-0.000 000 000 000 176 557 401 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 401(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
-0.000 000 000 000 176 557 401(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:

|-0.000 000 000 000 176 557 401| = 0.000 000 000 000 176 557 401


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: 0.000 000 000 000 176 557 401.

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) 0.000 000 000 000 176 557 401 × 2 = 0 + 0.000 000 000 000 353 114 802;
  • 2) 0.000 000 000 000 353 114 802 × 2 = 0 + 0.000 000 000 000 706 229 604;
  • 3) 0.000 000 000 000 706 229 604 × 2 = 0 + 0.000 000 000 001 412 459 208;
  • 4) 0.000 000 000 001 412 459 208 × 2 = 0 + 0.000 000 000 002 824 918 416;
  • 5) 0.000 000 000 002 824 918 416 × 2 = 0 + 0.000 000 000 005 649 836 832;
  • 6) 0.000 000 000 005 649 836 832 × 2 = 0 + 0.000 000 000 011 299 673 664;
  • 7) 0.000 000 000 011 299 673 664 × 2 = 0 + 0.000 000 000 022 599 347 328;
  • 8) 0.000 000 000 022 599 347 328 × 2 = 0 + 0.000 000 000 045 198 694 656;
  • 9) 0.000 000 000 045 198 694 656 × 2 = 0 + 0.000 000 000 090 397 389 312;
  • 10) 0.000 000 000 090 397 389 312 × 2 = 0 + 0.000 000 000 180 794 778 624;
  • 11) 0.000 000 000 180 794 778 624 × 2 = 0 + 0.000 000 000 361 589 557 248;
  • 12) 0.000 000 000 361 589 557 248 × 2 = 0 + 0.000 000 000 723 179 114 496;
  • 13) 0.000 000 000 723 179 114 496 × 2 = 0 + 0.000 000 001 446 358 228 992;
  • 14) 0.000 000 001 446 358 228 992 × 2 = 0 + 0.000 000 002 892 716 457 984;
  • 15) 0.000 000 002 892 716 457 984 × 2 = 0 + 0.000 000 005 785 432 915 968;
  • 16) 0.000 000 005 785 432 915 968 × 2 = 0 + 0.000 000 011 570 865 831 936;
  • 17) 0.000 000 011 570 865 831 936 × 2 = 0 + 0.000 000 023 141 731 663 872;
  • 18) 0.000 000 023 141 731 663 872 × 2 = 0 + 0.000 000 046 283 463 327 744;
  • 19) 0.000 000 046 283 463 327 744 × 2 = 0 + 0.000 000 092 566 926 655 488;
  • 20) 0.000 000 092 566 926 655 488 × 2 = 0 + 0.000 000 185 133 853 310 976;
  • 21) 0.000 000 185 133 853 310 976 × 2 = 0 + 0.000 000 370 267 706 621 952;
  • 22) 0.000 000 370 267 706 621 952 × 2 = 0 + 0.000 000 740 535 413 243 904;
  • 23) 0.000 000 740 535 413 243 904 × 2 = 0 + 0.000 001 481 070 826 487 808;
  • 24) 0.000 001 481 070 826 487 808 × 2 = 0 + 0.000 002 962 141 652 975 616;
  • 25) 0.000 002 962 141 652 975 616 × 2 = 0 + 0.000 005 924 283 305 951 232;
  • 26) 0.000 005 924 283 305 951 232 × 2 = 0 + 0.000 011 848 566 611 902 464;
  • 27) 0.000 011 848 566 611 902 464 × 2 = 0 + 0.000 023 697 133 223 804 928;
  • 28) 0.000 023 697 133 223 804 928 × 2 = 0 + 0.000 047 394 266 447 609 856;
  • 29) 0.000 047 394 266 447 609 856 × 2 = 0 + 0.000 094 788 532 895 219 712;
  • 30) 0.000 094 788 532 895 219 712 × 2 = 0 + 0.000 189 577 065 790 439 424;
  • 31) 0.000 189 577 065 790 439 424 × 2 = 0 + 0.000 379 154 131 580 878 848;
  • 32) 0.000 379 154 131 580 878 848 × 2 = 0 + 0.000 758 308 263 161 757 696;
  • 33) 0.000 758 308 263 161 757 696 × 2 = 0 + 0.001 516 616 526 323 515 392;
  • 34) 0.001 516 616 526 323 515 392 × 2 = 0 + 0.003 033 233 052 647 030 784;
  • 35) 0.003 033 233 052 647 030 784 × 2 = 0 + 0.006 066 466 105 294 061 568;
  • 36) 0.006 066 466 105 294 061 568 × 2 = 0 + 0.012 132 932 210 588 123 136;
  • 37) 0.012 132 932 210 588 123 136 × 2 = 0 + 0.024 265 864 421 176 246 272;
  • 38) 0.024 265 864 421 176 246 272 × 2 = 0 + 0.048 531 728 842 352 492 544;
  • 39) 0.048 531 728 842 352 492 544 × 2 = 0 + 0.097 063 457 684 704 985 088;
  • 40) 0.097 063 457 684 704 985 088 × 2 = 0 + 0.194 126 915 369 409 970 176;
  • 41) 0.194 126 915 369 409 970 176 × 2 = 0 + 0.388 253 830 738 819 940 352;
  • 42) 0.388 253 830 738 819 940 352 × 2 = 0 + 0.776 507 661 477 639 880 704;
  • 43) 0.776 507 661 477 639 880 704 × 2 = 1 + 0.553 015 322 955 279 761 408;
  • 44) 0.553 015 322 955 279 761 408 × 2 = 1 + 0.106 030 645 910 559 522 816;
  • 45) 0.106 030 645 910 559 522 816 × 2 = 0 + 0.212 061 291 821 119 045 632;
  • 46) 0.212 061 291 821 119 045 632 × 2 = 0 + 0.424 122 583 642 238 091 264;
  • 47) 0.424 122 583 642 238 091 264 × 2 = 0 + 0.848 245 167 284 476 182 528;
  • 48) 0.848 245 167 284 476 182 528 × 2 = 1 + 0.696 490 334 568 952 365 056;
  • 49) 0.696 490 334 568 952 365 056 × 2 = 1 + 0.392 980 669 137 904 730 112;
  • 50) 0.392 980 669 137 904 730 112 × 2 = 0 + 0.785 961 338 275 809 460 224;
  • 51) 0.785 961 338 275 809 460 224 × 2 = 1 + 0.571 922 676 551 618 920 448;
  • 52) 0.571 922 676 551 618 920 448 × 2 = 1 + 0.143 845 353 103 237 840 896;
  • 53) 0.143 845 353 103 237 840 896 × 2 = 0 + 0.287 690 706 206 475 681 792;
  • 54) 0.287 690 706 206 475 681 792 × 2 = 0 + 0.575 381 412 412 951 363 584;
  • 55) 0.575 381 412 412 951 363 584 × 2 = 1 + 0.150 762 824 825 902 727 168;
  • 56) 0.150 762 824 825 902 727 168 × 2 = 0 + 0.301 525 649 651 805 454 336;
  • 57) 0.301 525 649 651 805 454 336 × 2 = 0 + 0.603 051 299 303 610 908 672;
  • 58) 0.603 051 299 303 610 908 672 × 2 = 1 + 0.206 102 598 607 221 817 344;
  • 59) 0.206 102 598 607 221 817 344 × 2 = 0 + 0.412 205 197 214 443 634 688;
  • 60) 0.412 205 197 214 443 634 688 × 2 = 0 + 0.824 410 394 428 887 269 376;
  • 61) 0.824 410 394 428 887 269 376 × 2 = 1 + 0.648 820 788 857 774 538 752;
  • 62) 0.648 820 788 857 774 538 752 × 2 = 1 + 0.297 641 577 715 549 077 504;
  • 63) 0.297 641 577 715 549 077 504 × 2 = 0 + 0.595 283 155 431 098 155 008;
  • 64) 0.595 283 155 431 098 155 008 × 2 = 1 + 0.190 566 310 862 196 310 016;
  • 65) 0.190 566 310 862 196 310 016 × 2 = 0 + 0.381 132 621 724 392 620 032;
  • 66) 0.381 132 621 724 392 620 032 × 2 = 0 + 0.762 265 243 448 785 240 064;
  • 67) 0.762 265 243 448 785 240 064 × 2 = 1 + 0.524 530 486 897 570 480 128;
  • 68) 0.524 530 486 897 570 480 128 × 2 = 1 + 0.049 060 973 795 140 960 256;
  • 69) 0.049 060 973 795 140 960 256 × 2 = 0 + 0.098 121 947 590 281 920 512;
  • 70) 0.098 121 947 590 281 920 512 × 2 = 0 + 0.196 243 895 180 563 841 024;
  • 71) 0.196 243 895 180 563 841 024 × 2 = 0 + 0.392 487 790 361 127 682 048;
  • 72) 0.392 487 790 361 127 682 048 × 2 = 0 + 0.784 975 580 722 255 364 096;
  • 73) 0.784 975 580 722 255 364 096 × 2 = 1 + 0.569 951 161 444 510 728 192;
  • 74) 0.569 951 161 444 510 728 192 × 2 = 1 + 0.139 902 322 889 021 456 384;
  • 75) 0.139 902 322 889 021 456 384 × 2 = 0 + 0.279 804 645 778 042 912 768;
  • 76) 0.279 804 645 778 042 912 768 × 2 = 0 + 0.559 609 291 556 085 825 536;
  • 77) 0.559 609 291 556 085 825 536 × 2 = 1 + 0.119 218 583 112 171 651 072;
  • 78) 0.119 218 583 112 171 651 072 × 2 = 0 + 0.238 437 166 224 343 302 144;
  • 79) 0.238 437 166 224 343 302 144 × 2 = 0 + 0.476 874 332 448 686 604 288;
  • 80) 0.476 874 332 448 686 604 288 × 2 = 0 + 0.953 748 664 897 373 208 576;
  • 81) 0.953 748 664 897 373 208 576 × 2 = 1 + 0.907 497 329 794 746 417 152;
  • 82) 0.907 497 329 794 746 417 152 × 2 = 1 + 0.814 994 659 589 492 834 304;
  • 83) 0.814 994 659 589 492 834 304 × 2 = 1 + 0.629 989 319 178 985 668 608;
  • 84) 0.629 989 319 178 985 668 608 × 2 = 1 + 0.259 978 638 357 971 337 216;
  • 85) 0.259 978 638 357 971 337 216 × 2 = 0 + 0.519 957 276 715 942 674 432;
  • 86) 0.519 957 276 715 942 674 432 × 2 = 1 + 0.039 914 553 431 885 348 864;
  • 87) 0.039 914 553 431 885 348 864 × 2 = 0 + 0.079 829 106 863 770 697 728;
  • 88) 0.079 829 106 863 770 697 728 × 2 = 0 + 0.159 658 213 727 541 395 456;
  • 89) 0.159 658 213 727 541 395 456 × 2 = 0 + 0.319 316 427 455 082 790 912;
  • 90) 0.319 316 427 455 082 790 912 × 2 = 0 + 0.638 632 854 910 165 581 824;
  • 91) 0.638 632 854 910 165 581 824 × 2 = 1 + 0.277 265 709 820 331 163 648;
  • 92) 0.277 265 709 820 331 163 648 × 2 = 0 + 0.554 531 419 640 662 327 296;
  • 93) 0.554 531 419 640 662 327 296 × 2 = 1 + 0.109 062 839 281 324 654 592;
  • 94) 0.109 062 839 281 324 654 592 × 2 = 0 + 0.218 125 678 562 649 309 184;
  • 95) 0.218 125 678 562 649 309 184 × 2 = 0 + 0.436 251 357 125 298 618 368;

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:


0.000 000 000 000 176 557 401(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 0011 0000 1100 1000 1111 0100 0010 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 401(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 0011 0000 1100 1000 1111 0100 0010 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:


0.000 000 000 000 176 557 401(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 0011 0000 1100 1000 1111 0100 0010 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 0011 0000 1100 1000 1111 0100 0010 100(2) × 20 =

1.1000 1101 1001 0010 0110 1001 1000 0110 0100 0111 1010 0001 0100(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 1001 1000 0110 0100 0111 1010 0001 0100


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 1001 1000 0110 0100 0111 1010 0001 0100 =

1000 1101 1001 0010 0110 1001 1000 0110 0100 0111 1010 0001 0100


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 1001 1000 0110 0100 0111 1010 0001 0100


Decimal number -0.000 000 000 000 176 557 401 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0110 1001 1000 0110 0100 0111 1010 0001 0100

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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