 530 433 433 6 × 2 = 1 + 0.603 471 685 060 866 867 2;
  • 84) 0.603 471 685 060 866 867 2 × 2 = 1 + 0.206 943 370 121 733 734 4;
  • 85) 0.206 943 370 121 733 734 4 × 2 = 0 + 0.413 886 740 243 467 468 8;
  • 86) 0.413 886 740 243 467 468 8 × 2 = 0 + 0.827 773 480 486 934 937 6;
  • 87) 0.827 773 480 486 934 937 6 × 2 = 1 + 0.655 546 960 973 869 875 2;
  • 88) 0.655 546 960 973 869 875 2 × 2 = 1 + 0.311 093 921 947 739 750 4;
  • 89) 0.311 093 921 947 739 750 4 × 2 = 0 + 0.622 187 843 895 479 500 8;
  • 90) 0.622 187 843 895 479 500 8 × 2 = 1 + 0.244 375 687 790 959 001 6;
  • 91) 0.244 375 687 790 959 001 6 × 2 = 0 + 0.488 751 375 581 918 003 2;
  • 92) 0.488 751 375 581 918 003 2 × 2 = 0 + 0.977 502 751 163 836 006 4;
  • 93) 0.977 502 751 163 836 006 4 × 2 = 1 + 0.955 005 502 327 672 012 8;
  • 94) 0.955 005 502 327 672 012 8 × 2 = 1 + 0.910 011 004 655 344 025 6;
  • 95) 0.910 011 004 655 344 025 6 × 2 = 1 + 0.820 022 009 310 688 051 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:


0.000 000 000 000 176 550 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1101 0101 0100 0100 1010 1110 0011 0011 0100 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 550 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1101 0101 0100 0100 1010 1110 0011 0011 0100 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:


0.000 000 000 000 176 550 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1101 0101 0100 0100 1010 1110 0011 0011 0100 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1101 0101 0100 0100 1010 1110 0011 0011 0100 111(2) × 20 =

1.1000 1101 1000 1110 1010 1010 0010 0101 0111 0001 1001 1010 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 1110 1010 1010 0010 0101 0111 0001 1001 1010 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 1110 1010 1010 0010 0101 0111 0001 1001 1010 0111 =

1000 1101 1000 1110 1010 1010 0010 0101 0111 0001 1001 1010 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 1110 1010 1010 0010 0101 0111 0001 1001 1010 0111


Decimal number -0.000 000 000 000 176 550 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1110 1010 1010 0010 0101 0111 0001 1001 1010 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