-0.000 000 000 000 176 557 522 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 522(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
-0.000 000 000 000 176 557 522(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:

|-0.000 000 000 000 176 557 522| = 0.000 000 000 000 176 557 522


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: 0.000 000 000 000 176 557 522.

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) 0.000 000 000 000 176 557 522 × 2 = 0 + 0.000 000 000 000 353 115 044;
  • 2) 0.000 000 000 000 353 115 044 × 2 = 0 + 0.000 000 000 000 706 230 088;
  • 3) 0.000 000 000 000 706 230 088 × 2 = 0 + 0.000 000 000 001 412 460 176;
  • 4) 0.000 000 000 001 412 460 176 × 2 = 0 + 0.000 000 000 002 824 920 352;
  • 5) 0.000 000 000 002 824 920 352 × 2 = 0 + 0.000 000 000 005 649 840 704;
  • 6) 0.000 000 000 005 649 840 704 × 2 = 0 + 0.000 000 000 011 299 681 408;
  • 7) 0.000 000 000 011 299 681 408 × 2 = 0 + 0.000 000 000 022 599 362 816;
  • 8) 0.000 000 000 022 599 362 816 × 2 = 0 + 0.000 000 000 045 198 725 632;
  • 9) 0.000 000 000 045 198 725 632 × 2 = 0 + 0.000 000 000 090 397 451 264;
  • 10) 0.000 000 000 090 397 451 264 × 2 = 0 + 0.000 000 000 180 794 902 528;
  • 11) 0.000 000 000 180 794 902 528 × 2 = 0 + 0.000 000 000 361 589 805 056;
  • 12) 0.000 000 000 361 589 805 056 × 2 = 0 + 0.000 000 000 723 179 610 112;
  • 13) 0.000 000 000 723 179 610 112 × 2 = 0 + 0.000 000 001 446 359 220 224;
  • 14) 0.000 000 001 446 359 220 224 × 2 = 0 + 0.000 000 002 892 718 440 448;
  • 15) 0.000 000 002 892 718 440 448 × 2 = 0 + 0.000 000 005 785 436 880 896;
  • 16) 0.000 000 005 785 436 880 896 × 2 = 0 + 0.000 000 011 570 873 761 792;
  • 17) 0.000 000 011 570 873 761 792 × 2 = 0 + 0.000 000 023 141 747 523 584;
  • 18) 0.000 000 023 141 747 523 584 × 2 = 0 + 0.000 000 046 283 495 047 168;
  • 19) 0.000 000 046 283 495 047 168 × 2 = 0 + 0.000 000 092 566 990 094 336;
  • 20) 0.000 000 092 566 990 094 336 × 2 = 0 + 0.000 000 185 133 980 188 672;
  • 21) 0.000 000 185 133 980 188 672 × 2 = 0 + 0.000 000 370 267 960 377 344;
  • 22) 0.000 000 370 267 960 377 344 × 2 = 0 + 0.000 000 740 535 920 754 688;
  • 23) 0.000 000 740 535 920 754 688 × 2 = 0 + 0.000 001 481 071 841 509 376;
  • 24) 0.000 001 481 071 841 509 376 × 2 = 0 + 0.000 002 962 143 683 018 752;
  • 25) 0.000 002 962 143 683 018 752 × 2 = 0 + 0.000 005 924 287 366 037 504;
  • 26) 0.000 005 924 287 366 037 504 × 2 = 0 + 0.000 011 848 574 732 075 008;
  • 27) 0.000 011 848 574 732 075 008 × 2 = 0 + 0.000 023 697 149 464 150 016;
  • 28) 0.000 023 697 149 464 150 016 × 2 = 0 + 0.000 047 394 298 928 300 032;
  • 29) 0.000 047 394 298 928 300 032 × 2 = 0 + 0.000 094 788 597 856 600 064;
  • 30) 0.000 094 788 597 856 600 064 × 2 = 0 + 0.000 189 577 195 713 200 128;
  • 31) 0.000 189 577 195 713 200 128 × 2 = 0 + 0.000 379 154 391 426 400 256;
  • 32) 0.000 379 154 391 426 400 256 × 2 = 0 + 0.000 758 308 782 852 800 512;
  • 33) 0.000 758 308 782 852 800 512 × 2 = 0 + 0.001 516 617 565 705 601 024;
  • 34) 0.001 516 617 565 705 601 024 × 2 = 0 + 0.003 033 235 131 411 202 048;
  • 35) 0.003 033 235 131 411 202 048 × 2 = 0 + 0.006 066 470 262 822 404 096;
  • 36) 0.006 066 470 262 822 404 096 × 2 = 0 + 0.012 132 940 525 644 808 192;
  • 37) 0.012 132 940 525 644 808 192 × 2 = 0 + 0.024 265 881 051 289 616 384;
  • 38) 0.024 265 881 051 289 616 384 × 2 = 0 + 0.048 531 762 102 579 232 768;
  • 39) 0.048 531 762 102 579 232 768 × 2 = 0 + 0.097 063 524 205 158 465 536;
  • 40) 0.097 063 524 205 158 465 536 × 2 = 0 + 0.194 127 048 410 316 931 072;
  • 41) 0.194 127 048 410 316 931 072 × 2 = 0 + 0.388 254 096 820 633 862 144;
  • 42) 0.388 254 096 820 633 862 144 × 2 = 0 + 0.776 508 193 641 267 724 288;
  • 43) 0.776 508 193 641 267 724 288 × 2 = 1 + 0.553 016 387 282 535 448 576;
  • 44) 0.553 016 387 282 535 448 576 × 2 = 1 + 0.106 032 774 565 070 897 152;
  • 45) 0.106 032 774 565 070 897 152 × 2 = 0 + 0.212 065 549 130 141 794 304;
  • 46) 0.212 065 549 130 141 794 304 × 2 = 0 + 0.424 131 098 260 283 588 608;
  • 47) 0.424 131 098 260 283 588 608 × 2 = 0 + 0.848 262 196 520 567 177 216;
  • 48) 0.848 262 196 520 567 177 216 × 2 = 1 + 0.696 524 393 041 134 354 432;
  • 49) 0.696 524 393 041 134 354 432 × 2 = 1 + 0.393 048 786 082 268 708 864;
  • 50) 0.393 048 786 082 268 708 864 × 2 = 0 + 0.786 097 572 164 537 417 728;
  • 51) 0.786 097 572 164 537 417 728 × 2 = 1 + 0.572 195 144 329 074 835 456;
  • 52) 0.572 195 144 329 074 835 456 × 2 = 1 + 0.144 390 288 658 149 670 912;
  • 53) 0.144 390 288 658 149 670 912 × 2 = 0 + 0.288 780 577 316 299 341 824;
  • 54) 0.288 780 577 316 299 341 824 × 2 = 0 + 0.577 561 154 632 598 683 648;
  • 55) 0.577 561 154 632 598 683 648 × 2 = 1 + 0.155 122 309 265 197 367 296;
  • 56) 0.155 122 309 265 197 367 296 × 2 = 0 + 0.310 244 618 530 394 734 592;
  • 57) 0.310 244 618 530 394 734 592 × 2 = 0 + 0.620 489 237 060 789 469 184;
  • 58) 0.620 489 237 060 789 469 184 × 2 = 1 + 0.240 978 474 121 578 938 368;
  • 59) 0.240 978 474 121 578 938 368 × 2 = 0 + 0.481 956 948 243 157 876 736;
  • 60) 0.481 956 948 243 157 876 736 × 2 = 0 + 0.963 913 896 486 315 753 472;
  • 61) 0.963 913 896 486 315 753 472 × 2 = 1 + 0.927 827 792 972 631 506 944;
  • 62) 0.927 827 792 972 631 506 944 × 2 = 1 + 0.855 655 585 945 263 013 888;
  • 63) 0.855 655 585 945 263 013 888 × 2 = 1 + 0.711 311 171 890 526 027 776;
  • 64) 0.711 311 171 890 526 027 776 × 2 = 1 + 0.422 622 343 781 052 055 552;
  • 65) 0.422 622 343 781 052 055 552 × 2 = 0 + 0.845 244 687 562 104 111 104;
  • 66) 0.845 244 687 562 104 111 104 × 2 = 1 + 0.690 489 375 124 208 222 208;
  • 67) 0.690 489 375 124 208 222 208 × 2 = 1 + 0.380 978 750 248 416 444 416;
  • 68) 0.380 978 750 248 416 444 416 × 2 = 0 + 0.761 957 500 496 832 888 832;
  • 69) 0.761 957 500 496 832 888 832 × 2 = 1 + 0.523 915 000 993 665 777 664;
  • 70) 0.523 915 000 993 665 777 664 × 2 = 1 + 0.047 830 001 987 331 555 328;
  • 71) 0.047 830 001 987 331 555 328 × 2 = 0 + 0.095 660 003 974 663 110 656;
  • 72) 0.095 660 003 974 663 110 656 × 2 = 0 + 0.191 320 007 949 326 221 312;
  • 73) 0.191 320 007 949 326 221 312 × 2 = 0 + 0.382 640 015 898 652 442 624;
  • 74) 0.382 640 015 898 652 442 624 × 2 = 0 + 0.765 280 031 797 304 885 248;
  • 75) 0.765 280 031 797 304 885 248 × 2 = 1 + 0.530 560 063 594 609 770 496;
  • 76) 0.530 560 063 594 609 770 496 × 2 = 1 + 0.061 120 127 189 219 540 992;
  • 77) 0.061 120 127 189 219 540 992 × 2 = 0 + 0.122 240 254 378 439 081 984;
  • 78) 0.122 240 254 378 439 081 984 × 2 = 0 + 0.244 480 508 756 878 163 968;
  • 79) 0.244 480 508 756 878 163 968 × 2 = 0 + 0.488 961 017 513 756 327 936;
  • 80) 0.488 961 017 513 756 327 936 × 2 = 0 + 0.977 922 035 027 512 655 872;
  • 81) 0.977 922 035 027 512 655 872 × 2 = 1 + 0.955 844 070 055 025 311 744;
  • 82) 0.955 844 070 055 025 311 744 × 2 = 1 + 0.911 688 140 110 050 623 488;
  • 83) 0.911 688 140 110 050 623 488 × 2 = 1 + 0.823 376 280 220 101 246 976;
  • 84) 0.823 376 280 220 101 246 976 × 2 = 1 + 0.646 752 560 440 202 493 952;
  • 85) 0.646 752 560 440 202 493 952 × 2 = 1 + 0.293 505 120 880 404 987 904;
  • 86) 0.293 505 120 880 404 987 904 × 2 = 0 + 0.587 010 241 760 809 975 808;
  • 87) 0.587 010 241 760 809 975 808 × 2 = 1 + 0.174 020 483 521 619 951 616;
  • 88) 0.174 020 483 521 619 951 616 × 2 = 0 + 0.348 040 967 043 239 903 232;
  • 89) 0.348 040 967 043 239 903 232 × 2 = 0 + 0.696 081 934 086 479 806 464;
  • 90) 0.696 081 934 086 479 806 464 × 2 = 1 + 0.392 163 868 172 959 612 928;
  • 91) 0.392 163 868 172 959 612 928 × 2 = 0 + 0.784 327 736 345 919 225 856;
  • 92) 0.784 327 736 345 919 225 856 × 2 = 1 + 0.568 655 472 691 838 451 712;
  • 93) 0.568 655 472 691 838 451 712 × 2 = 1 + 0.137 310 945 383 676 903 424;
  • 94) 0.137 310 945 383 676 903 424 × 2 = 0 + 0.274 621 890 767 353 806 848;
  • 95) 0.274 621 890 767 353 806 848 × 2 = 0 + 0.549 243 781 534 707 613 696;

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:


0.000 000 000 000 176 557 522(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0110 1100 0011 0000 1111 1010 0101 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 522(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0110 1100 0011 0000 1111 1010 0101 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:


0.000 000 000 000 176 557 522(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0110 1100 0011 0000 1111 1010 0101 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1111 0110 1100 0011 0000 1111 1010 0101 100(2) × 20 =

1.1000 1101 1001 0010 0111 1011 0110 0001 1000 0111 1101 0010 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 0010 0111 1011 0110 0001 1000 0111 1101 0010 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 0010 0111 1011 0110 0001 1000 0111 1101 0010 1100 =

1000 1101 1001 0010 0111 1011 0110 0001 1000 0111 1101 0010 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 0010 0111 1011 0110 0001 1000 0111 1101 0010 1100


Decimal number -0.000 000 000 000 176 557 522 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0111 1011 0110 0001 1000 0111 1101 0010 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