-0.000 000 000 000 000 000 23 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 000 23(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 23(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:

|-0.000 000 000 000 000 000 23| = 0.000 000 000 000 000 000 23


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: 0.000 000 000 000 000 000 23.

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 23 × 2 = 0 + 0.000 000 000 000 000 000 46;
  • 2) 0.000 000 000 000 000 000 46 × 2 = 0 + 0.000 000 000 000 000 000 92;
  • 3) 0.000 000 000 000 000 000 92 × 2 = 0 + 0.000 000 000 000 000 001 84;
  • 4) 0.000 000 000 000 000 001 84 × 2 = 0 + 0.000 000 000 000 000 003 68;
  • 5) 0.000 000 000 000 000 003 68 × 2 = 0 + 0.000 000 000 000 000 007 36;
  • 6) 0.000 000 000 000 000 007 36 × 2 = 0 + 0.000 000 000 000 000 014 72;
  • 7) 0.000 000 000 000 000 014 72 × 2 = 0 + 0.000 000 000 000 000 029 44;
  • 8) 0.000 000 000 000 000 029 44 × 2 = 0 + 0.000 000 000 000 000 058 88;
  • 9) 0.000 000 000 000 000 058 88 × 2 = 0 + 0.000 000 000 000 000 117 76;
  • 10) 0.000 000 000 000 000 117 76 × 2 = 0 + 0.000 000 000 000 000 235 52;
  • 11) 0.000 000 000 000 000 235 52 × 2 = 0 + 0.000 000 000 000 000 471 04;
  • 12) 0.000 000 000 000 000 471 04 × 2 = 0 + 0.000 000 000 000 000 942 08;
  • 13) 0.000 000 000 000 000 942 08 × 2 = 0 + 0.000 000 000 000 001 884 16;
  • 14) 0.000 000 000 000 001 884 16 × 2 = 0 + 0.000 000 000 000 003 768 32;
  • 15) 0.000 000 000 000 003 768 32 × 2 = 0 + 0.000 000 000 000 007 536 64;
  • 16) 0.000 000 000 000 007 536 64 × 2 = 0 + 0.000 000 000 000 015 073 28;
  • 17) 0.000 000 000 000 015 073 28 × 2 = 0 + 0.000 000 000 000 030 146 56;
  • 18) 0.000 000 000 000 030 146 56 × 2 = 0 + 0.000 000 000 000 060 293 12;
  • 19) 0.000 000 000 000 060 293 12 × 2 = 0 + 0.000 000 000 000 120 586 24;
  • 20) 0.000 000 000 000 120 586 24 × 2 = 0 + 0.000 000 000 000 241 172 48;
  • 21) 0.000 000 000 000 241 172 48 × 2 = 0 + 0.000 000 000 000 482 344 96;
  • 22) 0.000 000 000 000 482 344 96 × 2 = 0 + 0.000 000 000 000 964 689 92;
  • 23) 0.000 000 000 000 964 689 92 × 2 = 0 + 0.000 000 000 001 929 379 84;
  • 24) 0.000 000 000 001 929 379 84 × 2 = 0 + 0.000 000 000 003 858 759 68;
  • 25) 0.000 000 000 003 858 759 68 × 2 = 0 + 0.000 000 000 007 717 519 36;
  • 26) 0.000 000 000 007 717 519 36 × 2 = 0 + 0.000 000 000 015 435 038 72;
  • 27) 0.000 000 000 015 435 038 72 × 2 = 0 + 0.000 000 000 030 870 077 44;
  • 28) 0.000 000 000 030 870 077 44 × 2 = 0 + 0.000 000 000 061 740 154 88;
  • 29) 0.000 000 000 061 740 154 88 × 2 = 0 + 0.000 000 000 123 480 309 76;
  • 30) 0.000 000 000 123 480 309 76 × 2 = 0 + 0.000 000 000 246 960 619 52;
  • 31) 0.000 000 000 246 960 619 52 × 2 = 0 + 0.000 000 000 493 921 239 04;
  • 32) 0.000 000 000 493 921 239 04 × 2 = 0 + 0.000 000 000 987 842 478 08;
  • 33) 0.000 000 000 987 842 478 08 × 2 = 0 + 0.000 000 001 975 684 956 16;
  • 34) 0.000 000 001 975 684 956 16 × 2 = 0 + 0.000 000 003 951 369 912 32;
  • 35) 0.000 000 003 951 369 912 32 × 2 = 0 + 0.000 000 007 902 739 824 64;
  • 36) 0.000 000 007 902 739 824 64 × 2 = 0 + 0.000 000 015 805 479 649 28;
  • 37) 0.000 000 015 805 479 649 28 × 2 = 0 + 0.000 000 031 610 959 298 56;
  • 38) 0.000 000 031 610 959 298 56 × 2 = 0 + 0.000 000 063 221 918 597 12;
  • 39) 0.000 000 063 221 918 597 12 × 2 = 0 + 0.000 000 126 443 837 194 24;
  • 40) 0.000 000 126 443 837 194 24 × 2 = 0 + 0.000 000 252 887 674 388 48;
  • 41) 0.000 000 252 887 674 388 48 × 2 = 0 + 0.000 000 505 775 348 776 96;
  • 42) 0.000 000 505 775 348 776 96 × 2 = 0 + 0.000 001 011 550 697 553 92;
  • 43) 0.000 001 011 550 697 553 92 × 2 = 0 + 0.000 002 023 101 395 107 84;
  • 44) 0.000 002 023 101 395 107 84 × 2 = 0 + 0.000 004 046 202 790 215 68;
  • 45) 0.000 004 046 202 790 215 68 × 2 = 0 + 0.000 008 092 405 580 431 36;
  • 46) 0.000 008 092 405 580 431 36 × 2 = 0 + 0.000 016 184 811 160 862 72;
  • 47) 0.000 016 184 811 160 862 72 × 2 = 0 + 0.000 032 369 622 321 725 44;
  • 48) 0.000 032 369 622 321 725 44 × 2 = 0 + 0.000 064 739 244 643 450 88;
  • 49) 0.000 064 739 244 643 450 88 × 2 = 0 + 0.000 129 478 489 286 901 76;
  • 50) 0.000 129 478 489 286 901 76 × 2 = 0 + 0.000 258 956 978 573 803 52;
  • 51) 0.000 258 956 978 573 803 52 × 2 = 0 + 0.000 517 913 957 147 607 04;
  • 52) 0.000 517 913 957 147 607 04 × 2 = 0 + 0.001 035 827 914 295 214 08;
  • 53) 0.001 035 827 914 295 214 08 × 2 = 0 + 0.002 071 655 828 590 428 16;
  • 54) 0.002 071 655 828 590 428 16 × 2 = 0 + 0.004 143 311 657 180 856 32;
  • 55) 0.004 143 311 657 180 856 32 × 2 = 0 + 0.008 286 623 314 361 712 64;
  • 56) 0.008 286 623 314 361 712 64 × 2 = 0 + 0.016 573 246 628 723 425 28;
  • 57) 0.016 573 246 628 723 425 28 × 2 = 0 + 0.033 146 493 257 446 850 56;
  • 58) 0.033 146 493 257 446 850 56 × 2 = 0 + 0.066 292 986 514 893 701 12;
  • 59) 0.066 292 986 514 893 701 12 × 2 = 0 + 0.132 585 973 029 787 402 24;
  • 60) 0.132 585 973 029 787 402 24 × 2 = 0 + 0.265 171 946 059 574 804 48;
  • 61) 0.265 171 946 059 574 804 48 × 2 = 0 + 0.530 343 892 119 149 608 96;
  • 62) 0.530 343 892 119 149 608 96 × 2 = 1 + 0.060 687 784 238 299 217 92;
  • 63) 0.060 687 784 238 299 217 92 × 2 = 0 + 0.121 375 568 476 598 435 84;
  • 64) 0.121 375 568 476 598 435 84 × 2 = 0 + 0.242 751 136 953 196 871 68;
  • 65) 0.242 751 136 953 196 871 68 × 2 = 0 + 0.485 502 273 906 393 743 36;
  • 66) 0.485 502 273 906 393 743 36 × 2 = 0 + 0.971 004 547 812 787 486 72;
  • 67) 0.971 004 547 812 787 486 72 × 2 = 1 + 0.942 009 095 625 574 973 44;
  • 68) 0.942 009 095 625 574 973 44 × 2 = 1 + 0.884 018 191 251 149 946 88;
  • 69) 0.884 018 191 251 149 946 88 × 2 = 1 + 0.768 036 382 502 299 893 76;
  • 70) 0.768 036 382 502 299 893 76 × 2 = 1 + 0.536 072 765 004 599 787 52;
  • 71) 0.536 072 765 004 599 787 52 × 2 = 1 + 0.072 145 530 009 199 575 04;
  • 72) 0.072 145 530 009 199 575 04 × 2 = 0 + 0.144 291 060 018 399 150 08;
  • 73) 0.144 291 060 018 399 150 08 × 2 = 0 + 0.288 582 120 036 798 300 16;
  • 74) 0.288 582 120 036 798 300 16 × 2 = 0 + 0.577 164 240 073 596 600 32;
  • 75) 0.577 164 240 073 596 600 32 × 2 = 1 + 0.154 328 480 147 193 200 64;
  • 76) 0.154 328 480 147 193 200 64 × 2 = 0 + 0.308 656 960 294 386 401 28;
  • 77) 0.308 656 960 294 386 401 28 × 2 = 0 + 0.617 313 920 588 772 802 56;
  • 78) 0.617 313 920 588 772 802 56 × 2 = 1 + 0.234 627 841 177 545 605 12;
  • 79) 0.234 627 841 177 545 605 12 × 2 = 0 + 0.469 255 682 355 091 210 24;
  • 80) 0.469 255 682 355 091 210 24 × 2 = 0 + 0.938 511 364 710 182 420 48;
  • 81) 0.938 511 364 710 182 420 48 × 2 = 1 + 0.877 022 729 420 364 840 96;
  • 82) 0.877 022 729 420 364 840 96 × 2 = 1 + 0.754 045 458 840 729 681 92;
  • 83) 0.754 045 458 840 729 681 92 × 2 = 1 + 0.508 090 917 681 459 363 84;
  • 84) 0.508 090 917 681 459 363 84 × 2 = 1 + 0.016 181 835 362 918 727 68;
  • 85) 0.016 181 835 362 918 727 68 × 2 = 0 + 0.032 363 670 725 837 455 36;
  • 86) 0.032 363 670 725 837 455 36 × 2 = 0 + 0.064 727 341 451 674 910 72;
  • 87) 0.064 727 341 451 674 910 72 × 2 = 0 + 0.129 454 682 903 349 821 44;
  • 88) 0.129 454 682 903 349 821 44 × 2 = 0 + 0.258 909 365 806 699 642 88;
  • 89) 0.258 909 365 806 699 642 88 × 2 = 0 + 0.517 818 731 613 399 285 76;
  • 90) 0.517 818 731 613 399 285 76 × 2 = 1 + 0.035 637 463 226 798 571 52;
  • 91) 0.035 637 463 226 798 571 52 × 2 = 0 + 0.071 274 926 453 597 143 04;
  • 92) 0.071 274 926 453 597 143 04 × 2 = 0 + 0.142 549 852 907 194 286 08;
  • 93) 0.142 549 852 907 194 286 08 × 2 = 0 + 0.285 099 705 814 388 572 16;
  • 94) 0.285 099 705 814 388 572 16 × 2 = 0 + 0.570 199 411 628 777 144 32;
  • 95) 0.570 199 411 628 777 144 32 × 2 = 1 + 0.140 398 823 257 554 288 64;
  • 96) 0.140 398 823 257 554 288 64 × 2 = 0 + 0.280 797 646 515 108 577 28;
  • 97) 0.280 797 646 515 108 577 28 × 2 = 0 + 0.561 595 293 030 217 154 56;
  • 98) 0.561 595 293 030 217 154 56 × 2 = 1 + 0.123 190 586 060 434 309 12;
  • 99) 0.123 190 586 060 434 309 12 × 2 = 0 + 0.246 381 172 120 868 618 24;
  • 100) 0.246 381 172 120 868 618 24 × 2 = 0 + 0.492 762 344 241 737 236 48;
  • 101) 0.492 762 344 241 737 236 48 × 2 = 0 + 0.985 524 688 483 474 472 96;
  • 102) 0.985 524 688 483 474 472 96 × 2 = 1 + 0.971 049 376 966 948 945 92;
  • 103) 0.971 049 376 966 948 945 92 × 2 = 1 + 0.942 098 753 933 897 891 84;
  • 104) 0.942 098 753 933 897 891 84 × 2 = 1 + 0.884 197 507 867 795 783 68;
  • 105) 0.884 197 507 867 795 783 68 × 2 = 1 + 0.768 395 015 735 591 567 36;
  • 106) 0.768 395 015 735 591 567 36 × 2 = 1 + 0.536 790 031 471 183 134 72;
  • 107) 0.536 790 031 471 183 134 72 × 2 = 1 + 0.073 580 062 942 366 269 44;
  • 108) 0.073 580 062 942 366 269 44 × 2 = 0 + 0.147 160 125 884 732 538 88;
  • 109) 0.147 160 125 884 732 538 88 × 2 = 0 + 0.294 320 251 769 465 077 76;
  • 110) 0.294 320 251 769 465 077 76 × 2 = 0 + 0.588 640 503 538 930 155 52;
  • 111) 0.588 640 503 538 930 155 52 × 2 = 1 + 0.177 281 007 077 860 311 04;
  • 112) 0.177 281 007 077 860 311 04 × 2 = 0 + 0.354 562 014 155 720 622 08;
  • 113) 0.354 562 014 155 720 622 08 × 2 = 0 + 0.709 124 028 311 441 244 16;
  • 114) 0.709 124 028 311 441 244 16 × 2 = 1 + 0.418 248 056 622 882 488 32;

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:


0.000 000 000 000 000 000 23(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0011 1110 0010 0100 1111 0000 0100 0010 0100 0111 1110 0010 01(2)

6. Positive number before normalization:

0.000 000 000 000 000 000 23(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0011 1110 0010 0100 1111 0000 0100 0010 0100 0111 1110 0010 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 62 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 23(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0011 1110 0010 0100 1111 0000 0100 0010 0100 0111 1110 0010 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0011 1110 0010 0100 1111 0000 0100 0010 0100 0111 1110 0010 01(2) × 20 =

1.0000 1111 1000 1001 0011 1100 0001 0000 1001 0001 1111 1000 1001(2) × 2-62


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): -62

Mantissa (not normalized):
1.0000 1111 1000 1001 0011 1100 0001 0000 1001 0001 1111 1000 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-62 + 2(11-1) - 1 =

(-62 + 1 023)(10) =

961(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;
  • 961 ÷ 2 = 480 + 1;
  • 480 ÷ 2 = 240 + 0;
  • 240 ÷ 2 = 120 + 0;
  • 120 ÷ 2 = 60 + 0;
  • 60 ÷ 2 = 30 + 0;
  • 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) =

961(10) =

011 1100 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. 0000 1111 1000 1001 0011 1100 0001 0000 1001 0001 1111 1000 1001 =

0000 1111 1000 1001 0011 1100 0001 0000 1001 0001 1111 1000 1001


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 1100 0001

Mantissa (52 bits) =
0000 1111 1000 1001 0011 1100 0001 0000 1001 0001 1111 1000 1001


Decimal number -0.000 000 000 000 000 000 23 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1100 0001 - 0000 1111 1000 1001 0011 1100 0001 0000 1001 0001 1111 1000 1001

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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