-0.000 000 000 000 176 557 239 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 239(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
-0.000 000 000 000 176 557 239(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:

|-0.000 000 000 000 176 557 239| = 0.000 000 000 000 176 557 239


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: 0.000 000 000 000 176 557 239.

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) 0.000 000 000 000 176 557 239 × 2 = 0 + 0.000 000 000 000 353 114 478;
  • 2) 0.000 000 000 000 353 114 478 × 2 = 0 + 0.000 000 000 000 706 228 956;
  • 3) 0.000 000 000 000 706 228 956 × 2 = 0 + 0.000 000 000 001 412 457 912;
  • 4) 0.000 000 000 001 412 457 912 × 2 = 0 + 0.000 000 000 002 824 915 824;
  • 5) 0.000 000 000 002 824 915 824 × 2 = 0 + 0.000 000 000 005 649 831 648;
  • 6) 0.000 000 000 005 649 831 648 × 2 = 0 + 0.000 000 000 011 299 663 296;
  • 7) 0.000 000 000 011 299 663 296 × 2 = 0 + 0.000 000 000 022 599 326 592;
  • 8) 0.000 000 000 022 599 326 592 × 2 = 0 + 0.000 000 000 045 198 653 184;
  • 9) 0.000 000 000 045 198 653 184 × 2 = 0 + 0.000 000 000 090 397 306 368;
  • 10) 0.000 000 000 090 397 306 368 × 2 = 0 + 0.000 000 000 180 794 612 736;
  • 11) 0.000 000 000 180 794 612 736 × 2 = 0 + 0.000 000 000 361 589 225 472;
  • 12) 0.000 000 000 361 589 225 472 × 2 = 0 + 0.000 000 000 723 178 450 944;
  • 13) 0.000 000 000 723 178 450 944 × 2 = 0 + 0.000 000 001 446 356 901 888;
  • 14) 0.000 000 001 446 356 901 888 × 2 = 0 + 0.000 000 002 892 713 803 776;
  • 15) 0.000 000 002 892 713 803 776 × 2 = 0 + 0.000 000 005 785 427 607 552;
  • 16) 0.000 000 005 785 427 607 552 × 2 = 0 + 0.000 000 011 570 855 215 104;
  • 17) 0.000 000 011 570 855 215 104 × 2 = 0 + 0.000 000 023 141 710 430 208;
  • 18) 0.000 000 023 141 710 430 208 × 2 = 0 + 0.000 000 046 283 420 860 416;
  • 19) 0.000 000 046 283 420 860 416 × 2 = 0 + 0.000 000 092 566 841 720 832;
  • 20) 0.000 000 092 566 841 720 832 × 2 = 0 + 0.000 000 185 133 683 441 664;
  • 21) 0.000 000 185 133 683 441 664 × 2 = 0 + 0.000 000 370 267 366 883 328;
  • 22) 0.000 000 370 267 366 883 328 × 2 = 0 + 0.000 000 740 534 733 766 656;
  • 23) 0.000 000 740 534 733 766 656 × 2 = 0 + 0.000 001 481 069 467 533 312;
  • 24) 0.000 001 481 069 467 533 312 × 2 = 0 + 0.000 002 962 138 935 066 624;
  • 25) 0.000 002 962 138 935 066 624 × 2 = 0 + 0.000 005 924 277 870 133 248;
  • 26) 0.000 005 924 277 870 133 248 × 2 = 0 + 0.000 011 848 555 740 266 496;
  • 27) 0.000 011 848 555 740 266 496 × 2 = 0 + 0.000 023 697 111 480 532 992;
  • 28) 0.000 023 697 111 480 532 992 × 2 = 0 + 0.000 047 394 222 961 065 984;
  • 29) 0.000 047 394 222 961 065 984 × 2 = 0 + 0.000 094 788 445 922 131 968;
  • 30) 0.000 094 788 445 922 131 968 × 2 = 0 + 0.000 189 576 891 844 263 936;
  • 31) 0.000 189 576 891 844 263 936 × 2 = 0 + 0.000 379 153 783 688 527 872;
  • 32) 0.000 379 153 783 688 527 872 × 2 = 0 + 0.000 758 307 567 377 055 744;
  • 33) 0.000 758 307 567 377 055 744 × 2 = 0 + 0.001 516 615 134 754 111 488;
  • 34) 0.001 516 615 134 754 111 488 × 2 = 0 + 0.003 033 230 269 508 222 976;
  • 35) 0.003 033 230 269 508 222 976 × 2 = 0 + 0.006 066 460 539 016 445 952;
  • 36) 0.006 066 460 539 016 445 952 × 2 = 0 + 0.012 132 921 078 032 891 904;
  • 37) 0.012 132 921 078 032 891 904 × 2 = 0 + 0.024 265 842 156 065 783 808;
  • 38) 0.024 265 842 156 065 783 808 × 2 = 0 + 0.048 531 684 312 131 567 616;
  • 39) 0.048 531 684 312 131 567 616 × 2 = 0 + 0.097 063 368 624 263 135 232;
  • 40) 0.097 063 368 624 263 135 232 × 2 = 0 + 0.194 126 737 248 526 270 464;
  • 41) 0.194 126 737 248 526 270 464 × 2 = 0 + 0.388 253 474 497 052 540 928;
  • 42) 0.388 253 474 497 052 540 928 × 2 = 0 + 0.776 506 948 994 105 081 856;
  • 43) 0.776 506 948 994 105 081 856 × 2 = 1 + 0.553 013 897 988 210 163 712;
  • 44) 0.553 013 897 988 210 163 712 × 2 = 1 + 0.106 027 795 976 420 327 424;
  • 45) 0.106 027 795 976 420 327 424 × 2 = 0 + 0.212 055 591 952 840 654 848;
  • 46) 0.212 055 591 952 840 654 848 × 2 = 0 + 0.424 111 183 905 681 309 696;
  • 47) 0.424 111 183 905 681 309 696 × 2 = 0 + 0.848 222 367 811 362 619 392;
  • 48) 0.848 222 367 811 362 619 392 × 2 = 1 + 0.696 444 735 622 725 238 784;
  • 49) 0.696 444 735 622 725 238 784 × 2 = 1 + 0.392 889 471 245 450 477 568;
  • 50) 0.392 889 471 245 450 477 568 × 2 = 0 + 0.785 778 942 490 900 955 136;
  • 51) 0.785 778 942 490 900 955 136 × 2 = 1 + 0.571 557 884 981 801 910 272;
  • 52) 0.571 557 884 981 801 910 272 × 2 = 1 + 0.143 115 769 963 603 820 544;
  • 53) 0.143 115 769 963 603 820 544 × 2 = 0 + 0.286 231 539 927 207 641 088;
  • 54) 0.286 231 539 927 207 641 088 × 2 = 0 + 0.572 463 079 854 415 282 176;
  • 55) 0.572 463 079 854 415 282 176 × 2 = 1 + 0.144 926 159 708 830 564 352;
  • 56) 0.144 926 159 708 830 564 352 × 2 = 0 + 0.289 852 319 417 661 128 704;
  • 57) 0.289 852 319 417 661 128 704 × 2 = 0 + 0.579 704 638 835 322 257 408;
  • 58) 0.579 704 638 835 322 257 408 × 2 = 1 + 0.159 409 277 670 644 514 816;
  • 59) 0.159 409 277 670 644 514 816 × 2 = 0 + 0.318 818 555 341 289 029 632;
  • 60) 0.318 818 555 341 289 029 632 × 2 = 0 + 0.637 637 110 682 578 059 264;
  • 61) 0.637 637 110 682 578 059 264 × 2 = 1 + 0.275 274 221 365 156 118 528;
  • 62) 0.275 274 221 365 156 118 528 × 2 = 0 + 0.550 548 442 730 312 237 056;
  • 63) 0.550 548 442 730 312 237 056 × 2 = 1 + 0.101 096 885 460 624 474 112;
  • 64) 0.101 096 885 460 624 474 112 × 2 = 0 + 0.202 193 770 921 248 948 224;
  • 65) 0.202 193 770 921 248 948 224 × 2 = 0 + 0.404 387 541 842 497 896 448;
  • 66) 0.404 387 541 842 497 896 448 × 2 = 0 + 0.808 775 083 684 995 792 896;
  • 67) 0.808 775 083 684 995 792 896 × 2 = 1 + 0.617 550 167 369 991 585 792;
  • 68) 0.617 550 167 369 991 585 792 × 2 = 1 + 0.235 100 334 739 983 171 584;
  • 69) 0.235 100 334 739 983 171 584 × 2 = 0 + 0.470 200 669 479 966 343 168;
  • 70) 0.470 200 669 479 966 343 168 × 2 = 0 + 0.940 401 338 959 932 686 336;
  • 71) 0.940 401 338 959 932 686 336 × 2 = 1 + 0.880 802 677 919 865 372 672;
  • 72) 0.880 802 677 919 865 372 672 × 2 = 1 + 0.761 605 355 839 730 745 344;
  • 73) 0.761 605 355 839 730 745 344 × 2 = 1 + 0.523 210 711 679 461 490 688;
  • 74) 0.523 210 711 679 461 490 688 × 2 = 1 + 0.046 421 423 358 922 981 376;
  • 75) 0.046 421 423 358 922 981 376 × 2 = 0 + 0.092 842 846 717 845 962 752;
  • 76) 0.092 842 846 717 845 962 752 × 2 = 0 + 0.185 685 693 435 691 925 504;
  • 77) 0.185 685 693 435 691 925 504 × 2 = 0 + 0.371 371 386 871 383 851 008;
  • 78) 0.371 371 386 871 383 851 008 × 2 = 0 + 0.742 742 773 742 767 702 016;
  • 79) 0.742 742 773 742 767 702 016 × 2 = 1 + 0.485 485 547 485 535 404 032;
  • 80) 0.485 485 547 485 535 404 032 × 2 = 0 + 0.970 971 094 971 070 808 064;
  • 81) 0.970 971 094 971 070 808 064 × 2 = 1 + 0.941 942 189 942 141 616 128;
  • 82) 0.941 942 189 942 141 616 128 × 2 = 1 + 0.883 884 379 884 283 232 256;
  • 83) 0.883 884 379 884 283 232 256 × 2 = 1 + 0.767 768 759 768 566 464 512;
  • 84) 0.767 768 759 768 566 464 512 × 2 = 1 + 0.535 537 519 537 132 929 024;
  • 85) 0.535 537 519 537 132 929 024 × 2 = 1 + 0.071 075 039 074 265 858 048;
  • 86) 0.071 075 039 074 265 858 048 × 2 = 0 + 0.142 150 078 148 531 716 096;
  • 87) 0.142 150 078 148 531 716 096 × 2 = 0 + 0.284 300 156 297 063 432 192;
  • 88) 0.284 300 156 297 063 432 192 × 2 = 0 + 0.568 600 312 594 126 864 384;
  • 89) 0.568 600 312 594 126 864 384 × 2 = 1 + 0.137 200 625 188 253 728 768;
  • 90) 0.137 200 625 188 253 728 768 × 2 = 0 + 0.274 401 250 376 507 457 536;
  • 91) 0.274 401 250 376 507 457 536 × 2 = 0 + 0.548 802 500 753 014 915 072;
  • 92) 0.548 802 500 753 014 915 072 × 2 = 1 + 0.097 605 001 506 029 830 144;
  • 93) 0.097 605 001 506 029 830 144 × 2 = 0 + 0.195 210 003 012 059 660 288;
  • 94) 0.195 210 003 012 059 660 288 × 2 = 0 + 0.390 420 006 024 119 320 576;
  • 95) 0.390 420 006 024 119 320 576 × 2 = 0 + 0.780 840 012 048 238 641 152;

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:


0.000 000 000 000 176 557 239(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0011 0011 1100 0010 1111 1000 1001 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 239(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0011 0011 1100 0010 1111 1000 1001 000(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:


0.000 000 000 000 176 557 239(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0011 0011 1100 0010 1111 1000 1001 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0011 0011 1100 0010 1111 1000 1001 000(2) × 20 =

1.1000 1101 1001 0010 0101 0001 1001 1110 0001 0111 1100 0100 1000(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 0101 0001 1001 1110 0001 0111 1100 0100 1000


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 0101 0001 1001 1110 0001 0111 1100 0100 1000 =

1000 1101 1001 0010 0101 0001 1001 1110 0001 0111 1100 0100 1000


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 0101 0001 1001 1110 0001 0111 1100 0100 1000


Decimal number -0.000 000 000 000 176 557 239 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0101 0001 1001 1110 0001 0111 1100 0100 1000

Share

» Convert decimal -0.000 000 000 000 176 557 24 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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