-0.000 000 000 000 176 557 383 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 383(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
-0.000 000 000 000 176 557 383(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:

|-0.000 000 000 000 176 557 383| = 0.000 000 000 000 176 557 383


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: 0.000 000 000 000 176 557 383.

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) 0.000 000 000 000 176 557 383 × 2 = 0 + 0.000 000 000 000 353 114 766;
  • 2) 0.000 000 000 000 353 114 766 × 2 = 0 + 0.000 000 000 000 706 229 532;
  • 3) 0.000 000 000 000 706 229 532 × 2 = 0 + 0.000 000 000 001 412 459 064;
  • 4) 0.000 000 000 001 412 459 064 × 2 = 0 + 0.000 000 000 002 824 918 128;
  • 5) 0.000 000 000 002 824 918 128 × 2 = 0 + 0.000 000 000 005 649 836 256;
  • 6) 0.000 000 000 005 649 836 256 × 2 = 0 + 0.000 000 000 011 299 672 512;
  • 7) 0.000 000 000 011 299 672 512 × 2 = 0 + 0.000 000 000 022 599 345 024;
  • 8) 0.000 000 000 022 599 345 024 × 2 = 0 + 0.000 000 000 045 198 690 048;
  • 9) 0.000 000 000 045 198 690 048 × 2 = 0 + 0.000 000 000 090 397 380 096;
  • 10) 0.000 000 000 090 397 380 096 × 2 = 0 + 0.000 000 000 180 794 760 192;
  • 11) 0.000 000 000 180 794 760 192 × 2 = 0 + 0.000 000 000 361 589 520 384;
  • 12) 0.000 000 000 361 589 520 384 × 2 = 0 + 0.000 000 000 723 179 040 768;
  • 13) 0.000 000 000 723 179 040 768 × 2 = 0 + 0.000 000 001 446 358 081 536;
  • 14) 0.000 000 001 446 358 081 536 × 2 = 0 + 0.000 000 002 892 716 163 072;
  • 15) 0.000 000 002 892 716 163 072 × 2 = 0 + 0.000 000 005 785 432 326 144;
  • 16) 0.000 000 005 785 432 326 144 × 2 = 0 + 0.000 000 011 570 864 652 288;
  • 17) 0.000 000 011 570 864 652 288 × 2 = 0 + 0.000 000 023 141 729 304 576;
  • 18) 0.000 000 023 141 729 304 576 × 2 = 0 + 0.000 000 046 283 458 609 152;
  • 19) 0.000 000 046 283 458 609 152 × 2 = 0 + 0.000 000 092 566 917 218 304;
  • 20) 0.000 000 092 566 917 218 304 × 2 = 0 + 0.000 000 185 133 834 436 608;
  • 21) 0.000 000 185 133 834 436 608 × 2 = 0 + 0.000 000 370 267 668 873 216;
  • 22) 0.000 000 370 267 668 873 216 × 2 = 0 + 0.000 000 740 535 337 746 432;
  • 23) 0.000 000 740 535 337 746 432 × 2 = 0 + 0.000 001 481 070 675 492 864;
  • 24) 0.000 001 481 070 675 492 864 × 2 = 0 + 0.000 002 962 141 350 985 728;
  • 25) 0.000 002 962 141 350 985 728 × 2 = 0 + 0.000 005 924 282 701 971 456;
  • 26) 0.000 005 924 282 701 971 456 × 2 = 0 + 0.000 011 848 565 403 942 912;
  • 27) 0.000 011 848 565 403 942 912 × 2 = 0 + 0.000 023 697 130 807 885 824;
  • 28) 0.000 023 697 130 807 885 824 × 2 = 0 + 0.000 047 394 261 615 771 648;
  • 29) 0.000 047 394 261 615 771 648 × 2 = 0 + 0.000 094 788 523 231 543 296;
  • 30) 0.000 094 788 523 231 543 296 × 2 = 0 + 0.000 189 577 046 463 086 592;
  • 31) 0.000 189 577 046 463 086 592 × 2 = 0 + 0.000 379 154 092 926 173 184;
  • 32) 0.000 379 154 092 926 173 184 × 2 = 0 + 0.000 758 308 185 852 346 368;
  • 33) 0.000 758 308 185 852 346 368 × 2 = 0 + 0.001 516 616 371 704 692 736;
  • 34) 0.001 516 616 371 704 692 736 × 2 = 0 + 0.003 033 232 743 409 385 472;
  • 35) 0.003 033 232 743 409 385 472 × 2 = 0 + 0.006 066 465 486 818 770 944;
  • 36) 0.006 066 465 486 818 770 944 × 2 = 0 + 0.012 132 930 973 637 541 888;
  • 37) 0.012 132 930 973 637 541 888 × 2 = 0 + 0.024 265 861 947 275 083 776;
  • 38) 0.024 265 861 947 275 083 776 × 2 = 0 + 0.048 531 723 894 550 167 552;
  • 39) 0.048 531 723 894 550 167 552 × 2 = 0 + 0.097 063 447 789 100 335 104;
  • 40) 0.097 063 447 789 100 335 104 × 2 = 0 + 0.194 126 895 578 200 670 208;
  • 41) 0.194 126 895 578 200 670 208 × 2 = 0 + 0.388 253 791 156 401 340 416;
  • 42) 0.388 253 791 156 401 340 416 × 2 = 0 + 0.776 507 582 312 802 680 832;
  • 43) 0.776 507 582 312 802 680 832 × 2 = 1 + 0.553 015 164 625 605 361 664;
  • 44) 0.553 015 164 625 605 361 664 × 2 = 1 + 0.106 030 329 251 210 723 328;
  • 45) 0.106 030 329 251 210 723 328 × 2 = 0 + 0.212 060 658 502 421 446 656;
  • 46) 0.212 060 658 502 421 446 656 × 2 = 0 + 0.424 121 317 004 842 893 312;
  • 47) 0.424 121 317 004 842 893 312 × 2 = 0 + 0.848 242 634 009 685 786 624;
  • 48) 0.848 242 634 009 685 786 624 × 2 = 1 + 0.696 485 268 019 371 573 248;
  • 49) 0.696 485 268 019 371 573 248 × 2 = 1 + 0.392 970 536 038 743 146 496;
  • 50) 0.392 970 536 038 743 146 496 × 2 = 0 + 0.785 941 072 077 486 292 992;
  • 51) 0.785 941 072 077 486 292 992 × 2 = 1 + 0.571 882 144 154 972 585 984;
  • 52) 0.571 882 144 154 972 585 984 × 2 = 1 + 0.143 764 288 309 945 171 968;
  • 53) 0.143 764 288 309 945 171 968 × 2 = 0 + 0.287 528 576 619 890 343 936;
  • 54) 0.287 528 576 619 890 343 936 × 2 = 0 + 0.575 057 153 239 780 687 872;
  • 55) 0.575 057 153 239 780 687 872 × 2 = 1 + 0.150 114 306 479 561 375 744;
  • 56) 0.150 114 306 479 561 375 744 × 2 = 0 + 0.300 228 612 959 122 751 488;
  • 57) 0.300 228 612 959 122 751 488 × 2 = 0 + 0.600 457 225 918 245 502 976;
  • 58) 0.600 457 225 918 245 502 976 × 2 = 1 + 0.200 914 451 836 491 005 952;
  • 59) 0.200 914 451 836 491 005 952 × 2 = 0 + 0.401 828 903 672 982 011 904;
  • 60) 0.401 828 903 672 982 011 904 × 2 = 0 + 0.803 657 807 345 964 023 808;
  • 61) 0.803 657 807 345 964 023 808 × 2 = 1 + 0.607 315 614 691 928 047 616;
  • 62) 0.607 315 614 691 928 047 616 × 2 = 1 + 0.214 631 229 383 856 095 232;
  • 63) 0.214 631 229 383 856 095 232 × 2 = 0 + 0.429 262 458 767 712 190 464;
  • 64) 0.429 262 458 767 712 190 464 × 2 = 0 + 0.858 524 917 535 424 380 928;
  • 65) 0.858 524 917 535 424 380 928 × 2 = 1 + 0.717 049 835 070 848 761 856;
  • 66) 0.717 049 835 070 848 761 856 × 2 = 1 + 0.434 099 670 141 697 523 712;
  • 67) 0.434 099 670 141 697 523 712 × 2 = 0 + 0.868 199 340 283 395 047 424;
  • 68) 0.868 199 340 283 395 047 424 × 2 = 1 + 0.736 398 680 566 790 094 848;
  • 69) 0.736 398 680 566 790 094 848 × 2 = 1 + 0.472 797 361 133 580 189 696;
  • 70) 0.472 797 361 133 580 189 696 × 2 = 0 + 0.945 594 722 267 160 379 392;
  • 71) 0.945 594 722 267 160 379 392 × 2 = 1 + 0.891 189 444 534 320 758 784;
  • 72) 0.891 189 444 534 320 758 784 × 2 = 1 + 0.782 378 889 068 641 517 568;
  • 73) 0.782 378 889 068 641 517 568 × 2 = 1 + 0.564 757 778 137 283 035 136;
  • 74) 0.564 757 778 137 283 035 136 × 2 = 1 + 0.129 515 556 274 566 070 272;
  • 75) 0.129 515 556 274 566 070 272 × 2 = 0 + 0.259 031 112 549 132 140 544;
  • 76) 0.259 031 112 549 132 140 544 × 2 = 0 + 0.518 062 225 098 264 281 088;
  • 77) 0.518 062 225 098 264 281 088 × 2 = 1 + 0.036 124 450 196 528 562 176;
  • 78) 0.036 124 450 196 528 562 176 × 2 = 0 + 0.072 248 900 393 057 124 352;
  • 79) 0.072 248 900 393 057 124 352 × 2 = 0 + 0.144 497 800 786 114 248 704;
  • 80) 0.144 497 800 786 114 248 704 × 2 = 0 + 0.288 995 601 572 228 497 408;
  • 81) 0.288 995 601 572 228 497 408 × 2 = 0 + 0.577 991 203 144 456 994 816;
  • 82) 0.577 991 203 144 456 994 816 × 2 = 1 + 0.155 982 406 288 913 989 632;
  • 83) 0.155 982 406 288 913 989 632 × 2 = 0 + 0.311 964 812 577 827 979 264;
  • 84) 0.311 964 812 577 827 979 264 × 2 = 0 + 0.623 929 625 155 655 958 528;
  • 85) 0.623 929 625 155 655 958 528 × 2 = 1 + 0.247 859 250 311 311 917 056;
  • 86) 0.247 859 250 311 311 917 056 × 2 = 0 + 0.495 718 500 622 623 834 112;
  • 87) 0.495 718 500 622 623 834 112 × 2 = 0 + 0.991 437 001 245 247 668 224;
  • 88) 0.991 437 001 245 247 668 224 × 2 = 1 + 0.982 874 002 490 495 336 448;
  • 89) 0.982 874 002 490 495 336 448 × 2 = 1 + 0.965 748 004 980 990 672 896;
  • 90) 0.965 748 004 980 990 672 896 × 2 = 1 + 0.931 496 009 961 981 345 792;
  • 91) 0.931 496 009 961 981 345 792 × 2 = 1 + 0.862 992 019 923 962 691 584;
  • 92) 0.862 992 019 923 962 691 584 × 2 = 1 + 0.725 984 039 847 925 383 168;
  • 93) 0.725 984 039 847 925 383 168 × 2 = 1 + 0.451 968 079 695 850 766 336;
  • 94) 0.451 968 079 695 850 766 336 × 2 = 0 + 0.903 936 159 391 701 532 672;
  • 95) 0.903 936 159 391 701 532 672 × 2 = 1 + 0.807 872 318 783 403 065 344;

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:


0.000 000 000 000 176 557 383(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1101 1011 1100 1000 0100 1001 1111 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 383(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1101 1011 1100 1000 0100 1001 1111 101(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:


0.000 000 000 000 176 557 383(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1101 1011 1100 1000 0100 1001 1111 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1101 1011 1100 1000 0100 1001 1111 101(2) × 20 =

1.1000 1101 1001 0010 0110 0110 1101 1110 0100 0010 0100 1111 1101(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 0110 1101 1110 0100 0010 0100 1111 1101


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 0110 1101 1110 0100 0010 0100 1111 1101 =

1000 1101 1001 0010 0110 0110 1101 1110 0100 0010 0100 1111 1101


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 0110 1101 1110 0100 0010 0100 1111 1101


Decimal number -0.000 000 000 000 176 557 383 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0110 0110 1101 1110 0100 0010 0100 1111 1101

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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