-0.000 000 000 000 176 557 574 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 574(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
-0.000 000 000 000 176 557 574(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:

|-0.000 000 000 000 176 557 574| = 0.000 000 000 000 176 557 574


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: 0.000 000 000 000 176 557 574.

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) 0.000 000 000 000 176 557 574 × 2 = 0 + 0.000 000 000 000 353 115 148;
  • 2) 0.000 000 000 000 353 115 148 × 2 = 0 + 0.000 000 000 000 706 230 296;
  • 3) 0.000 000 000 000 706 230 296 × 2 = 0 + 0.000 000 000 001 412 460 592;
  • 4) 0.000 000 000 001 412 460 592 × 2 = 0 + 0.000 000 000 002 824 921 184;
  • 5) 0.000 000 000 002 824 921 184 × 2 = 0 + 0.000 000 000 005 649 842 368;
  • 6) 0.000 000 000 005 649 842 368 × 2 = 0 + 0.000 000 000 011 299 684 736;
  • 7) 0.000 000 000 011 299 684 736 × 2 = 0 + 0.000 000 000 022 599 369 472;
  • 8) 0.000 000 000 022 599 369 472 × 2 = 0 + 0.000 000 000 045 198 738 944;
  • 9) 0.000 000 000 045 198 738 944 × 2 = 0 + 0.000 000 000 090 397 477 888;
  • 10) 0.000 000 000 090 397 477 888 × 2 = 0 + 0.000 000 000 180 794 955 776;
  • 11) 0.000 000 000 180 794 955 776 × 2 = 0 + 0.000 000 000 361 589 911 552;
  • 12) 0.000 000 000 361 589 911 552 × 2 = 0 + 0.000 000 000 723 179 823 104;
  • 13) 0.000 000 000 723 179 823 104 × 2 = 0 + 0.000 000 001 446 359 646 208;
  • 14) 0.000 000 001 446 359 646 208 × 2 = 0 + 0.000 000 002 892 719 292 416;
  • 15) 0.000 000 002 892 719 292 416 × 2 = 0 + 0.000 000 005 785 438 584 832;
  • 16) 0.000 000 005 785 438 584 832 × 2 = 0 + 0.000 000 011 570 877 169 664;
  • 17) 0.000 000 011 570 877 169 664 × 2 = 0 + 0.000 000 023 141 754 339 328;
  • 18) 0.000 000 023 141 754 339 328 × 2 = 0 + 0.000 000 046 283 508 678 656;
  • 19) 0.000 000 046 283 508 678 656 × 2 = 0 + 0.000 000 092 567 017 357 312;
  • 20) 0.000 000 092 567 017 357 312 × 2 = 0 + 0.000 000 185 134 034 714 624;
  • 21) 0.000 000 185 134 034 714 624 × 2 = 0 + 0.000 000 370 268 069 429 248;
  • 22) 0.000 000 370 268 069 429 248 × 2 = 0 + 0.000 000 740 536 138 858 496;
  • 23) 0.000 000 740 536 138 858 496 × 2 = 0 + 0.000 001 481 072 277 716 992;
  • 24) 0.000 001 481 072 277 716 992 × 2 = 0 + 0.000 002 962 144 555 433 984;
  • 25) 0.000 002 962 144 555 433 984 × 2 = 0 + 0.000 005 924 289 110 867 968;
  • 26) 0.000 005 924 289 110 867 968 × 2 = 0 + 0.000 011 848 578 221 735 936;
  • 27) 0.000 011 848 578 221 735 936 × 2 = 0 + 0.000 023 697 156 443 471 872;
  • 28) 0.000 023 697 156 443 471 872 × 2 = 0 + 0.000 047 394 312 886 943 744;
  • 29) 0.000 047 394 312 886 943 744 × 2 = 0 + 0.000 094 788 625 773 887 488;
  • 30) 0.000 094 788 625 773 887 488 × 2 = 0 + 0.000 189 577 251 547 774 976;
  • 31) 0.000 189 577 251 547 774 976 × 2 = 0 + 0.000 379 154 503 095 549 952;
  • 32) 0.000 379 154 503 095 549 952 × 2 = 0 + 0.000 758 309 006 191 099 904;
  • 33) 0.000 758 309 006 191 099 904 × 2 = 0 + 0.001 516 618 012 382 199 808;
  • 34) 0.001 516 618 012 382 199 808 × 2 = 0 + 0.003 033 236 024 764 399 616;
  • 35) 0.003 033 236 024 764 399 616 × 2 = 0 + 0.006 066 472 049 528 799 232;
  • 36) 0.006 066 472 049 528 799 232 × 2 = 0 + 0.012 132 944 099 057 598 464;
  • 37) 0.012 132 944 099 057 598 464 × 2 = 0 + 0.024 265 888 198 115 196 928;
  • 38) 0.024 265 888 198 115 196 928 × 2 = 0 + 0.048 531 776 396 230 393 856;
  • 39) 0.048 531 776 396 230 393 856 × 2 = 0 + 0.097 063 552 792 460 787 712;
  • 40) 0.097 063 552 792 460 787 712 × 2 = 0 + 0.194 127 105 584 921 575 424;
  • 41) 0.194 127 105 584 921 575 424 × 2 = 0 + 0.388 254 211 169 843 150 848;
  • 42) 0.388 254 211 169 843 150 848 × 2 = 0 + 0.776 508 422 339 686 301 696;
  • 43) 0.776 508 422 339 686 301 696 × 2 = 1 + 0.553 016 844 679 372 603 392;
  • 44) 0.553 016 844 679 372 603 392 × 2 = 1 + 0.106 033 689 358 745 206 784;
  • 45) 0.106 033 689 358 745 206 784 × 2 = 0 + 0.212 067 378 717 490 413 568;
  • 46) 0.212 067 378 717 490 413 568 × 2 = 0 + 0.424 134 757 434 980 827 136;
  • 47) 0.424 134 757 434 980 827 136 × 2 = 0 + 0.848 269 514 869 961 654 272;
  • 48) 0.848 269 514 869 961 654 272 × 2 = 1 + 0.696 539 029 739 923 308 544;
  • 49) 0.696 539 029 739 923 308 544 × 2 = 1 + 0.393 078 059 479 846 617 088;
  • 50) 0.393 078 059 479 846 617 088 × 2 = 0 + 0.786 156 118 959 693 234 176;
  • 51) 0.786 156 118 959 693 234 176 × 2 = 1 + 0.572 312 237 919 386 468 352;
  • 52) 0.572 312 237 919 386 468 352 × 2 = 1 + 0.144 624 475 838 772 936 704;
  • 53) 0.144 624 475 838 772 936 704 × 2 = 0 + 0.289 248 951 677 545 873 408;
  • 54) 0.289 248 951 677 545 873 408 × 2 = 0 + 0.578 497 903 355 091 746 816;
  • 55) 0.578 497 903 355 091 746 816 × 2 = 1 + 0.156 995 806 710 183 493 632;
  • 56) 0.156 995 806 710 183 493 632 × 2 = 0 + 0.313 991 613 420 366 987 264;
  • 57) 0.313 991 613 420 366 987 264 × 2 = 0 + 0.627 983 226 840 733 974 528;
  • 58) 0.627 983 226 840 733 974 528 × 2 = 1 + 0.255 966 453 681 467 949 056;
  • 59) 0.255 966 453 681 467 949 056 × 2 = 0 + 0.511 932 907 362 935 898 112;
  • 60) 0.511 932 907 362 935 898 112 × 2 = 1 + 0.023 865 814 725 871 796 224;
  • 61) 0.023 865 814 725 871 796 224 × 2 = 0 + 0.047 731 629 451 743 592 448;
  • 62) 0.047 731 629 451 743 592 448 × 2 = 0 + 0.095 463 258 903 487 184 896;
  • 63) 0.095 463 258 903 487 184 896 × 2 = 0 + 0.190 926 517 806 974 369 792;
  • 64) 0.190 926 517 806 974 369 792 × 2 = 0 + 0.381 853 035 613 948 739 584;
  • 65) 0.381 853 035 613 948 739 584 × 2 = 0 + 0.763 706 071 227 897 479 168;
  • 66) 0.763 706 071 227 897 479 168 × 2 = 1 + 0.527 412 142 455 794 958 336;
  • 67) 0.527 412 142 455 794 958 336 × 2 = 1 + 0.054 824 284 911 589 916 672;
  • 68) 0.054 824 284 911 589 916 672 × 2 = 0 + 0.109 648 569 823 179 833 344;
  • 69) 0.109 648 569 823 179 833 344 × 2 = 0 + 0.219 297 139 646 359 666 688;
  • 70) 0.219 297 139 646 359 666 688 × 2 = 0 + 0.438 594 279 292 719 333 376;
  • 71) 0.438 594 279 292 719 333 376 × 2 = 0 + 0.877 188 558 585 438 666 752;
  • 72) 0.877 188 558 585 438 666 752 × 2 = 1 + 0.754 377 117 170 877 333 504;
  • 73) 0.754 377 117 170 877 333 504 × 2 = 1 + 0.508 754 234 341 754 667 008;
  • 74) 0.508 754 234 341 754 667 008 × 2 = 1 + 0.017 508 468 683 509 334 016;
  • 75) 0.017 508 468 683 509 334 016 × 2 = 0 + 0.035 016 937 367 018 668 032;
  • 76) 0.035 016 937 367 018 668 032 × 2 = 0 + 0.070 033 874 734 037 336 064;
  • 77) 0.070 033 874 734 037 336 064 × 2 = 0 + 0.140 067 749 468 074 672 128;
  • 78) 0.140 067 749 468 074 672 128 × 2 = 0 + 0.280 135 498 936 149 344 256;
  • 79) 0.280 135 498 936 149 344 256 × 2 = 0 + 0.560 270 997 872 298 688 512;
  • 80) 0.560 270 997 872 298 688 512 × 2 = 1 + 0.120 541 995 744 597 377 024;
  • 81) 0.120 541 995 744 597 377 024 × 2 = 0 + 0.241 083 991 489 194 754 048;
  • 82) 0.241 083 991 489 194 754 048 × 2 = 0 + 0.482 167 982 978 389 508 096;
  • 83) 0.482 167 982 978 389 508 096 × 2 = 0 + 0.964 335 965 956 779 016 192;
  • 84) 0.964 335 965 956 779 016 192 × 2 = 1 + 0.928 671 931 913 558 032 384;
  • 85) 0.928 671 931 913 558 032 384 × 2 = 1 + 0.857 343 863 827 116 064 768;
  • 86) 0.857 343 863 827 116 064 768 × 2 = 1 + 0.714 687 727 654 232 129 536;
  • 87) 0.714 687 727 654 232 129 536 × 2 = 1 + 0.429 375 455 308 464 259 072;
  • 88) 0.429 375 455 308 464 259 072 × 2 = 0 + 0.858 750 910 616 928 518 144;
  • 89) 0.858 750 910 616 928 518 144 × 2 = 1 + 0.717 501 821 233 857 036 288;
  • 90) 0.717 501 821 233 857 036 288 × 2 = 1 + 0.435 003 642 467 714 072 576;
  • 91) 0.435 003 642 467 714 072 576 × 2 = 0 + 0.870 007 284 935 428 145 152;
  • 92) 0.870 007 284 935 428 145 152 × 2 = 1 + 0.740 014 569 870 856 290 304;
  • 93) 0.740 014 569 870 856 290 304 × 2 = 1 + 0.480 029 139 741 712 580 608;
  • 94) 0.480 029 139 741 712 580 608 × 2 = 0 + 0.960 058 279 483 425 161 216;
  • 95) 0.960 058 279 483 425 161 216 × 2 = 1 + 0.920 116 558 966 850 322 432;

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:


0.000 000 000 000 176 557 574(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0000 0110 0001 1100 0001 0001 1110 1101 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 574(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0000 0110 0001 1100 0001 0001 1110 1101 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:


0.000 000 000 000 176 557 574(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0000 0110 0001 1100 0001 0001 1110 1101 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0000 0110 0001 1100 0001 0001 1110 1101 101(2) × 20 =

1.1000 1101 1001 0010 1000 0011 0000 1110 0000 1000 1111 0110 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 1000 0011 0000 1110 0000 1000 1111 0110 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 1000 0011 0000 1110 0000 1000 1111 0110 1101 =

1000 1101 1001 0010 1000 0011 0000 1110 0000 1000 1111 0110 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 1000 0011 0000 1110 0000 1000 1111 0110 1101


Decimal number -0.000 000 000 000 176 557 574 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1000 0011 0000 1110 0000 1000 1111 0110 1101

Share

» Convert decimal -0.000 000 000 000 176 557 665 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: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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