-0.000 000 000 000 176 557 442 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 442(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
-0.000 000 000 000 176 557 442(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:

|-0.000 000 000 000 176 557 442| = 0.000 000 000 000 176 557 442


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: 0.000 000 000 000 176 557 442.

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) 0.000 000 000 000 176 557 442 × 2 = 0 + 0.000 000 000 000 353 114 884;
  • 2) 0.000 000 000 000 353 114 884 × 2 = 0 + 0.000 000 000 000 706 229 768;
  • 3) 0.000 000 000 000 706 229 768 × 2 = 0 + 0.000 000 000 001 412 459 536;
  • 4) 0.000 000 000 001 412 459 536 × 2 = 0 + 0.000 000 000 002 824 919 072;
  • 5) 0.000 000 000 002 824 919 072 × 2 = 0 + 0.000 000 000 005 649 838 144;
  • 6) 0.000 000 000 005 649 838 144 × 2 = 0 + 0.000 000 000 011 299 676 288;
  • 7) 0.000 000 000 011 299 676 288 × 2 = 0 + 0.000 000 000 022 599 352 576;
  • 8) 0.000 000 000 022 599 352 576 × 2 = 0 + 0.000 000 000 045 198 705 152;
  • 9) 0.000 000 000 045 198 705 152 × 2 = 0 + 0.000 000 000 090 397 410 304;
  • 10) 0.000 000 000 090 397 410 304 × 2 = 0 + 0.000 000 000 180 794 820 608;
  • 11) 0.000 000 000 180 794 820 608 × 2 = 0 + 0.000 000 000 361 589 641 216;
  • 12) 0.000 000 000 361 589 641 216 × 2 = 0 + 0.000 000 000 723 179 282 432;
  • 13) 0.000 000 000 723 179 282 432 × 2 = 0 + 0.000 000 001 446 358 564 864;
  • 14) 0.000 000 001 446 358 564 864 × 2 = 0 + 0.000 000 002 892 717 129 728;
  • 15) 0.000 000 002 892 717 129 728 × 2 = 0 + 0.000 000 005 785 434 259 456;
  • 16) 0.000 000 005 785 434 259 456 × 2 = 0 + 0.000 000 011 570 868 518 912;
  • 17) 0.000 000 011 570 868 518 912 × 2 = 0 + 0.000 000 023 141 737 037 824;
  • 18) 0.000 000 023 141 737 037 824 × 2 = 0 + 0.000 000 046 283 474 075 648;
  • 19) 0.000 000 046 283 474 075 648 × 2 = 0 + 0.000 000 092 566 948 151 296;
  • 20) 0.000 000 092 566 948 151 296 × 2 = 0 + 0.000 000 185 133 896 302 592;
  • 21) 0.000 000 185 133 896 302 592 × 2 = 0 + 0.000 000 370 267 792 605 184;
  • 22) 0.000 000 370 267 792 605 184 × 2 = 0 + 0.000 000 740 535 585 210 368;
  • 23) 0.000 000 740 535 585 210 368 × 2 = 0 + 0.000 001 481 071 170 420 736;
  • 24) 0.000 001 481 071 170 420 736 × 2 = 0 + 0.000 002 962 142 340 841 472;
  • 25) 0.000 002 962 142 340 841 472 × 2 = 0 + 0.000 005 924 284 681 682 944;
  • 26) 0.000 005 924 284 681 682 944 × 2 = 0 + 0.000 011 848 569 363 365 888;
  • 27) 0.000 011 848 569 363 365 888 × 2 = 0 + 0.000 023 697 138 726 731 776;
  • 28) 0.000 023 697 138 726 731 776 × 2 = 0 + 0.000 047 394 277 453 463 552;
  • 29) 0.000 047 394 277 453 463 552 × 2 = 0 + 0.000 094 788 554 906 927 104;
  • 30) 0.000 094 788 554 906 927 104 × 2 = 0 + 0.000 189 577 109 813 854 208;
  • 31) 0.000 189 577 109 813 854 208 × 2 = 0 + 0.000 379 154 219 627 708 416;
  • 32) 0.000 379 154 219 627 708 416 × 2 = 0 + 0.000 758 308 439 255 416 832;
  • 33) 0.000 758 308 439 255 416 832 × 2 = 0 + 0.001 516 616 878 510 833 664;
  • 34) 0.001 516 616 878 510 833 664 × 2 = 0 + 0.003 033 233 757 021 667 328;
  • 35) 0.003 033 233 757 021 667 328 × 2 = 0 + 0.006 066 467 514 043 334 656;
  • 36) 0.006 066 467 514 043 334 656 × 2 = 0 + 0.012 132 935 028 086 669 312;
  • 37) 0.012 132 935 028 086 669 312 × 2 = 0 + 0.024 265 870 056 173 338 624;
  • 38) 0.024 265 870 056 173 338 624 × 2 = 0 + 0.048 531 740 112 346 677 248;
  • 39) 0.048 531 740 112 346 677 248 × 2 = 0 + 0.097 063 480 224 693 354 496;
  • 40) 0.097 063 480 224 693 354 496 × 2 = 0 + 0.194 126 960 449 386 708 992;
  • 41) 0.194 126 960 449 386 708 992 × 2 = 0 + 0.388 253 920 898 773 417 984;
  • 42) 0.388 253 920 898 773 417 984 × 2 = 0 + 0.776 507 841 797 546 835 968;
  • 43) 0.776 507 841 797 546 835 968 × 2 = 1 + 0.553 015 683 595 093 671 936;
  • 44) 0.553 015 683 595 093 671 936 × 2 = 1 + 0.106 031 367 190 187 343 872;
  • 45) 0.106 031 367 190 187 343 872 × 2 = 0 + 0.212 062 734 380 374 687 744;
  • 46) 0.212 062 734 380 374 687 744 × 2 = 0 + 0.424 125 468 760 749 375 488;
  • 47) 0.424 125 468 760 749 375 488 × 2 = 0 + 0.848 250 937 521 498 750 976;
  • 48) 0.848 250 937 521 498 750 976 × 2 = 1 + 0.696 501 875 042 997 501 952;
  • 49) 0.696 501 875 042 997 501 952 × 2 = 1 + 0.393 003 750 085 995 003 904;
  • 50) 0.393 003 750 085 995 003 904 × 2 = 0 + 0.786 007 500 171 990 007 808;
  • 51) 0.786 007 500 171 990 007 808 × 2 = 1 + 0.572 015 000 343 980 015 616;
  • 52) 0.572 015 000 343 980 015 616 × 2 = 1 + 0.144 030 000 687 960 031 232;
  • 53) 0.144 030 000 687 960 031 232 × 2 = 0 + 0.288 060 001 375 920 062 464;
  • 54) 0.288 060 001 375 920 062 464 × 2 = 0 + 0.576 120 002 751 840 124 928;
  • 55) 0.576 120 002 751 840 124 928 × 2 = 1 + 0.152 240 005 503 680 249 856;
  • 56) 0.152 240 005 503 680 249 856 × 2 = 0 + 0.304 480 011 007 360 499 712;
  • 57) 0.304 480 011 007 360 499 712 × 2 = 0 + 0.608 960 022 014 720 999 424;
  • 58) 0.608 960 022 014 720 999 424 × 2 = 1 + 0.217 920 044 029 441 998 848;
  • 59) 0.217 920 044 029 441 998 848 × 2 = 0 + 0.435 840 088 058 883 997 696;
  • 60) 0.435 840 088 058 883 997 696 × 2 = 0 + 0.871 680 176 117 767 995 392;
  • 61) 0.871 680 176 117 767 995 392 × 2 = 1 + 0.743 360 352 235 535 990 784;
  • 62) 0.743 360 352 235 535 990 784 × 2 = 1 + 0.486 720 704 471 071 981 568;
  • 63) 0.486 720 704 471 071 981 568 × 2 = 0 + 0.973 441 408 942 143 963 136;
  • 64) 0.973 441 408 942 143 963 136 × 2 = 1 + 0.946 882 817 884 287 926 272;
  • 65) 0.946 882 817 884 287 926 272 × 2 = 1 + 0.893 765 635 768 575 852 544;
  • 66) 0.893 765 635 768 575 852 544 × 2 = 1 + 0.787 531 271 537 151 705 088;
  • 67) 0.787 531 271 537 151 705 088 × 2 = 1 + 0.575 062 543 074 303 410 176;
  • 68) 0.575 062 543 074 303 410 176 × 2 = 1 + 0.150 125 086 148 606 820 352;
  • 69) 0.150 125 086 148 606 820 352 × 2 = 0 + 0.300 250 172 297 213 640 704;
  • 70) 0.300 250 172 297 213 640 704 × 2 = 0 + 0.600 500 344 594 427 281 408;
  • 71) 0.600 500 344 594 427 281 408 × 2 = 1 + 0.201 000 689 188 854 562 816;
  • 72) 0.201 000 689 188 854 562 816 × 2 = 0 + 0.402 001 378 377 709 125 632;
  • 73) 0.402 001 378 377 709 125 632 × 2 = 0 + 0.804 002 756 755 418 251 264;
  • 74) 0.804 002 756 755 418 251 264 × 2 = 1 + 0.608 005 513 510 836 502 528;
  • 75) 0.608 005 513 510 836 502 528 × 2 = 1 + 0.216 011 027 021 673 005 056;
  • 76) 0.216 011 027 021 673 005 056 × 2 = 0 + 0.432 022 054 043 346 010 112;
  • 77) 0.432 022 054 043 346 010 112 × 2 = 0 + 0.864 044 108 086 692 020 224;
  • 78) 0.864 044 108 086 692 020 224 × 2 = 1 + 0.728 088 216 173 384 040 448;
  • 79) 0.728 088 216 173 384 040 448 × 2 = 1 + 0.456 176 432 346 768 080 896;
  • 80) 0.456 176 432 346 768 080 896 × 2 = 0 + 0.912 352 864 693 536 161 792;
  • 81) 0.912 352 864 693 536 161 792 × 2 = 1 + 0.824 705 729 387 072 323 584;
  • 82) 0.824 705 729 387 072 323 584 × 2 = 1 + 0.649 411 458 774 144 647 168;
  • 83) 0.649 411 458 774 144 647 168 × 2 = 1 + 0.298 822 917 548 289 294 336;
  • 84) 0.298 822 917 548 289 294 336 × 2 = 0 + 0.597 645 835 096 578 588 672;
  • 85) 0.597 645 835 096 578 588 672 × 2 = 1 + 0.195 291 670 193 157 177 344;
  • 86) 0.195 291 670 193 157 177 344 × 2 = 0 + 0.390 583 340 386 314 354 688;
  • 87) 0.390 583 340 386 314 354 688 × 2 = 0 + 0.781 166 680 772 628 709 376;
  • 88) 0.781 166 680 772 628 709 376 × 2 = 1 + 0.562 333 361 545 257 418 752;
  • 89) 0.562 333 361 545 257 418 752 × 2 = 1 + 0.124 666 723 090 514 837 504;
  • 90) 0.124 666 723 090 514 837 504 × 2 = 0 + 0.249 333 446 181 029 675 008;
  • 91) 0.249 333 446 181 029 675 008 × 2 = 0 + 0.498 666 892 362 059 350 016;
  • 92) 0.498 666 892 362 059 350 016 × 2 = 0 + 0.997 333 784 724 118 700 032;
  • 93) 0.997 333 784 724 118 700 032 × 2 = 1 + 0.994 667 569 448 237 400 064;
  • 94) 0.994 667 569 448 237 400 064 × 2 = 1 + 0.989 335 138 896 474 800 128;
  • 95) 0.989 335 138 896 474 800 128 × 2 = 1 + 0.978 670 277 792 949 600 256;

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:


0.000 000 000 000 176 557 442(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1111 0010 0110 0110 1110 1001 1000 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 442(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1111 0010 0110 0110 1110 1001 1000 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:


0.000 000 000 000 176 557 442(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1111 0010 0110 0110 1110 1001 1000 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1111 0010 0110 0110 1110 1001 1000 111(2) × 20 =

1.1000 1101 1001 0010 0110 1111 1001 0011 0011 0111 0100 1100 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 1001 0010 0110 1111 1001 0011 0011 0111 0100 1100 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 1001 0010 0110 1111 1001 0011 0011 0111 0100 1100 0111 =

1000 1101 1001 0010 0110 1111 1001 0011 0011 0111 0100 1100 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 1001 0010 0110 1111 1001 0011 0011 0111 0100 1100 0111


Decimal number -0.000 000 000 000 176 557 442 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0110 1111 1001 0011 0011 0111 0100 1100 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