-0.000 000 000 000 028 421 709 383 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 028 421 709 383(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
-0.000 000 000 000 028 421 709 383(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:

|-0.000 000 000 000 028 421 709 383| = 0.000 000 000 000 028 421 709 383


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: 0.000 000 000 000 028 421 709 383.

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) 0.000 000 000 000 028 421 709 383 × 2 = 0 + 0.000 000 000 000 056 843 418 766;
  • 2) 0.000 000 000 000 056 843 418 766 × 2 = 0 + 0.000 000 000 000 113 686 837 532;
  • 3) 0.000 000 000 000 113 686 837 532 × 2 = 0 + 0.000 000 000 000 227 373 675 064;
  • 4) 0.000 000 000 000 227 373 675 064 × 2 = 0 + 0.000 000 000 000 454 747 350 128;
  • 5) 0.000 000 000 000 454 747 350 128 × 2 = 0 + 0.000 000 000 000 909 494 700 256;
  • 6) 0.000 000 000 000 909 494 700 256 × 2 = 0 + 0.000 000 000 001 818 989 400 512;
  • 7) 0.000 000 000 001 818 989 400 512 × 2 = 0 + 0.000 000 000 003 637 978 801 024;
  • 8) 0.000 000 000 003 637 978 801 024 × 2 = 0 + 0.000 000 000 007 275 957 602 048;
  • 9) 0.000 000 000 007 275 957 602 048 × 2 = 0 + 0.000 000 000 014 551 915 204 096;
  • 10) 0.000 000 000 014 551 915 204 096 × 2 = 0 + 0.000 000 000 029 103 830 408 192;
  • 11) 0.000 000 000 029 103 830 408 192 × 2 = 0 + 0.000 000 000 058 207 660 816 384;
  • 12) 0.000 000 000 058 207 660 816 384 × 2 = 0 + 0.000 000 000 116 415 321 632 768;
  • 13) 0.000 000 000 116 415 321 632 768 × 2 = 0 + 0.000 000 000 232 830 643 265 536;
  • 14) 0.000 000 000 232 830 643 265 536 × 2 = 0 + 0.000 000 000 465 661 286 531 072;
  • 15) 0.000 000 000 465 661 286 531 072 × 2 = 0 + 0.000 000 000 931 322 573 062 144;
  • 16) 0.000 000 000 931 322 573 062 144 × 2 = 0 + 0.000 000 001 862 645 146 124 288;
  • 17) 0.000 000 001 862 645 146 124 288 × 2 = 0 + 0.000 000 003 725 290 292 248 576;
  • 18) 0.000 000 003 725 290 292 248 576 × 2 = 0 + 0.000 000 007 450 580 584 497 152;
  • 19) 0.000 000 007 450 580 584 497 152 × 2 = 0 + 0.000 000 014 901 161 168 994 304;
  • 20) 0.000 000 014 901 161 168 994 304 × 2 = 0 + 0.000 000 029 802 322 337 988 608;
  • 21) 0.000 000 029 802 322 337 988 608 × 2 = 0 + 0.000 000 059 604 644 675 977 216;
  • 22) 0.000 000 059 604 644 675 977 216 × 2 = 0 + 0.000 000 119 209 289 351 954 432;
  • 23) 0.000 000 119 209 289 351 954 432 × 2 = 0 + 0.000 000 238 418 578 703 908 864;
  • 24) 0.000 000 238 418 578 703 908 864 × 2 = 0 + 0.000 000 476 837 157 407 817 728;
  • 25) 0.000 000 476 837 157 407 817 728 × 2 = 0 + 0.000 000 953 674 314 815 635 456;
  • 26) 0.000 000 953 674 314 815 635 456 × 2 = 0 + 0.000 001 907 348 629 631 270 912;
  • 27) 0.000 001 907 348 629 631 270 912 × 2 = 0 + 0.000 003 814 697 259 262 541 824;
  • 28) 0.000 003 814 697 259 262 541 824 × 2 = 0 + 0.000 007 629 394 518 525 083 648;
  • 29) 0.000 007 629 394 518 525 083 648 × 2 = 0 + 0.000 015 258 789 037 050 167 296;
  • 30) 0.000 015 258 789 037 050 167 296 × 2 = 0 + 0.000 030 517 578 074 100 334 592;
  • 31) 0.000 030 517 578 074 100 334 592 × 2 = 0 + 0.000 061 035 156 148 200 669 184;
  • 32) 0.000 061 035 156 148 200 669 184 × 2 = 0 + 0.000 122 070 312 296 401 338 368;
  • 33) 0.000 122 070 312 296 401 338 368 × 2 = 0 + 0.000 244 140 624 592 802 676 736;
  • 34) 0.000 244 140 624 592 802 676 736 × 2 = 0 + 0.000 488 281 249 185 605 353 472;
  • 35) 0.000 488 281 249 185 605 353 472 × 2 = 0 + 0.000 976 562 498 371 210 706 944;
  • 36) 0.000 976 562 498 371 210 706 944 × 2 = 0 + 0.001 953 124 996 742 421 413 888;
  • 37) 0.001 953 124 996 742 421 413 888 × 2 = 0 + 0.003 906 249 993 484 842 827 776;
  • 38) 0.003 906 249 993 484 842 827 776 × 2 = 0 + 0.007 812 499 986 969 685 655 552;
  • 39) 0.007 812 499 986 969 685 655 552 × 2 = 0 + 0.015 624 999 973 939 371 311 104;
  • 40) 0.015 624 999 973 939 371 311 104 × 2 = 0 + 0.031 249 999 947 878 742 622 208;
  • 41) 0.031 249 999 947 878 742 622 208 × 2 = 0 + 0.062 499 999 895 757 485 244 416;
  • 42) 0.062 499 999 895 757 485 244 416 × 2 = 0 + 0.124 999 999 791 514 970 488 832;
  • 43) 0.124 999 999 791 514 970 488 832 × 2 = 0 + 0.249 999 999 583 029 940 977 664;
  • 44) 0.249 999 999 583 029 940 977 664 × 2 = 0 + 0.499 999 999 166 059 881 955 328;
  • 45) 0.499 999 999 166 059 881 955 328 × 2 = 0 + 0.999 999 998 332 119 763 910 656;
  • 46) 0.999 999 998 332 119 763 910 656 × 2 = 1 + 0.999 999 996 664 239 527 821 312;
  • 47) 0.999 999 996 664 239 527 821 312 × 2 = 1 + 0.999 999 993 328 479 055 642 624;
  • 48) 0.999 999 993 328 479 055 642 624 × 2 = 1 + 0.999 999 986 656 958 111 285 248;
  • 49) 0.999 999 986 656 958 111 285 248 × 2 = 1 + 0.999 999 973 313 916 222 570 496;
  • 50) 0.999 999 973 313 916 222 570 496 × 2 = 1 + 0.999 999 946 627 832 445 140 992;
  • 51) 0.999 999 946 627 832 445 140 992 × 2 = 1 + 0.999 999 893 255 664 890 281 984;
  • 52) 0.999 999 893 255 664 890 281 984 × 2 = 1 + 0.999 999 786 511 329 780 563 968;
  • 53) 0.999 999 786 511 329 780 563 968 × 2 = 1 + 0.999 999 573 022 659 561 127 936;
  • 54) 0.999 999 573 022 659 561 127 936 × 2 = 1 + 0.999 999 146 045 319 122 255 872;
  • 55) 0.999 999 146 045 319 122 255 872 × 2 = 1 + 0.999 998 292 090 638 244 511 744;
  • 56) 0.999 998 292 090 638 244 511 744 × 2 = 1 + 0.999 996 584 181 276 489 023 488;
  • 57) 0.999 996 584 181 276 489 023 488 × 2 = 1 + 0.999 993 168 362 552 978 046 976;
  • 58) 0.999 993 168 362 552 978 046 976 × 2 = 1 + 0.999 986 336 725 105 956 093 952;
  • 59) 0.999 986 336 725 105 956 093 952 × 2 = 1 + 0.999 972 673 450 211 912 187 904;
  • 60) 0.999 972 673 450 211 912 187 904 × 2 = 1 + 0.999 945 346 900 423 824 375 808;
  • 61) 0.999 945 346 900 423 824 375 808 × 2 = 1 + 0.999 890 693 800 847 648 751 616;
  • 62) 0.999 890 693 800 847 648 751 616 × 2 = 1 + 0.999 781 387 601 695 297 503 232;
  • 63) 0.999 781 387 601 695 297 503 232 × 2 = 1 + 0.999 562 775 203 390 595 006 464;
  • 64) 0.999 562 775 203 390 595 006 464 × 2 = 1 + 0.999 125 550 406 781 190 012 928;
  • 65) 0.999 125 550 406 781 190 012 928 × 2 = 1 + 0.998 251 100 813 562 380 025 856;
  • 66) 0.998 251 100 813 562 380 025 856 × 2 = 1 + 0.996 502 201 627 124 760 051 712;
  • 67) 0.996 502 201 627 124 760 051 712 × 2 = 1 + 0.993 004 403 254 249 520 103 424;
  • 68) 0.993 004 403 254 249 520 103 424 × 2 = 1 + 0.986 008 806 508 499 040 206 848;
  • 69) 0.986 008 806 508 499 040 206 848 × 2 = 1 + 0.972 017 613 016 998 080 413 696;
  • 70) 0.972 017 613 016 998 080 413 696 × 2 = 1 + 0.944 035 226 033 996 160 827 392;
  • 71) 0.944 035 226 033 996 160 827 392 × 2 = 1 + 0.888 070 452 067 992 321 654 784;
  • 72) 0.888 070 452 067 992 321 654 784 × 2 = 1 + 0.776 140 904 135 984 643 309 568;
  • 73) 0.776 140 904 135 984 643 309 568 × 2 = 1 + 0.552 281 808 271 969 286 619 136;
  • 74) 0.552 281 808 271 969 286 619 136 × 2 = 1 + 0.104 563 616 543 938 573 238 272;
  • 75) 0.104 563 616 543 938 573 238 272 × 2 = 0 + 0.209 127 233 087 877 146 476 544;
  • 76) 0.209 127 233 087 877 146 476 544 × 2 = 0 + 0.418 254 466 175 754 292 953 088;
  • 77) 0.418 254 466 175 754 292 953 088 × 2 = 0 + 0.836 508 932 351 508 585 906 176;
  • 78) 0.836 508 932 351 508 585 906 176 × 2 = 1 + 0.673 017 864 703 017 171 812 352;
  • 79) 0.673 017 864 703 017 171 812 352 × 2 = 1 + 0.346 035 729 406 034 343 624 704;
  • 80) 0.346 035 729 406 034 343 624 704 × 2 = 0 + 0.692 071 458 812 068 687 249 408;
  • 81) 0.692 071 458 812 068 687 249 408 × 2 = 1 + 0.384 142 917 624 137 374 498 816;
  • 82) 0.384 142 917 624 137 374 498 816 × 2 = 0 + 0.768 285 835 248 274 748 997 632;
  • 83) 0.768 285 835 248 274 748 997 632 × 2 = 1 + 0.536 571 670 496 549 497 995 264;
  • 84) 0.536 571 670 496 549 497 995 264 × 2 = 1 + 0.073 143 340 993 098 995 990 528;
  • 85) 0.073 143 340 993 098 995 990 528 × 2 = 0 + 0.146 286 681 986 197 991 981 056;
  • 86) 0.146 286 681 986 197 991 981 056 × 2 = 0 + 0.292 573 363 972 395 983 962 112;
  • 87) 0.292 573 363 972 395 983 962 112 × 2 = 0 + 0.585 146 727 944 791 967 924 224;
  • 88) 0.585 146 727 944 791 967 924 224 × 2 = 1 + 0.170 293 455 889 583 935 848 448;
  • 89) 0.170 293 455 889 583 935 848 448 × 2 = 0 + 0.340 586 911 779 167 871 696 896;
  • 90) 0.340 586 911 779 167 871 696 896 × 2 = 0 + 0.681 173 823 558 335 743 393 792;
  • 91) 0.681 173 823 558 335 743 393 792 × 2 = 1 + 0.362 347 647 116 671 486 787 584;
  • 92) 0.362 347 647 116 671 486 787 584 × 2 = 0 + 0.724 695 294 233 342 973 575 168;
  • 93) 0.724 695 294 233 342 973 575 168 × 2 = 1 + 0.449 390 588 466 685 947 150 336;
  • 94) 0.449 390 588 466 685 947 150 336 × 2 = 0 + 0.898 781 176 933 371 894 300 672;
  • 95) 0.898 781 176 933 371 894 300 672 × 2 = 1 + 0.797 562 353 866 743 788 601 344;
  • 96) 0.797 562 353 866 743 788 601 344 × 2 = 1 + 0.595 124 707 733 487 577 202 688;
  • 97) 0.595 124 707 733 487 577 202 688 × 2 = 1 + 0.190 249 415 466 975 154 405 376;
  • 98) 0.190 249 415 466 975 154 405 376 × 2 = 0 + 0.380 498 830 933 950 308 810 752;

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:


0.000 000 000 000 028 421 709 383(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 1100 0110 1011 0001 0010 1011 10(2)

6. Positive number before normalization:

0.000 000 000 000 028 421 709 383(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 1100 0110 1011 0001 0010 1011 10(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:


0.000 000 000 000 028 421 709 383(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 1100 0110 1011 0001 0010 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 1100 0110 1011 0001 0010 1011 10(2) × 20 =

1.1111 1111 1111 1111 1111 1111 1111 0001 1010 1100 0100 1010 1110(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 1111 0001 1010 1100 0100 1010 1110


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 1111 0001 1010 1100 0100 1010 1110 =

1111 1111 1111 1111 1111 1111 1111 0001 1010 1100 0100 1010 1110


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 1111 0001 1010 1100 0100 1010 1110


Decimal number -0.000 000 000 000 028 421 709 383 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 1111 1111 1111 1111 1111 1111 1111 0001 1010 1100 0100 1010 1110

» Convert decimal -0.000 000 000 000 028 421 709 388 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 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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