-0.000 000 000 000 176 569 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 569(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
-0.000 000 000 000 176 569(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:

|-0.000 000 000 000 176 569| = 0.000 000 000 000 176 569


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: 0.000 000 000 000 176 569.

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) 0.000 000 000 000 176 569 × 2 = 0 + 0.000 000 000 000 353 138;
  • 2) 0.000 000 000 000 353 138 × 2 = 0 + 0.000 000 000 000 706 276;
  • 3) 0.000 000 000 000 706 276 × 2 = 0 + 0.000 000 000 001 412 552;
  • 4) 0.000 000 000 001 412 552 × 2 = 0 + 0.000 000 000 002 825 104;
  • 5) 0.000 000 000 002 825 104 × 2 = 0 + 0.000 000 000 005 650 208;
  • 6) 0.000 000 000 005 650 208 × 2 = 0 + 0.000 000 000 011 300 416;
  • 7) 0.000 000 000 011 300 416 × 2 = 0 + 0.000 000 000 022 600 832;
  • 8) 0.000 000 000 022 600 832 × 2 = 0 + 0.000 000 000 045 201 664;
  • 9) 0.000 000 000 045 201 664 × 2 = 0 + 0.000 000 000 090 403 328;
  • 10) 0.000 000 000 090 403 328 × 2 = 0 + 0.000 000 000 180 806 656;
  • 11) 0.000 000 000 180 806 656 × 2 = 0 + 0.000 000 000 361 613 312;
  • 12) 0.000 000 000 361 613 312 × 2 = 0 + 0.000 000 000 723 226 624;
  • 13) 0.000 000 000 723 226 624 × 2 = 0 + 0.000 000 001 446 453 248;
  • 14) 0.000 000 001 446 453 248 × 2 = 0 + 0.000 000 002 892 906 496;
  • 15) 0.000 000 002 892 906 496 × 2 = 0 + 0.000 000 005 785 812 992;
  • 16) 0.000 000 005 785 812 992 × 2 = 0 + 0.000 000 011 571 625 984;
  • 17) 0.000 000 011 571 625 984 × 2 = 0 + 0.000 000 023 143 251 968;
  • 18) 0.000 000 023 143 251 968 × 2 = 0 + 0.000 000 046 286 503 936;
  • 19) 0.000 000 046 286 503 936 × 2 = 0 + 0.000 000 092 573 007 872;
  • 20) 0.000 000 092 573 007 872 × 2 = 0 + 0.000 000 185 146 015 744;
  • 21) 0.000 000 185 146 015 744 × 2 = 0 + 0.000 000 370 292 031 488;
  • 22) 0.000 000 370 292 031 488 × 2 = 0 + 0.000 000 740 584 062 976;
  • 23) 0.000 000 740 584 062 976 × 2 = 0 + 0.000 001 481 168 125 952;
  • 24) 0.000 001 481 168 125 952 × 2 = 0 + 0.000 002 962 336 251 904;
  • 25) 0.000 002 962 336 251 904 × 2 = 0 + 0.000 005 924 672 503 808;
  • 26) 0.000 005 924 672 503 808 × 2 = 0 + 0.000 011 849 345 007 616;
  • 27) 0.000 011 849 345 007 616 × 2 = 0 + 0.000 023 698 690 015 232;
  • 28) 0.000 023 698 690 015 232 × 2 = 0 + 0.000 047 397 380 030 464;
  • 29) 0.000 047 397 380 030 464 × 2 = 0 + 0.000 094 794 760 060 928;
  • 30) 0.000 094 794 760 060 928 × 2 = 0 + 0.000 189 589 520 121 856;
  • 31) 0.000 189 589 520 121 856 × 2 = 0 + 0.000 379 179 040 243 712;
  • 32) 0.000 379 179 040 243 712 × 2 = 0 + 0.000 758 358 080 487 424;
  • 33) 0.000 758 358 080 487 424 × 2 = 0 + 0.001 516 716 160 974 848;
  • 34) 0.001 516 716 160 974 848 × 2 = 0 + 0.003 033 432 321 949 696;
  • 35) 0.003 033 432 321 949 696 × 2 = 0 + 0.006 066 864 643 899 392;
  • 36) 0.006 066 864 643 899 392 × 2 = 0 + 0.012 133 729 287 798 784;
  • 37) 0.012 133 729 287 798 784 × 2 = 0 + 0.024 267 458 575 597 568;
  • 38) 0.024 267 458 575 597 568 × 2 = 0 + 0.048 534 917 151 195 136;
  • 39) 0.048 534 917 151 195 136 × 2 = 0 + 0.097 069 834 302 390 272;
  • 40) 0.097 069 834 302 390 272 × 2 = 0 + 0.194 139 668 604 780 544;
  • 41) 0.194 139 668 604 780 544 × 2 = 0 + 0.388 279 337 209 561 088;
  • 42) 0.388 279 337 209 561 088 × 2 = 0 + 0.776 558 674 419 122 176;
  • 43) 0.776 558 674 419 122 176 × 2 = 1 + 0.553 117 348 838 244 352;
  • 44) 0.553 117 348 838 244 352 × 2 = 1 + 0.106 234 697 676 488 704;
  • 45) 0.106 234 697 676 488 704 × 2 = 0 + 0.212 469 395 352 977 408;
  • 46) 0.212 469 395 352 977 408 × 2 = 0 + 0.424 938 790 705 954 816;
  • 47) 0.424 938 790 705 954 816 × 2 = 0 + 0.849 877 581 411 909 632;
  • 48) 0.849 877 581 411 909 632 × 2 = 1 + 0.699 755 162 823 819 264;
  • 49) 0.699 755 162 823 819 264 × 2 = 1 + 0.399 510 325 647 638 528;
  • 50) 0.399 510 325 647 638 528 × 2 = 0 + 0.799 020 651 295 277 056;
  • 51) 0.799 020 651 295 277 056 × 2 = 1 + 0.598 041 302 590 554 112;
  • 52) 0.598 041 302 590 554 112 × 2 = 1 + 0.196 082 605 181 108 224;
  • 53) 0.196 082 605 181 108 224 × 2 = 0 + 0.392 165 210 362 216 448;
  • 54) 0.392 165 210 362 216 448 × 2 = 0 + 0.784 330 420 724 432 896;
  • 55) 0.784 330 420 724 432 896 × 2 = 1 + 0.568 660 841 448 865 792;
  • 56) 0.568 660 841 448 865 792 × 2 = 1 + 0.137 321 682 897 731 584;
  • 57) 0.137 321 682 897 731 584 × 2 = 0 + 0.274 643 365 795 463 168;
  • 58) 0.274 643 365 795 463 168 × 2 = 0 + 0.549 286 731 590 926 336;
  • 59) 0.549 286 731 590 926 336 × 2 = 1 + 0.098 573 463 181 852 672;
  • 60) 0.098 573 463 181 852 672 × 2 = 0 + 0.197 146 926 363 705 344;
  • 61) 0.197 146 926 363 705 344 × 2 = 0 + 0.394 293 852 727 410 688;
  • 62) 0.394 293 852 727 410 688 × 2 = 0 + 0.788 587 705 454 821 376;
  • 63) 0.788 587 705 454 821 376 × 2 = 1 + 0.577 175 410 909 642 752;
  • 64) 0.577 175 410 909 642 752 × 2 = 1 + 0.154 350 821 819 285 504;
  • 65) 0.154 350 821 819 285 504 × 2 = 0 + 0.308 701 643 638 571 008;
  • 66) 0.308 701 643 638 571 008 × 2 = 0 + 0.617 403 287 277 142 016;
  • 67) 0.617 403 287 277 142 016 × 2 = 1 + 0.234 806 574 554 284 032;
  • 68) 0.234 806 574 554 284 032 × 2 = 0 + 0.469 613 149 108 568 064;
  • 69) 0.469 613 149 108 568 064 × 2 = 0 + 0.939 226 298 217 136 128;
  • 70) 0.939 226 298 217 136 128 × 2 = 1 + 0.878 452 596 434 272 256;
  • 71) 0.878 452 596 434 272 256 × 2 = 1 + 0.756 905 192 868 544 512;
  • 72) 0.756 905 192 868 544 512 × 2 = 1 + 0.513 810 385 737 089 024;
  • 73) 0.513 810 385 737 089 024 × 2 = 1 + 0.027 620 771 474 178 048;
  • 74) 0.027 620 771 474 178 048 × 2 = 0 + 0.055 241 542 948 356 096;
  • 75) 0.055 241 542 948 356 096 × 2 = 0 + 0.110 483 085 896 712 192;
  • 76) 0.110 483 085 896 712 192 × 2 = 0 + 0.220 966 171 793 424 384;
  • 77) 0.220 966 171 793 424 384 × 2 = 0 + 0.441 932 343 586 848 768;
  • 78) 0.441 932 343 586 848 768 × 2 = 0 + 0.883 864 687 173 697 536;
  • 79) 0.883 864 687 173 697 536 × 2 = 1 + 0.767 729 374 347 395 072;
  • 80) 0.767 729 374 347 395 072 × 2 = 1 + 0.535 458 748 694 790 144;
  • 81) 0.535 458 748 694 790 144 × 2 = 1 + 0.070 917 497 389 580 288;
  • 82) 0.070 917 497 389 580 288 × 2 = 0 + 0.141 834 994 779 160 576;
  • 83) 0.141 834 994 779 160 576 × 2 = 0 + 0.283 669 989 558 321 152;
  • 84) 0.283 669 989 558 321 152 × 2 = 0 + 0.567 339 979 116 642 304;
  • 85) 0.567 339 979 116 642 304 × 2 = 1 + 0.134 679 958 233 284 608;
  • 86) 0.134 679 958 233 284 608 × 2 = 0 + 0.269 359 916 466 569 216;
  • 87) 0.269 359 916 466 569 216 × 2 = 0 + 0.538 719 832 933 138 432;
  • 88) 0.538 719 832 933 138 432 × 2 = 1 + 0.077 439 665 866 276 864;
  • 89) 0.077 439 665 866 276 864 × 2 = 0 + 0.154 879 331 732 553 728;
  • 90) 0.154 879 331 732 553 728 × 2 = 0 + 0.309 758 663 465 107 456;
  • 91) 0.309 758 663 465 107 456 × 2 = 0 + 0.619 517 326 930 214 912;
  • 92) 0.619 517 326 930 214 912 × 2 = 1 + 0.239 034 653 860 429 824;
  • 93) 0.239 034 653 860 429 824 × 2 = 0 + 0.478 069 307 720 859 648;
  • 94) 0.478 069 307 720 859 648 × 2 = 0 + 0.956 138 615 441 719 296;
  • 95) 0.956 138 615 441 719 296 × 2 = 1 + 0.912 277 230 883 438 592;

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:


0.000 000 000 000 176 569(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0010 0011 0010 0111 1000 0011 1000 1001 0001 001(2)

6. Positive number before normalization:

0.000 000 000 000 176 569(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0010 0011 0010 0111 1000 0011 1000 1001 0001 001(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:


0.000 000 000 000 176 569(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0010 0011 0010 0111 1000 0011 1000 1001 0001 001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0010 0011 0010 0111 1000 0011 1000 1001 0001 001(2) × 20 =

1.1000 1101 1001 1001 0001 1001 0011 1100 0001 1100 0100 1000 1001(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 1001 0001 1001 0011 1100 0001 1100 0100 1000 1001


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 1001 0001 1001 0011 1100 0001 1100 0100 1000 1001 =

1000 1101 1001 1001 0001 1001 0011 1100 0001 1100 0100 1000 1001


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 1001 0001 1001 0011 1100 0001 1100 0100 1000 1001


Decimal number -0.000 000 000 000 176 569 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 1001 0001 1001 0011 1100 0001 1100 0100 1000 1001

Share

» Convert decimal -0.000 000 000 000 176 567 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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