0.000 000 000 000 000 000 010 33 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 010 33(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
0.000 000 000 000 000 000 010 33(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: 0.000 000 000 000 000 000 010 33.

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) 0.000 000 000 000 000 000 010 33 × 2 = 0 + 0.000 000 000 000 000 000 020 66;
  • 2) 0.000 000 000 000 000 000 020 66 × 2 = 0 + 0.000 000 000 000 000 000 041 32;
  • 3) 0.000 000 000 000 000 000 041 32 × 2 = 0 + 0.000 000 000 000 000 000 082 64;
  • 4) 0.000 000 000 000 000 000 082 64 × 2 = 0 + 0.000 000 000 000 000 000 165 28;
  • 5) 0.000 000 000 000 000 000 165 28 × 2 = 0 + 0.000 000 000 000 000 000 330 56;
  • 6) 0.000 000 000 000 000 000 330 56 × 2 = 0 + 0.000 000 000 000 000 000 661 12;
  • 7) 0.000 000 000 000 000 000 661 12 × 2 = 0 + 0.000 000 000 000 000 001 322 24;
  • 8) 0.000 000 000 000 000 001 322 24 × 2 = 0 + 0.000 000 000 000 000 002 644 48;
  • 9) 0.000 000 000 000 000 002 644 48 × 2 = 0 + 0.000 000 000 000 000 005 288 96;
  • 10) 0.000 000 000 000 000 005 288 96 × 2 = 0 + 0.000 000 000 000 000 010 577 92;
  • 11) 0.000 000 000 000 000 010 577 92 × 2 = 0 + 0.000 000 000 000 000 021 155 84;
  • 12) 0.000 000 000 000 000 021 155 84 × 2 = 0 + 0.000 000 000 000 000 042 311 68;
  • 13) 0.000 000 000 000 000 042 311 68 × 2 = 0 + 0.000 000 000 000 000 084 623 36;
  • 14) 0.000 000 000 000 000 084 623 36 × 2 = 0 + 0.000 000 000 000 000 169 246 72;
  • 15) 0.000 000 000 000 000 169 246 72 × 2 = 0 + 0.000 000 000 000 000 338 493 44;
  • 16) 0.000 000 000 000 000 338 493 44 × 2 = 0 + 0.000 000 000 000 000 676 986 88;
  • 17) 0.000 000 000 000 000 676 986 88 × 2 = 0 + 0.000 000 000 000 001 353 973 76;
  • 18) 0.000 000 000 000 001 353 973 76 × 2 = 0 + 0.000 000 000 000 002 707 947 52;
  • 19) 0.000 000 000 000 002 707 947 52 × 2 = 0 + 0.000 000 000 000 005 415 895 04;
  • 20) 0.000 000 000 000 005 415 895 04 × 2 = 0 + 0.000 000 000 000 010 831 790 08;
  • 21) 0.000 000 000 000 010 831 790 08 × 2 = 0 + 0.000 000 000 000 021 663 580 16;
  • 22) 0.000 000 000 000 021 663 580 16 × 2 = 0 + 0.000 000 000 000 043 327 160 32;
  • 23) 0.000 000 000 000 043 327 160 32 × 2 = 0 + 0.000 000 000 000 086 654 320 64;
  • 24) 0.000 000 000 000 086 654 320 64 × 2 = 0 + 0.000 000 000 000 173 308 641 28;
  • 25) 0.000 000 000 000 173 308 641 28 × 2 = 0 + 0.000 000 000 000 346 617 282 56;
  • 26) 0.000 000 000 000 346 617 282 56 × 2 = 0 + 0.000 000 000 000 693 234 565 12;
  • 27) 0.000 000 000 000 693 234 565 12 × 2 = 0 + 0.000 000 000 001 386 469 130 24;
  • 28) 0.000 000 000 001 386 469 130 24 × 2 = 0 + 0.000 000 000 002 772 938 260 48;
  • 29) 0.000 000 000 002 772 938 260 48 × 2 = 0 + 0.000 000 000 005 545 876 520 96;
  • 30) 0.000 000 000 005 545 876 520 96 × 2 = 0 + 0.000 000 000 011 091 753 041 92;
  • 31) 0.000 000 000 011 091 753 041 92 × 2 = 0 + 0.000 000 000 022 183 506 083 84;
  • 32) 0.000 000 000 022 183 506 083 84 × 2 = 0 + 0.000 000 000 044 367 012 167 68;
  • 33) 0.000 000 000 044 367 012 167 68 × 2 = 0 + 0.000 000 000 088 734 024 335 36;
  • 34) 0.000 000 000 088 734 024 335 36 × 2 = 0 + 0.000 000 000 177 468 048 670 72;
  • 35) 0.000 000 000 177 468 048 670 72 × 2 = 0 + 0.000 000 000 354 936 097 341 44;
  • 36) 0.000 000 000 354 936 097 341 44 × 2 = 0 + 0.000 000 000 709 872 194 682 88;
  • 37) 0.000 000 000 709 872 194 682 88 × 2 = 0 + 0.000 000 001 419 744 389 365 76;
  • 38) 0.000 000 001 419 744 389 365 76 × 2 = 0 + 0.000 000 002 839 488 778 731 52;
  • 39) 0.000 000 002 839 488 778 731 52 × 2 = 0 + 0.000 000 005 678 977 557 463 04;
  • 40) 0.000 000 005 678 977 557 463 04 × 2 = 0 + 0.000 000 011 357 955 114 926 08;
  • 41) 0.000 000 011 357 955 114 926 08 × 2 = 0 + 0.000 000 022 715 910 229 852 16;
  • 42) 0.000 000 022 715 910 229 852 16 × 2 = 0 + 0.000 000 045 431 820 459 704 32;
  • 43) 0.000 000 045 431 820 459 704 32 × 2 = 0 + 0.000 000 090 863 640 919 408 64;
  • 44) 0.000 000 090 863 640 919 408 64 × 2 = 0 + 0.000 000 181 727 281 838 817 28;
  • 45) 0.000 000 181 727 281 838 817 28 × 2 = 0 + 0.000 000 363 454 563 677 634 56;
  • 46) 0.000 000 363 454 563 677 634 56 × 2 = 0 + 0.000 000 726 909 127 355 269 12;
  • 47) 0.000 000 726 909 127 355 269 12 × 2 = 0 + 0.000 001 453 818 254 710 538 24;
  • 48) 0.000 001 453 818 254 710 538 24 × 2 = 0 + 0.000 002 907 636 509 421 076 48;
  • 49) 0.000 002 907 636 509 421 076 48 × 2 = 0 + 0.000 005 815 273 018 842 152 96;
  • 50) 0.000 005 815 273 018 842 152 96 × 2 = 0 + 0.000 011 630 546 037 684 305 92;
  • 51) 0.000 011 630 546 037 684 305 92 × 2 = 0 + 0.000 023 261 092 075 368 611 84;
  • 52) 0.000 023 261 092 075 368 611 84 × 2 = 0 + 0.000 046 522 184 150 737 223 68;
  • 53) 0.000 046 522 184 150 737 223 68 × 2 = 0 + 0.000 093 044 368 301 474 447 36;
  • 54) 0.000 093 044 368 301 474 447 36 × 2 = 0 + 0.000 186 088 736 602 948 894 72;
  • 55) 0.000 186 088 736 602 948 894 72 × 2 = 0 + 0.000 372 177 473 205 897 789 44;
  • 56) 0.000 372 177 473 205 897 789 44 × 2 = 0 + 0.000 744 354 946 411 795 578 88;
  • 57) 0.000 744 354 946 411 795 578 88 × 2 = 0 + 0.001 488 709 892 823 591 157 76;
  • 58) 0.001 488 709 892 823 591 157 76 × 2 = 0 + 0.002 977 419 785 647 182 315 52;
  • 59) 0.002 977 419 785 647 182 315 52 × 2 = 0 + 0.005 954 839 571 294 364 631 04;
  • 60) 0.005 954 839 571 294 364 631 04 × 2 = 0 + 0.011 909 679 142 588 729 262 08;
  • 61) 0.011 909 679 142 588 729 262 08 × 2 = 0 + 0.023 819 358 285 177 458 524 16;
  • 62) 0.023 819 358 285 177 458 524 16 × 2 = 0 + 0.047 638 716 570 354 917 048 32;
  • 63) 0.047 638 716 570 354 917 048 32 × 2 = 0 + 0.095 277 433 140 709 834 096 64;
  • 64) 0.095 277 433 140 709 834 096 64 × 2 = 0 + 0.190 554 866 281 419 668 193 28;
  • 65) 0.190 554 866 281 419 668 193 28 × 2 = 0 + 0.381 109 732 562 839 336 386 56;
  • 66) 0.381 109 732 562 839 336 386 56 × 2 = 0 + 0.762 219 465 125 678 672 773 12;
  • 67) 0.762 219 465 125 678 672 773 12 × 2 = 1 + 0.524 438 930 251 357 345 546 24;
  • 68) 0.524 438 930 251 357 345 546 24 × 2 = 1 + 0.048 877 860 502 714 691 092 48;
  • 69) 0.048 877 860 502 714 691 092 48 × 2 = 0 + 0.097 755 721 005 429 382 184 96;
  • 70) 0.097 755 721 005 429 382 184 96 × 2 = 0 + 0.195 511 442 010 858 764 369 92;
  • 71) 0.195 511 442 010 858 764 369 92 × 2 = 0 + 0.391 022 884 021 717 528 739 84;
  • 72) 0.391 022 884 021 717 528 739 84 × 2 = 0 + 0.782 045 768 043 435 057 479 68;
  • 73) 0.782 045 768 043 435 057 479 68 × 2 = 1 + 0.564 091 536 086 870 114 959 36;
  • 74) 0.564 091 536 086 870 114 959 36 × 2 = 1 + 0.128 183 072 173 740 229 918 72;
  • 75) 0.128 183 072 173 740 229 918 72 × 2 = 0 + 0.256 366 144 347 480 459 837 44;
  • 76) 0.256 366 144 347 480 459 837 44 × 2 = 0 + 0.512 732 288 694 960 919 674 88;
  • 77) 0.512 732 288 694 960 919 674 88 × 2 = 1 + 0.025 464 577 389 921 839 349 76;
  • 78) 0.025 464 577 389 921 839 349 76 × 2 = 0 + 0.050 929 154 779 843 678 699 52;
  • 79) 0.050 929 154 779 843 678 699 52 × 2 = 0 + 0.101 858 309 559 687 357 399 04;
  • 80) 0.101 858 309 559 687 357 399 04 × 2 = 0 + 0.203 716 619 119 374 714 798 08;
  • 81) 0.203 716 619 119 374 714 798 08 × 2 = 0 + 0.407 433 238 238 749 429 596 16;
  • 82) 0.407 433 238 238 749 429 596 16 × 2 = 0 + 0.814 866 476 477 498 859 192 32;
  • 83) 0.814 866 476 477 498 859 192 32 × 2 = 1 + 0.629 732 952 954 997 718 384 64;
  • 84) 0.629 732 952 954 997 718 384 64 × 2 = 1 + 0.259 465 905 909 995 436 769 28;
  • 85) 0.259 465 905 909 995 436 769 28 × 2 = 0 + 0.518 931 811 819 990 873 538 56;
  • 86) 0.518 931 811 819 990 873 538 56 × 2 = 1 + 0.037 863 623 639 981 747 077 12;
  • 87) 0.037 863 623 639 981 747 077 12 × 2 = 0 + 0.075 727 247 279 963 494 154 24;
  • 88) 0.075 727 247 279 963 494 154 24 × 2 = 0 + 0.151 454 494 559 926 988 308 48;
  • 89) 0.151 454 494 559 926 988 308 48 × 2 = 0 + 0.302 908 989 119 853 976 616 96;
  • 90) 0.302 908 989 119 853 976 616 96 × 2 = 0 + 0.605 817 978 239 707 953 233 92;
  • 91) 0.605 817 978 239 707 953 233 92 × 2 = 1 + 0.211 635 956 479 415 906 467 84;
  • 92) 0.211 635 956 479 415 906 467 84 × 2 = 0 + 0.423 271 912 958 831 812 935 68;
  • 93) 0.423 271 912 958 831 812 935 68 × 2 = 0 + 0.846 543 825 917 663 625 871 36;
  • 94) 0.846 543 825 917 663 625 871 36 × 2 = 1 + 0.693 087 651 835 327 251 742 72;
  • 95) 0.693 087 651 835 327 251 742 72 × 2 = 1 + 0.386 175 303 670 654 503 485 44;
  • 96) 0.386 175 303 670 654 503 485 44 × 2 = 0 + 0.772 350 607 341 309 006 970 88;
  • 97) 0.772 350 607 341 309 006 970 88 × 2 = 1 + 0.544 701 214 682 618 013 941 76;
  • 98) 0.544 701 214 682 618 013 941 76 × 2 = 1 + 0.089 402 429 365 236 027 883 52;
  • 99) 0.089 402 429 365 236 027 883 52 × 2 = 0 + 0.178 804 858 730 472 055 767 04;
  • 100) 0.178 804 858 730 472 055 767 04 × 2 = 0 + 0.357 609 717 460 944 111 534 08;
  • 101) 0.357 609 717 460 944 111 534 08 × 2 = 0 + 0.715 219 434 921 888 223 068 16;
  • 102) 0.715 219 434 921 888 223 068 16 × 2 = 1 + 0.430 438 869 843 776 446 136 32;
  • 103) 0.430 438 869 843 776 446 136 32 × 2 = 0 + 0.860 877 739 687 552 892 272 64;
  • 104) 0.860 877 739 687 552 892 272 64 × 2 = 1 + 0.721 755 479 375 105 784 545 28;
  • 105) 0.721 755 479 375 105 784 545 28 × 2 = 1 + 0.443 510 958 750 211 569 090 56;
  • 106) 0.443 510 958 750 211 569 090 56 × 2 = 0 + 0.887 021 917 500 423 138 181 12;
  • 107) 0.887 021 917 500 423 138 181 12 × 2 = 1 + 0.774 043 835 000 846 276 362 24;
  • 108) 0.774 043 835 000 846 276 362 24 × 2 = 1 + 0.548 087 670 001 692 552 724 48;
  • 109) 0.548 087 670 001 692 552 724 48 × 2 = 1 + 0.096 175 340 003 385 105 448 96;
  • 110) 0.096 175 340 003 385 105 448 96 × 2 = 0 + 0.192 350 680 006 770 210 897 92;
  • 111) 0.192 350 680 006 770 210 897 92 × 2 = 0 + 0.384 701 360 013 540 421 795 84;
  • 112) 0.384 701 360 013 540 421 795 84 × 2 = 0 + 0.769 402 720 027 080 843 591 68;
  • 113) 0.769 402 720 027 080 843 591 68 × 2 = 1 + 0.538 805 440 054 161 687 183 36;
  • 114) 0.538 805 440 054 161 687 183 36 × 2 = 1 + 0.077 610 880 108 323 374 366 72;
  • 115) 0.077 610 880 108 323 374 366 72 × 2 = 0 + 0.155 221 760 216 646 748 733 44;
  • 116) 0.155 221 760 216 646 748 733 44 × 2 = 0 + 0.310 443 520 433 293 497 466 88;
  • 117) 0.310 443 520 433 293 497 466 88 × 2 = 0 + 0.620 887 040 866 586 994 933 76;
  • 118) 0.620 887 040 866 586 994 933 76 × 2 = 1 + 0.241 774 081 733 173 989 867 52;
  • 119) 0.241 774 081 733 173 989 867 52 × 2 = 0 + 0.483 548 163 466 347 979 735 04;

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:


0.000 000 000 000 000 000 010 33(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 1100 1000 0011 0100 0010 0110 1100 0101 1011 1000 1100 010(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 010 33(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 1100 1000 0011 0100 0010 0110 1100 0101 1011 1000 1100 010(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:


0.000 000 000 000 000 000 010 33(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 1100 1000 0011 0100 0010 0110 1100 0101 1011 1000 1100 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 1100 1000 0011 0100 0010 0110 1100 0101 1011 1000 1100 010(2) × 20 =

1.1000 0110 0100 0001 1010 0001 0011 0110 0010 1101 1100 0110 0010(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):
1.1000 0110 0100 0001 1010 0001 0011 0110 0010 1101 1100 0110 0010


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. 1000 0110 0100 0001 1010 0001 0011 0110 0010 1101 1100 0110 0010 =

1000 0110 0100 0001 1010 0001 0011 0110 0010 1101 1100 0110 0010


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) =
1000 0110 0100 0001 1010 0001 0011 0110 0010 1101 1100 0110 0010


Decimal number 0.000 000 000 000 000 000 010 33 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1011 1100 - 1000 0110 0100 0001 1010 0001 0011 0110 0010 1101 1100 0110 0010

Share

» Convert decimal 0.000 000 000 000 000 000 009 73 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