0.000 000 000 000 000 000 009 11 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 009 11(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
0.000 000 000 000 000 000 009 11(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 000 009 11.

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) 0.000 000 000 000 000 000 009 11 × 2 = 0 + 0.000 000 000 000 000 000 018 22;
  • 2) 0.000 000 000 000 000 000 018 22 × 2 = 0 + 0.000 000 000 000 000 000 036 44;
  • 3) 0.000 000 000 000 000 000 036 44 × 2 = 0 + 0.000 000 000 000 000 000 072 88;
  • 4) 0.000 000 000 000 000 000 072 88 × 2 = 0 + 0.000 000 000 000 000 000 145 76;
  • 5) 0.000 000 000 000 000 000 145 76 × 2 = 0 + 0.000 000 000 000 000 000 291 52;
  • 6) 0.000 000 000 000 000 000 291 52 × 2 = 0 + 0.000 000 000 000 000 000 583 04;
  • 7) 0.000 000 000 000 000 000 583 04 × 2 = 0 + 0.000 000 000 000 000 001 166 08;
  • 8) 0.000 000 000 000 000 001 166 08 × 2 = 0 + 0.000 000 000 000 000 002 332 16;
  • 9) 0.000 000 000 000 000 002 332 16 × 2 = 0 + 0.000 000 000 000 000 004 664 32;
  • 10) 0.000 000 000 000 000 004 664 32 × 2 = 0 + 0.000 000 000 000 000 009 328 64;
  • 11) 0.000 000 000 000 000 009 328 64 × 2 = 0 + 0.000 000 000 000 000 018 657 28;
  • 12) 0.000 000 000 000 000 018 657 28 × 2 = 0 + 0.000 000 000 000 000 037 314 56;
  • 13) 0.000 000 000 000 000 037 314 56 × 2 = 0 + 0.000 000 000 000 000 074 629 12;
  • 14) 0.000 000 000 000 000 074 629 12 × 2 = 0 + 0.000 000 000 000 000 149 258 24;
  • 15) 0.000 000 000 000 000 149 258 24 × 2 = 0 + 0.000 000 000 000 000 298 516 48;
  • 16) 0.000 000 000 000 000 298 516 48 × 2 = 0 + 0.000 000 000 000 000 597 032 96;
  • 17) 0.000 000 000 000 000 597 032 96 × 2 = 0 + 0.000 000 000 000 001 194 065 92;
  • 18) 0.000 000 000 000 001 194 065 92 × 2 = 0 + 0.000 000 000 000 002 388 131 84;
  • 19) 0.000 000 000 000 002 388 131 84 × 2 = 0 + 0.000 000 000 000 004 776 263 68;
  • 20) 0.000 000 000 000 004 776 263 68 × 2 = 0 + 0.000 000 000 000 009 552 527 36;
  • 21) 0.000 000 000 000 009 552 527 36 × 2 = 0 + 0.000 000 000 000 019 105 054 72;
  • 22) 0.000 000 000 000 019 105 054 72 × 2 = 0 + 0.000 000 000 000 038 210 109 44;
  • 23) 0.000 000 000 000 038 210 109 44 × 2 = 0 + 0.000 000 000 000 076 420 218 88;
  • 24) 0.000 000 000 000 076 420 218 88 × 2 = 0 + 0.000 000 000 000 152 840 437 76;
  • 25) 0.000 000 000 000 152 840 437 76 × 2 = 0 + 0.000 000 000 000 305 680 875 52;
  • 26) 0.000 000 000 000 305 680 875 52 × 2 = 0 + 0.000 000 000 000 611 361 751 04;
  • 27) 0.000 000 000 000 611 361 751 04 × 2 = 0 + 0.000 000 000 001 222 723 502 08;
  • 28) 0.000 000 000 001 222 723 502 08 × 2 = 0 + 0.000 000 000 002 445 447 004 16;
  • 29) 0.000 000 000 002 445 447 004 16 × 2 = 0 + 0.000 000 000 004 890 894 008 32;
  • 30) 0.000 000 000 004 890 894 008 32 × 2 = 0 + 0.000 000 000 009 781 788 016 64;
  • 31) 0.000 000 000 009 781 788 016 64 × 2 = 0 + 0.000 000 000 019 563 576 033 28;
  • 32) 0.000 000 000 019 563 576 033 28 × 2 = 0 + 0.000 000 000 039 127 152 066 56;
  • 33) 0.000 000 000 039 127 152 066 56 × 2 = 0 + 0.000 000 000 078 254 304 133 12;
  • 34) 0.000 000 000 078 254 304 133 12 × 2 = 0 + 0.000 000 000 156 508 608 266 24;
  • 35) 0.000 000 000 156 508 608 266 24 × 2 = 0 + 0.000 000 000 313 017 216 532 48;
  • 36) 0.000 000 000 313 017 216 532 48 × 2 = 0 + 0.000 000 000 626 034 433 064 96;
  • 37) 0.000 000 000 626 034 433 064 96 × 2 = 0 + 0.000 000 001 252 068 866 129 92;
  • 38) 0.000 000 001 252 068 866 129 92 × 2 = 0 + 0.000 000 002 504 137 732 259 84;
  • 39) 0.000 000 002 504 137 732 259 84 × 2 = 0 + 0.000 000 005 008 275 464 519 68;
  • 40) 0.000 000 005 008 275 464 519 68 × 2 = 0 + 0.000 000 010 016 550 929 039 36;
  • 41) 0.000 000 010 016 550 929 039 36 × 2 = 0 + 0.000 000 020 033 101 858 078 72;
  • 42) 0.000 000 020 033 101 858 078 72 × 2 = 0 + 0.000 000 040 066 203 716 157 44;
  • 43) 0.000 000 040 066 203 716 157 44 × 2 = 0 + 0.000 000 080 132 407 432 314 88;
  • 44) 0.000 000 080 132 407 432 314 88 × 2 = 0 + 0.000 000 160 264 814 864 629 76;
  • 45) 0.000 000 160 264 814 864 629 76 × 2 = 0 + 0.000 000 320 529 629 729 259 52;
  • 46) 0.000 000 320 529 629 729 259 52 × 2 = 0 + 0.000 000 641 059 259 458 519 04;
  • 47) 0.000 000 641 059 259 458 519 04 × 2 = 0 + 0.000 001 282 118 518 917 038 08;
  • 48) 0.000 001 282 118 518 917 038 08 × 2 = 0 + 0.000 002 564 237 037 834 076 16;
  • 49) 0.000 002 564 237 037 834 076 16 × 2 = 0 + 0.000 005 128 474 075 668 152 32;
  • 50) 0.000 005 128 474 075 668 152 32 × 2 = 0 + 0.000 010 256 948 151 336 304 64;
  • 51) 0.000 010 256 948 151 336 304 64 × 2 = 0 + 0.000 020 513 896 302 672 609 28;
  • 52) 0.000 020 513 896 302 672 609 28 × 2 = 0 + 0.000 041 027 792 605 345 218 56;
  • 53) 0.000 041 027 792 605 345 218 56 × 2 = 0 + 0.000 082 055 585 210 690 437 12;
  • 54) 0.000 082 055 585 210 690 437 12 × 2 = 0 + 0.000 164 111 170 421 380 874 24;
  • 55) 0.000 164 111 170 421 380 874 24 × 2 = 0 + 0.000 328 222 340 842 761 748 48;
  • 56) 0.000 328 222 340 842 761 748 48 × 2 = 0 + 0.000 656 444 681 685 523 496 96;
  • 57) 0.000 656 444 681 685 523 496 96 × 2 = 0 + 0.001 312 889 363 371 046 993 92;
  • 58) 0.001 312 889 363 371 046 993 92 × 2 = 0 + 0.002 625 778 726 742 093 987 84;
  • 59) 0.002 625 778 726 742 093 987 84 × 2 = 0 + 0.005 251 557 453 484 187 975 68;
  • 60) 0.005 251 557 453 484 187 975 68 × 2 = 0 + 0.010 503 114 906 968 375 951 36;
  • 61) 0.010 503 114 906 968 375 951 36 × 2 = 0 + 0.021 006 229 813 936 751 902 72;
  • 62) 0.021 006 229 813 936 751 902 72 × 2 = 0 + 0.042 012 459 627 873 503 805 44;
  • 63) 0.042 012 459 627 873 503 805 44 × 2 = 0 + 0.084 024 919 255 747 007 610 88;
  • 64) 0.084 024 919 255 747 007 610 88 × 2 = 0 + 0.168 049 838 511 494 015 221 76;
  • 65) 0.168 049 838 511 494 015 221 76 × 2 = 0 + 0.336 099 677 022 988 030 443 52;
  • 66) 0.336 099 677 022 988 030 443 52 × 2 = 0 + 0.672 199 354 045 976 060 887 04;
  • 67) 0.672 199 354 045 976 060 887 04 × 2 = 1 + 0.344 398 708 091 952 121 774 08;
  • 68) 0.344 398 708 091 952 121 774 08 × 2 = 0 + 0.688 797 416 183 904 243 548 16;
  • 69) 0.688 797 416 183 904 243 548 16 × 2 = 1 + 0.377 594 832 367 808 487 096 32;
  • 70) 0.377 594 832 367 808 487 096 32 × 2 = 0 + 0.755 189 664 735 616 974 192 64;
  • 71) 0.755 189 664 735 616 974 192 64 × 2 = 1 + 0.510 379 329 471 233 948 385 28;
  • 72) 0.510 379 329 471 233 948 385 28 × 2 = 1 + 0.020 758 658 942 467 896 770 56;
  • 73) 0.020 758 658 942 467 896 770 56 × 2 = 0 + 0.041 517 317 884 935 793 541 12;
  • 74) 0.041 517 317 884 935 793 541 12 × 2 = 0 + 0.083 034 635 769 871 587 082 24;
  • 75) 0.083 034 635 769 871 587 082 24 × 2 = 0 + 0.166 069 271 539 743 174 164 48;
  • 76) 0.166 069 271 539 743 174 164 48 × 2 = 0 + 0.332 138 543 079 486 348 328 96;
  • 77) 0.332 138 543 079 486 348 328 96 × 2 = 0 + 0.664 277 086 158 972 696 657 92;
  • 78) 0.664 277 086 158 972 696 657 92 × 2 = 1 + 0.328 554 172 317 945 393 315 84;
  • 79) 0.328 554 172 317 945 393 315 84 × 2 = 0 + 0.657 108 344 635 890 786 631 68;
  • 80) 0.657 108 344 635 890 786 631 68 × 2 = 1 + 0.314 216 689 271 781 573 263 36;
  • 81) 0.314 216 689 271 781 573 263 36 × 2 = 0 + 0.628 433 378 543 563 146 526 72;
  • 82) 0.628 433 378 543 563 146 526 72 × 2 = 1 + 0.256 866 757 087 126 293 053 44;
  • 83) 0.256 866 757 087 126 293 053 44 × 2 = 0 + 0.513 733 514 174 252 586 106 88;
  • 84) 0.513 733 514 174 252 586 106 88 × 2 = 1 + 0.027 467 028 348 505 172 213 76;
  • 85) 0.027 467 028 348 505 172 213 76 × 2 = 0 + 0.054 934 056 697 010 344 427 52;
  • 86) 0.054 934 056 697 010 344 427 52 × 2 = 0 + 0.109 868 113 394 020 688 855 04;
  • 87) 0.109 868 113 394 020 688 855 04 × 2 = 0 + 0.219 736 226 788 041 377 710 08;
  • 88) 0.219 736 226 788 041 377 710 08 × 2 = 0 + 0.439 472 453 576 082 755 420 16;
  • 89) 0.439 472 453 576 082 755 420 16 × 2 = 0 + 0.878 944 907 152 165 510 840 32;
  • 90) 0.878 944 907 152 165 510 840 32 × 2 = 1 + 0.757 889 814 304 331 021 680 64;
  • 91) 0.757 889 814 304 331 021 680 64 × 2 = 1 + 0.515 779 628 608 662 043 361 28;
  • 92) 0.515 779 628 608 662 043 361 28 × 2 = 1 + 0.031 559 257 217 324 086 722 56;
  • 93) 0.031 559 257 217 324 086 722 56 × 2 = 0 + 0.063 118 514 434 648 173 445 12;
  • 94) 0.063 118 514 434 648 173 445 12 × 2 = 0 + 0.126 237 028 869 296 346 890 24;
  • 95) 0.126 237 028 869 296 346 890 24 × 2 = 0 + 0.252 474 057 738 592 693 780 48;
  • 96) 0.252 474 057 738 592 693 780 48 × 2 = 0 + 0.504 948 115 477 185 387 560 96;
  • 97) 0.504 948 115 477 185 387 560 96 × 2 = 1 + 0.009 896 230 954 370 775 121 92;
  • 98) 0.009 896 230 954 370 775 121 92 × 2 = 0 + 0.019 792 461 908 741 550 243 84;
  • 99) 0.019 792 461 908 741 550 243 84 × 2 = 0 + 0.039 584 923 817 483 100 487 68;
  • 100) 0.039 584 923 817 483 100 487 68 × 2 = 0 + 0.079 169 847 634 966 200 975 36;
  • 101) 0.079 169 847 634 966 200 975 36 × 2 = 0 + 0.158 339 695 269 932 401 950 72;
  • 102) 0.158 339 695 269 932 401 950 72 × 2 = 0 + 0.316 679 390 539 864 803 901 44;
  • 103) 0.316 679 390 539 864 803 901 44 × 2 = 0 + 0.633 358 781 079 729 607 802 88;
  • 104) 0.633 358 781 079 729 607 802 88 × 2 = 1 + 0.266 717 562 159 459 215 605 76;
  • 105) 0.266 717 562 159 459 215 605 76 × 2 = 0 + 0.533 435 124 318 918 431 211 52;
  • 106) 0.533 435 124 318 918 431 211 52 × 2 = 1 + 0.066 870 248 637 836 862 423 04;
  • 107) 0.066 870 248 637 836 862 423 04 × 2 = 0 + 0.133 740 497 275 673 724 846 08;
  • 108) 0.133 740 497 275 673 724 846 08 × 2 = 0 + 0.267 480 994 551 347 449 692 16;
  • 109) 0.267 480 994 551 347 449 692 16 × 2 = 0 + 0.534 961 989 102 694 899 384 32;
  • 110) 0.534 961 989 102 694 899 384 32 × 2 = 1 + 0.069 923 978 205 389 798 768 64;
  • 111) 0.069 923 978 205 389 798 768 64 × 2 = 0 + 0.139 847 956 410 779 597 537 28;
  • 112) 0.139 847 956 410 779 597 537 28 × 2 = 0 + 0.279 695 912 821 559 195 074 56;
  • 113) 0.279 695 912 821 559 195 074 56 × 2 = 0 + 0.559 391 825 643 118 390 149 12;
  • 114) 0.559 391 825 643 118 390 149 12 × 2 = 1 + 0.118 783 651 286 236 780 298 24;
  • 115) 0.118 783 651 286 236 780 298 24 × 2 = 0 + 0.237 567 302 572 473 560 596 48;
  • 116) 0.237 567 302 572 473 560 596 48 × 2 = 0 + 0.475 134 605 144 947 121 192 96;
  • 117) 0.475 134 605 144 947 121 192 96 × 2 = 0 + 0.950 269 210 289 894 242 385 92;
  • 118) 0.950 269 210 289 894 242 385 92 × 2 = 1 + 0.900 538 420 579 788 484 771 84;
  • 119) 0.900 538 420 579 788 484 771 84 × 2 = 1 + 0.801 076 841 159 576 969 543 68;

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).


4. 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:


0.000 000 000 000 000 000 009 11(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1011 0000 0101 0101 0000 0111 0000 1000 0001 0100 0100 0100 011(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 009 11(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1011 0000 0101 0101 0000 0111 0000 1000 0001 0100 0100 0100 011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 67 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 009 11(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1011 0000 0101 0101 0000 0111 0000 1000 0001 0100 0100 0100 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1011 0000 0101 0101 0000 0111 0000 1000 0001 0100 0100 0100 011(2) × 20 =

1.0101 1000 0010 1010 1000 0011 1000 0100 0000 1010 0010 0010 0011(2) × 2-67


7. 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 0 (a positive number)

Exponent (unadjusted): -67

Mantissa (not normalized):
1.0101 1000 0010 1010 1000 0011 1000 0100 0000 1010 0010 0010 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-67 + 2(11-1) - 1 =

(-67 + 1 023)(10) =

956(10)


9. 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;
  • 956 ÷ 2 = 478 + 0;
  • 478 ÷ 2 = 239 + 0;
  • 239 ÷ 2 = 119 + 1;
  • 119 ÷ 2 = 59 + 1;
  • 59 ÷ 2 = 29 + 1;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

956(10) =

011 1011 1100(2)


11. 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. 0101 1000 0010 1010 1000 0011 1000 0100 0000 1010 0010 0010 0011 =

0101 1000 0010 1010 1000 0011 1000 0100 0000 1010 0010 0010 0011


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1011 1100

Mantissa (52 bits) =
0101 1000 0010 1010 1000 0011 1000 0100 0000 1010 0010 0010 0011


Decimal number 0.000 000 000 000 000 000 009 11 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1011 1100 - 0101 1000 0010 1010 1000 0011 1000 0100 0000 1010 0010 0010 0011

Share

» Convert decimal 0.000 000 000 000 000 000 009 06 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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