-0.000 000 000 000 028 421 709 323 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 028 421 709 323(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
-0.000 000 000 000 028 421 709 323(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:

|-0.000 000 000 000 028 421 709 323| = 0.000 000 000 000 028 421 709 323


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: 0.000 000 000 000 028 421 709 323.

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) 0.000 000 000 000 028 421 709 323 × 2 = 0 + 0.000 000 000 000 056 843 418 646;
  • 2) 0.000 000 000 000 056 843 418 646 × 2 = 0 + 0.000 000 000 000 113 686 837 292;
  • 3) 0.000 000 000 000 113 686 837 292 × 2 = 0 + 0.000 000 000 000 227 373 674 584;
  • 4) 0.000 000 000 000 227 373 674 584 × 2 = 0 + 0.000 000 000 000 454 747 349 168;
  • 5) 0.000 000 000 000 454 747 349 168 × 2 = 0 + 0.000 000 000 000 909 494 698 336;
  • 6) 0.000 000 000 000 909 494 698 336 × 2 = 0 + 0.000 000 000 001 818 989 396 672;
  • 7) 0.000 000 000 001 818 989 396 672 × 2 = 0 + 0.000 000 000 003 637 978 793 344;
  • 8) 0.000 000 000 003 637 978 793 344 × 2 = 0 + 0.000 000 000 007 275 957 586 688;
  • 9) 0.000 000 000 007 275 957 586 688 × 2 = 0 + 0.000 000 000 014 551 915 173 376;
  • 10) 0.000 000 000 014 551 915 173 376 × 2 = 0 + 0.000 000 000 029 103 830 346 752;
  • 11) 0.000 000 000 029 103 830 346 752 × 2 = 0 + 0.000 000 000 058 207 660 693 504;
  • 12) 0.000 000 000 058 207 660 693 504 × 2 = 0 + 0.000 000 000 116 415 321 387 008;
  • 13) 0.000 000 000 116 415 321 387 008 × 2 = 0 + 0.000 000 000 232 830 642 774 016;
  • 14) 0.000 000 000 232 830 642 774 016 × 2 = 0 + 0.000 000 000 465 661 285 548 032;
  • 15) 0.000 000 000 465 661 285 548 032 × 2 = 0 + 0.000 000 000 931 322 571 096 064;
  • 16) 0.000 000 000 931 322 571 096 064 × 2 = 0 + 0.000 000 001 862 645 142 192 128;
  • 17) 0.000 000 001 862 645 142 192 128 × 2 = 0 + 0.000 000 003 725 290 284 384 256;
  • 18) 0.000 000 003 725 290 284 384 256 × 2 = 0 + 0.000 000 007 450 580 568 768 512;
  • 19) 0.000 000 007 450 580 568 768 512 × 2 = 0 + 0.000 000 014 901 161 137 537 024;
  • 20) 0.000 000 014 901 161 137 537 024 × 2 = 0 + 0.000 000 029 802 322 275 074 048;
  • 21) 0.000 000 029 802 322 275 074 048 × 2 = 0 + 0.000 000 059 604 644 550 148 096;
  • 22) 0.000 000 059 604 644 550 148 096 × 2 = 0 + 0.000 000 119 209 289 100 296 192;
  • 23) 0.000 000 119 209 289 100 296 192 × 2 = 0 + 0.000 000 238 418 578 200 592 384;
  • 24) 0.000 000 238 418 578 200 592 384 × 2 = 0 + 0.000 000 476 837 156 401 184 768;
  • 25) 0.000 000 476 837 156 401 184 768 × 2 = 0 + 0.000 000 953 674 312 802 369 536;
  • 26) 0.000 000 953 674 312 802 369 536 × 2 = 0 + 0.000 001 907 348 625 604 739 072;
  • 27) 0.000 001 907 348 625 604 739 072 × 2 = 0 + 0.000 003 814 697 251 209 478 144;
  • 28) 0.000 003 814 697 251 209 478 144 × 2 = 0 + 0.000 007 629 394 502 418 956 288;
  • 29) 0.000 007 629 394 502 418 956 288 × 2 = 0 + 0.000 015 258 789 004 837 912 576;
  • 30) 0.000 015 258 789 004 837 912 576 × 2 = 0 + 0.000 030 517 578 009 675 825 152;
  • 31) 0.000 030 517 578 009 675 825 152 × 2 = 0 + 0.000 061 035 156 019 351 650 304;
  • 32) 0.000 061 035 156 019 351 650 304 × 2 = 0 + 0.000 122 070 312 038 703 300 608;
  • 33) 0.000 122 070 312 038 703 300 608 × 2 = 0 + 0.000 244 140 624 077 406 601 216;
  • 34) 0.000 244 140 624 077 406 601 216 × 2 = 0 + 0.000 488 281 248 154 813 202 432;
  • 35) 0.000 488 281 248 154 813 202 432 × 2 = 0 + 0.000 976 562 496 309 626 404 864;
  • 36) 0.000 976 562 496 309 626 404 864 × 2 = 0 + 0.001 953 124 992 619 252 809 728;
  • 37) 0.001 953 124 992 619 252 809 728 × 2 = 0 + 0.003 906 249 985 238 505 619 456;
  • 38) 0.003 906 249 985 238 505 619 456 × 2 = 0 + 0.007 812 499 970 477 011 238 912;
  • 39) 0.007 812 499 970 477 011 238 912 × 2 = 0 + 0.015 624 999 940 954 022 477 824;
  • 40) 0.015 624 999 940 954 022 477 824 × 2 = 0 + 0.031 249 999 881 908 044 955 648;
  • 41) 0.031 249 999 881 908 044 955 648 × 2 = 0 + 0.062 499 999 763 816 089 911 296;
  • 42) 0.062 499 999 763 816 089 911 296 × 2 = 0 + 0.124 999 999 527 632 179 822 592;
  • 43) 0.124 999 999 527 632 179 822 592 × 2 = 0 + 0.249 999 999 055 264 359 645 184;
  • 44) 0.249 999 999 055 264 359 645 184 × 2 = 0 + 0.499 999 998 110 528 719 290 368;
  • 45) 0.499 999 998 110 528 719 290 368 × 2 = 0 + 0.999 999 996 221 057 438 580 736;
  • 46) 0.999 999 996 221 057 438 580 736 × 2 = 1 + 0.999 999 992 442 114 877 161 472;
  • 47) 0.999 999 992 442 114 877 161 472 × 2 = 1 + 0.999 999 984 884 229 754 322 944;
  • 48) 0.999 999 984 884 229 754 322 944 × 2 = 1 + 0.999 999 969 768 459 508 645 888;
  • 49) 0.999 999 969 768 459 508 645 888 × 2 = 1 + 0.999 999 939 536 919 017 291 776;
  • 50) 0.999 999 939 536 919 017 291 776 × 2 = 1 + 0.999 999 879 073 838 034 583 552;
  • 51) 0.999 999 879 073 838 034 583 552 × 2 = 1 + 0.999 999 758 147 676 069 167 104;
  • 52) 0.999 999 758 147 676 069 167 104 × 2 = 1 + 0.999 999 516 295 352 138 334 208;
  • 53) 0.999 999 516 295 352 138 334 208 × 2 = 1 + 0.999 999 032 590 704 276 668 416;
  • 54) 0.999 999 032 590 704 276 668 416 × 2 = 1 + 0.999 998 065 181 408 553 336 832;
  • 55) 0.999 998 065 181 408 553 336 832 × 2 = 1 + 0.999 996 130 362 817 106 673 664;
  • 56) 0.999 996 130 362 817 106 673 664 × 2 = 1 + 0.999 992 260 725 634 213 347 328;
  • 57) 0.999 992 260 725 634 213 347 328 × 2 = 1 + 0.999 984 521 451 268 426 694 656;
  • 58) 0.999 984 521 451 268 426 694 656 × 2 = 1 + 0.999 969 042 902 536 853 389 312;
  • 59) 0.999 969 042 902 536 853 389 312 × 2 = 1 + 0.999 938 085 805 073 706 778 624;
  • 60) 0.999 938 085 805 073 706 778 624 × 2 = 1 + 0.999 876 171 610 147 413 557 248;
  • 61) 0.999 876 171 610 147 413 557 248 × 2 = 1 + 0.999 752 343 220 294 827 114 496;
  • 62) 0.999 752 343 220 294 827 114 496 × 2 = 1 + 0.999 504 686 440 589 654 228 992;
  • 63) 0.999 504 686 440 589 654 228 992 × 2 = 1 + 0.999 009 372 881 179 308 457 984;
  • 64) 0.999 009 372 881 179 308 457 984 × 2 = 1 + 0.998 018 745 762 358 616 915 968;
  • 65) 0.998 018 745 762 358 616 915 968 × 2 = 1 + 0.996 037 491 524 717 233 831 936;
  • 66) 0.996 037 491 524 717 233 831 936 × 2 = 1 + 0.992 074 983 049 434 467 663 872;
  • 67) 0.992 074 983 049 434 467 663 872 × 2 = 1 + 0.984 149 966 098 868 935 327 744;
  • 68) 0.984 149 966 098 868 935 327 744 × 2 = 1 + 0.968 299 932 197 737 870 655 488;
  • 69) 0.968 299 932 197 737 870 655 488 × 2 = 1 + 0.936 599 864 395 475 741 310 976;
  • 70) 0.936 599 864 395 475 741 310 976 × 2 = 1 + 0.873 199 728 790 951 482 621 952;
  • 71) 0.873 199 728 790 951 482 621 952 × 2 = 1 + 0.746 399 457 581 902 965 243 904;
  • 72) 0.746 399 457 581 902 965 243 904 × 2 = 1 + 0.492 798 915 163 805 930 487 808;
  • 73) 0.492 798 915 163 805 930 487 808 × 2 = 0 + 0.985 597 830 327 611 860 975 616;
  • 74) 0.985 597 830 327 611 860 975 616 × 2 = 1 + 0.971 195 660 655 223 721 951 232;
  • 75) 0.971 195 660 655 223 721 951 232 × 2 = 1 + 0.942 391 321 310 447 443 902 464;
  • 76) 0.942 391 321 310 447 443 902 464 × 2 = 1 + 0.884 782 642 620 894 887 804 928;
  • 77) 0.884 782 642 620 894 887 804 928 × 2 = 1 + 0.769 565 285 241 789 775 609 856;
  • 78) 0.769 565 285 241 789 775 609 856 × 2 = 1 + 0.539 130 570 483 579 551 219 712;
  • 79) 0.539 130 570 483 579 551 219 712 × 2 = 1 + 0.078 261 140 967 159 102 439 424;
  • 80) 0.078 261 140 967 159 102 439 424 × 2 = 0 + 0.156 522 281 934 318 204 878 848;
  • 81) 0.156 522 281 934 318 204 878 848 × 2 = 0 + 0.313 044 563 868 636 409 757 696;
  • 82) 0.313 044 563 868 636 409 757 696 × 2 = 0 + 0.626 089 127 737 272 819 515 392;
  • 83) 0.626 089 127 737 272 819 515 392 × 2 = 1 + 0.252 178 255 474 545 639 030 784;
  • 84) 0.252 178 255 474 545 639 030 784 × 2 = 0 + 0.504 356 510 949 091 278 061 568;
  • 85) 0.504 356 510 949 091 278 061 568 × 2 = 1 + 0.008 713 021 898 182 556 123 136;
  • 86) 0.008 713 021 898 182 556 123 136 × 2 = 0 + 0.017 426 043 796 365 112 246 272;
  • 87) 0.017 426 043 796 365 112 246 272 × 2 = 0 + 0.034 852 087 592 730 224 492 544;
  • 88) 0.034 852 087 592 730 224 492 544 × 2 = 0 + 0.069 704 175 185 460 448 985 088;
  • 89) 0.069 704 175 185 460 448 985 088 × 2 = 0 + 0.139 408 350 370 920 897 970 176;
  • 90) 0.139 408 350 370 920 897 970 176 × 2 = 0 + 0.278 816 700 741 841 795 940 352;
  • 91) 0.278 816 700 741 841 795 940 352 × 2 = 0 + 0.557 633 401 483 683 591 880 704;
  • 92) 0.557 633 401 483 683 591 880 704 × 2 = 1 + 0.115 266 802 967 367 183 761 408;
  • 93) 0.115 266 802 967 367 183 761 408 × 2 = 0 + 0.230 533 605 934 734 367 522 816;
  • 94) 0.230 533 605 934 734 367 522 816 × 2 = 0 + 0.461 067 211 869 468 735 045 632;
  • 95) 0.461 067 211 869 468 735 045 632 × 2 = 0 + 0.922 134 423 738 937 470 091 264;
  • 96) 0.922 134 423 738 937 470 091 264 × 2 = 1 + 0.844 268 847 477 874 940 182 528;
  • 97) 0.844 268 847 477 874 940 182 528 × 2 = 1 + 0.688 537 694 955 749 880 365 056;
  • 98) 0.688 537 694 955 749 880 365 056 × 2 = 1 + 0.377 075 389 911 499 760 730 112;

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:


0.000 000 000 000 028 421 709 323(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 0111 1110 0010 1000 0001 0001 11(2)

6. Positive number before normalization:

0.000 000 000 000 028 421 709 323(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 0111 1110 0010 1000 0001 0001 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 028 421 709 323(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 0111 1110 0010 1000 0001 0001 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 0111 1110 0010 1000 0001 0001 11(2) × 20 =

1.1111 1111 1111 1111 1111 1111 1101 1111 1000 1010 0000 0100 0111(2) × 2-46


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): -46

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 1101 1111 1000 1010 0000 0100 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 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) =

977(10) =

011 1101 0001(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. 1111 1111 1111 1111 1111 1111 1101 1111 1000 1010 0000 0100 0111 =

1111 1111 1111 1111 1111 1111 1101 1111 1000 1010 0000 0100 0111


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 0001

Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 1101 1111 1000 1010 0000 0100 0111


Decimal number -0.000 000 000 000 028 421 709 323 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 1111 1111 1111 1111 1111 1111 1101 1111 1000 1010 0000 0100 0111

» Convert decimal -0.000 000 000 000 028 421 709 399 to 64 bit double precision IEEE 754 binary floating point representation standard

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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