-0.000 000 000 000 000 000 122 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 000 122(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
-0.000 000 000 000 000 000 122(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:

|-0.000 000 000 000 000 000 122| = 0.000 000 000 000 000 000 122


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: 0.000 000 000 000 000 000 122.

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) 0.000 000 000 000 000 000 122 × 2 = 0 + 0.000 000 000 000 000 000 244;
  • 2) 0.000 000 000 000 000 000 244 × 2 = 0 + 0.000 000 000 000 000 000 488;
  • 3) 0.000 000 000 000 000 000 488 × 2 = 0 + 0.000 000 000 000 000 000 976;
  • 4) 0.000 000 000 000 000 000 976 × 2 = 0 + 0.000 000 000 000 000 001 952;
  • 5) 0.000 000 000 000 000 001 952 × 2 = 0 + 0.000 000 000 000 000 003 904;
  • 6) 0.000 000 000 000 000 003 904 × 2 = 0 + 0.000 000 000 000 000 007 808;
  • 7) 0.000 000 000 000 000 007 808 × 2 = 0 + 0.000 000 000 000 000 015 616;
  • 8) 0.000 000 000 000 000 015 616 × 2 = 0 + 0.000 000 000 000 000 031 232;
  • 9) 0.000 000 000 000 000 031 232 × 2 = 0 + 0.000 000 000 000 000 062 464;
  • 10) 0.000 000 000 000 000 062 464 × 2 = 0 + 0.000 000 000 000 000 124 928;
  • 11) 0.000 000 000 000 000 124 928 × 2 = 0 + 0.000 000 000 000 000 249 856;
  • 12) 0.000 000 000 000 000 249 856 × 2 = 0 + 0.000 000 000 000 000 499 712;
  • 13) 0.000 000 000 000 000 499 712 × 2 = 0 + 0.000 000 000 000 000 999 424;
  • 14) 0.000 000 000 000 000 999 424 × 2 = 0 + 0.000 000 000 000 001 998 848;
  • 15) 0.000 000 000 000 001 998 848 × 2 = 0 + 0.000 000 000 000 003 997 696;
  • 16) 0.000 000 000 000 003 997 696 × 2 = 0 + 0.000 000 000 000 007 995 392;
  • 17) 0.000 000 000 000 007 995 392 × 2 = 0 + 0.000 000 000 000 015 990 784;
  • 18) 0.000 000 000 000 015 990 784 × 2 = 0 + 0.000 000 000 000 031 981 568;
  • 19) 0.000 000 000 000 031 981 568 × 2 = 0 + 0.000 000 000 000 063 963 136;
  • 20) 0.000 000 000 000 063 963 136 × 2 = 0 + 0.000 000 000 000 127 926 272;
  • 21) 0.000 000 000 000 127 926 272 × 2 = 0 + 0.000 000 000 000 255 852 544;
  • 22) 0.000 000 000 000 255 852 544 × 2 = 0 + 0.000 000 000 000 511 705 088;
  • 23) 0.000 000 000 000 511 705 088 × 2 = 0 + 0.000 000 000 001 023 410 176;
  • 24) 0.000 000 000 001 023 410 176 × 2 = 0 + 0.000 000 000 002 046 820 352;
  • 25) 0.000 000 000 002 046 820 352 × 2 = 0 + 0.000 000 000 004 093 640 704;
  • 26) 0.000 000 000 004 093 640 704 × 2 = 0 + 0.000 000 000 008 187 281 408;
  • 27) 0.000 000 000 008 187 281 408 × 2 = 0 + 0.000 000 000 016 374 562 816;
  • 28) 0.000 000 000 016 374 562 816 × 2 = 0 + 0.000 000 000 032 749 125 632;
  • 29) 0.000 000 000 032 749 125 632 × 2 = 0 + 0.000 000 000 065 498 251 264;
  • 30) 0.000 000 000 065 498 251 264 × 2 = 0 + 0.000 000 000 130 996 502 528;
  • 31) 0.000 000 000 130 996 502 528 × 2 = 0 + 0.000 000 000 261 993 005 056;
  • 32) 0.000 000 000 261 993 005 056 × 2 = 0 + 0.000 000 000 523 986 010 112;
  • 33) 0.000 000 000 523 986 010 112 × 2 = 0 + 0.000 000 001 047 972 020 224;
  • 34) 0.000 000 001 047 972 020 224 × 2 = 0 + 0.000 000 002 095 944 040 448;
  • 35) 0.000 000 002 095 944 040 448 × 2 = 0 + 0.000 000 004 191 888 080 896;
  • 36) 0.000 000 004 191 888 080 896 × 2 = 0 + 0.000 000 008 383 776 161 792;
  • 37) 0.000 000 008 383 776 161 792 × 2 = 0 + 0.000 000 016 767 552 323 584;
  • 38) 0.000 000 016 767 552 323 584 × 2 = 0 + 0.000 000 033 535 104 647 168;
  • 39) 0.000 000 033 535 104 647 168 × 2 = 0 + 0.000 000 067 070 209 294 336;
  • 40) 0.000 000 067 070 209 294 336 × 2 = 0 + 0.000 000 134 140 418 588 672;
  • 41) 0.000 000 134 140 418 588 672 × 2 = 0 + 0.000 000 268 280 837 177 344;
  • 42) 0.000 000 268 280 837 177 344 × 2 = 0 + 0.000 000 536 561 674 354 688;
  • 43) 0.000 000 536 561 674 354 688 × 2 = 0 + 0.000 001 073 123 348 709 376;
  • 44) 0.000 001 073 123 348 709 376 × 2 = 0 + 0.000 002 146 246 697 418 752;
  • 45) 0.000 002 146 246 697 418 752 × 2 = 0 + 0.000 004 292 493 394 837 504;
  • 46) 0.000 004 292 493 394 837 504 × 2 = 0 + 0.000 008 584 986 789 675 008;
  • 47) 0.000 008 584 986 789 675 008 × 2 = 0 + 0.000 017 169 973 579 350 016;
  • 48) 0.000 017 169 973 579 350 016 × 2 = 0 + 0.000 034 339 947 158 700 032;
  • 49) 0.000 034 339 947 158 700 032 × 2 = 0 + 0.000 068 679 894 317 400 064;
  • 50) 0.000 068 679 894 317 400 064 × 2 = 0 + 0.000 137 359 788 634 800 128;
  • 51) 0.000 137 359 788 634 800 128 × 2 = 0 + 0.000 274 719 577 269 600 256;
  • 52) 0.000 274 719 577 269 600 256 × 2 = 0 + 0.000 549 439 154 539 200 512;
  • 53) 0.000 549 439 154 539 200 512 × 2 = 0 + 0.001 098 878 309 078 401 024;
  • 54) 0.001 098 878 309 078 401 024 × 2 = 0 + 0.002 197 756 618 156 802 048;
  • 55) 0.002 197 756 618 156 802 048 × 2 = 0 + 0.004 395 513 236 313 604 096;
  • 56) 0.004 395 513 236 313 604 096 × 2 = 0 + 0.008 791 026 472 627 208 192;
  • 57) 0.008 791 026 472 627 208 192 × 2 = 0 + 0.017 582 052 945 254 416 384;
  • 58) 0.017 582 052 945 254 416 384 × 2 = 0 + 0.035 164 105 890 508 832 768;
  • 59) 0.035 164 105 890 508 832 768 × 2 = 0 + 0.070 328 211 781 017 665 536;
  • 60) 0.070 328 211 781 017 665 536 × 2 = 0 + 0.140 656 423 562 035 331 072;
  • 61) 0.140 656 423 562 035 331 072 × 2 = 0 + 0.281 312 847 124 070 662 144;
  • 62) 0.281 312 847 124 070 662 144 × 2 = 0 + 0.562 625 694 248 141 324 288;
  • 63) 0.562 625 694 248 141 324 288 × 2 = 1 + 0.125 251 388 496 282 648 576;
  • 64) 0.125 251 388 496 282 648 576 × 2 = 0 + 0.250 502 776 992 565 297 152;
  • 65) 0.250 502 776 992 565 297 152 × 2 = 0 + 0.501 005 553 985 130 594 304;
  • 66) 0.501 005 553 985 130 594 304 × 2 = 1 + 0.002 011 107 970 261 188 608;
  • 67) 0.002 011 107 970 261 188 608 × 2 = 0 + 0.004 022 215 940 522 377 216;
  • 68) 0.004 022 215 940 522 377 216 × 2 = 0 + 0.008 044 431 881 044 754 432;
  • 69) 0.008 044 431 881 044 754 432 × 2 = 0 + 0.016 088 863 762 089 508 864;
  • 70) 0.016 088 863 762 089 508 864 × 2 = 0 + 0.032 177 727 524 179 017 728;
  • 71) 0.032 177 727 524 179 017 728 × 2 = 0 + 0.064 355 455 048 358 035 456;
  • 72) 0.064 355 455 048 358 035 456 × 2 = 0 + 0.128 710 910 096 716 070 912;
  • 73) 0.128 710 910 096 716 070 912 × 2 = 0 + 0.257 421 820 193 432 141 824;
  • 74) 0.257 421 820 193 432 141 824 × 2 = 0 + 0.514 843 640 386 864 283 648;
  • 75) 0.514 843 640 386 864 283 648 × 2 = 1 + 0.029 687 280 773 728 567 296;
  • 76) 0.029 687 280 773 728 567 296 × 2 = 0 + 0.059 374 561 547 457 134 592;
  • 77) 0.059 374 561 547 457 134 592 × 2 = 0 + 0.118 749 123 094 914 269 184;
  • 78) 0.118 749 123 094 914 269 184 × 2 = 0 + 0.237 498 246 189 828 538 368;
  • 79) 0.237 498 246 189 828 538 368 × 2 = 0 + 0.474 996 492 379 657 076 736;
  • 80) 0.474 996 492 379 657 076 736 × 2 = 0 + 0.949 992 984 759 314 153 472;
  • 81) 0.949 992 984 759 314 153 472 × 2 = 1 + 0.899 985 969 518 628 306 944;
  • 82) 0.899 985 969 518 628 306 944 × 2 = 1 + 0.799 971 939 037 256 613 888;
  • 83) 0.799 971 939 037 256 613 888 × 2 = 1 + 0.599 943 878 074 513 227 776;
  • 84) 0.599 943 878 074 513 227 776 × 2 = 1 + 0.199 887 756 149 026 455 552;
  • 85) 0.199 887 756 149 026 455 552 × 2 = 0 + 0.399 775 512 298 052 911 104;
  • 86) 0.399 775 512 298 052 911 104 × 2 = 0 + 0.799 551 024 596 105 822 208;
  • 87) 0.799 551 024 596 105 822 208 × 2 = 1 + 0.599 102 049 192 211 644 416;
  • 88) 0.599 102 049 192 211 644 416 × 2 = 1 + 0.198 204 098 384 423 288 832;
  • 89) 0.198 204 098 384 423 288 832 × 2 = 0 + 0.396 408 196 768 846 577 664;
  • 90) 0.396 408 196 768 846 577 664 × 2 = 0 + 0.792 816 393 537 693 155 328;
  • 91) 0.792 816 393 537 693 155 328 × 2 = 1 + 0.585 632 787 075 386 310 656;
  • 92) 0.585 632 787 075 386 310 656 × 2 = 1 + 0.171 265 574 150 772 621 312;
  • 93) 0.171 265 574 150 772 621 312 × 2 = 0 + 0.342 531 148 301 545 242 624;
  • 94) 0.342 531 148 301 545 242 624 × 2 = 0 + 0.685 062 296 603 090 485 248;
  • 95) 0.685 062 296 603 090 485 248 × 2 = 1 + 0.370 124 593 206 180 970 496;
  • 96) 0.370 124 593 206 180 970 496 × 2 = 0 + 0.740 249 186 412 361 940 992;
  • 97) 0.740 249 186 412 361 940 992 × 2 = 1 + 0.480 498 372 824 723 881 984;
  • 98) 0.480 498 372 824 723 881 984 × 2 = 0 + 0.960 996 745 649 447 763 968;
  • 99) 0.960 996 745 649 447 763 968 × 2 = 1 + 0.921 993 491 298 895 527 936;
  • 100) 0.921 993 491 298 895 527 936 × 2 = 1 + 0.843 986 982 597 791 055 872;
  • 101) 0.843 986 982 597 791 055 872 × 2 = 1 + 0.687 973 965 195 582 111 744;
  • 102) 0.687 973 965 195 582 111 744 × 2 = 1 + 0.375 947 930 391 164 223 488;
  • 103) 0.375 947 930 391 164 223 488 × 2 = 0 + 0.751 895 860 782 328 446 976;
  • 104) 0.751 895 860 782 328 446 976 × 2 = 1 + 0.503 791 721 564 656 893 952;
  • 105) 0.503 791 721 564 656 893 952 × 2 = 1 + 0.007 583 443 129 313 787 904;
  • 106) 0.007 583 443 129 313 787 904 × 2 = 0 + 0.015 166 886 258 627 575 808;
  • 107) 0.015 166 886 258 627 575 808 × 2 = 0 + 0.030 333 772 517 255 151 616;
  • 108) 0.030 333 772 517 255 151 616 × 2 = 0 + 0.060 667 545 034 510 303 232;
  • 109) 0.060 667 545 034 510 303 232 × 2 = 0 + 0.121 335 090 069 020 606 464;
  • 110) 0.121 335 090 069 020 606 464 × 2 = 0 + 0.242 670 180 138 041 212 928;
  • 111) 0.242 670 180 138 041 212 928 × 2 = 0 + 0.485 340 360 276 082 425 856;
  • 112) 0.485 340 360 276 082 425 856 × 2 = 0 + 0.970 680 720 552 164 851 712;
  • 113) 0.970 680 720 552 164 851 712 × 2 = 1 + 0.941 361 441 104 329 703 424;
  • 114) 0.941 361 441 104 329 703 424 × 2 = 1 + 0.882 722 882 208 659 406 848;
  • 115) 0.882 722 882 208 659 406 848 × 2 = 1 + 0.765 445 764 417 318 813 696;

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:


0.000 000 000 000 000 000 122(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0100 0000 0010 0000 1111 0011 0011 0010 1011 1101 1000 0000 111(2)

6. Positive number before normalization:

0.000 000 000 000 000 000 122(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0100 0000 0010 0000 1111 0011 0011 0010 1011 1101 1000 0000 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 63 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 122(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0100 0000 0010 0000 1111 0011 0011 0010 1011 1101 1000 0000 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0100 0000 0010 0000 1111 0011 0011 0010 1011 1101 1000 0000 111(2) × 20 =

1.0010 0000 0001 0000 0111 1001 1001 1001 0101 1110 1100 0000 0111(2) × 2-63


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): -63

Mantissa (not normalized):
1.0010 0000 0001 0000 0111 1001 1001 1001 0101 1110 1100 0000 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-63 + 2(11-1) - 1 =

(-63 + 1 023)(10) =

960(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;
  • 960 ÷ 2 = 480 + 0;
  • 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) =

960(10) =

011 1100 0000(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. 0010 0000 0001 0000 0111 1001 1001 1001 0101 1110 1100 0000 0111 =

0010 0000 0001 0000 0111 1001 1001 1001 0101 1110 1100 0000 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 1100 0000

Mantissa (52 bits) =
0010 0000 0001 0000 0111 1001 1001 1001 0101 1110 1100 0000 0111


Decimal number -0.000 000 000 000 000 000 122 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1100 0000 - 0010 0000 0001 0000 0111 1001 1001 1001 0101 1110 1100 0000 0111

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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