0.000 000 000 000 000 000 187 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 187(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
0.000 000 000 000 000 000 187(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: 0.000 000 000 000 000 000 187.

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) 0.000 000 000 000 000 000 187 × 2 = 0 + 0.000 000 000 000 000 000 374;
  • 2) 0.000 000 000 000 000 000 374 × 2 = 0 + 0.000 000 000 000 000 000 748;
  • 3) 0.000 000 000 000 000 000 748 × 2 = 0 + 0.000 000 000 000 000 001 496;
  • 4) 0.000 000 000 000 000 001 496 × 2 = 0 + 0.000 000 000 000 000 002 992;
  • 5) 0.000 000 000 000 000 002 992 × 2 = 0 + 0.000 000 000 000 000 005 984;
  • 6) 0.000 000 000 000 000 005 984 × 2 = 0 + 0.000 000 000 000 000 011 968;
  • 7) 0.000 000 000 000 000 011 968 × 2 = 0 + 0.000 000 000 000 000 023 936;
  • 8) 0.000 000 000 000 000 023 936 × 2 = 0 + 0.000 000 000 000 000 047 872;
  • 9) 0.000 000 000 000 000 047 872 × 2 = 0 + 0.000 000 000 000 000 095 744;
  • 10) 0.000 000 000 000 000 095 744 × 2 = 0 + 0.000 000 000 000 000 191 488;
  • 11) 0.000 000 000 000 000 191 488 × 2 = 0 + 0.000 000 000 000 000 382 976;
  • 12) 0.000 000 000 000 000 382 976 × 2 = 0 + 0.000 000 000 000 000 765 952;
  • 13) 0.000 000 000 000 000 765 952 × 2 = 0 + 0.000 000 000 000 001 531 904;
  • 14) 0.000 000 000 000 001 531 904 × 2 = 0 + 0.000 000 000 000 003 063 808;
  • 15) 0.000 000 000 000 003 063 808 × 2 = 0 + 0.000 000 000 000 006 127 616;
  • 16) 0.000 000 000 000 006 127 616 × 2 = 0 + 0.000 000 000 000 012 255 232;
  • 17) 0.000 000 000 000 012 255 232 × 2 = 0 + 0.000 000 000 000 024 510 464;
  • 18) 0.000 000 000 000 024 510 464 × 2 = 0 + 0.000 000 000 000 049 020 928;
  • 19) 0.000 000 000 000 049 020 928 × 2 = 0 + 0.000 000 000 000 098 041 856;
  • 20) 0.000 000 000 000 098 041 856 × 2 = 0 + 0.000 000 000 000 196 083 712;
  • 21) 0.000 000 000 000 196 083 712 × 2 = 0 + 0.000 000 000 000 392 167 424;
  • 22) 0.000 000 000 000 392 167 424 × 2 = 0 + 0.000 000 000 000 784 334 848;
  • 23) 0.000 000 000 000 784 334 848 × 2 = 0 + 0.000 000 000 001 568 669 696;
  • 24) 0.000 000 000 001 568 669 696 × 2 = 0 + 0.000 000 000 003 137 339 392;
  • 25) 0.000 000 000 003 137 339 392 × 2 = 0 + 0.000 000 000 006 274 678 784;
  • 26) 0.000 000 000 006 274 678 784 × 2 = 0 + 0.000 000 000 012 549 357 568;
  • 27) 0.000 000 000 012 549 357 568 × 2 = 0 + 0.000 000 000 025 098 715 136;
  • 28) 0.000 000 000 025 098 715 136 × 2 = 0 + 0.000 000 000 050 197 430 272;
  • 29) 0.000 000 000 050 197 430 272 × 2 = 0 + 0.000 000 000 100 394 860 544;
  • 30) 0.000 000 000 100 394 860 544 × 2 = 0 + 0.000 000 000 200 789 721 088;
  • 31) 0.000 000 000 200 789 721 088 × 2 = 0 + 0.000 000 000 401 579 442 176;
  • 32) 0.000 000 000 401 579 442 176 × 2 = 0 + 0.000 000 000 803 158 884 352;
  • 33) 0.000 000 000 803 158 884 352 × 2 = 0 + 0.000 000 001 606 317 768 704;
  • 34) 0.000 000 001 606 317 768 704 × 2 = 0 + 0.000 000 003 212 635 537 408;
  • 35) 0.000 000 003 212 635 537 408 × 2 = 0 + 0.000 000 006 425 271 074 816;
  • 36) 0.000 000 006 425 271 074 816 × 2 = 0 + 0.000 000 012 850 542 149 632;
  • 37) 0.000 000 012 850 542 149 632 × 2 = 0 + 0.000 000 025 701 084 299 264;
  • 38) 0.000 000 025 701 084 299 264 × 2 = 0 + 0.000 000 051 402 168 598 528;
  • 39) 0.000 000 051 402 168 598 528 × 2 = 0 + 0.000 000 102 804 337 197 056;
  • 40) 0.000 000 102 804 337 197 056 × 2 = 0 + 0.000 000 205 608 674 394 112;
  • 41) 0.000 000 205 608 674 394 112 × 2 = 0 + 0.000 000 411 217 348 788 224;
  • 42) 0.000 000 411 217 348 788 224 × 2 = 0 + 0.000 000 822 434 697 576 448;
  • 43) 0.000 000 822 434 697 576 448 × 2 = 0 + 0.000 001 644 869 395 152 896;
  • 44) 0.000 001 644 869 395 152 896 × 2 = 0 + 0.000 003 289 738 790 305 792;
  • 45) 0.000 003 289 738 790 305 792 × 2 = 0 + 0.000 006 579 477 580 611 584;
  • 46) 0.000 006 579 477 580 611 584 × 2 = 0 + 0.000 013 158 955 161 223 168;
  • 47) 0.000 013 158 955 161 223 168 × 2 = 0 + 0.000 026 317 910 322 446 336;
  • 48) 0.000 026 317 910 322 446 336 × 2 = 0 + 0.000 052 635 820 644 892 672;
  • 49) 0.000 052 635 820 644 892 672 × 2 = 0 + 0.000 105 271 641 289 785 344;
  • 50) 0.000 105 271 641 289 785 344 × 2 = 0 + 0.000 210 543 282 579 570 688;
  • 51) 0.000 210 543 282 579 570 688 × 2 = 0 + 0.000 421 086 565 159 141 376;
  • 52) 0.000 421 086 565 159 141 376 × 2 = 0 + 0.000 842 173 130 318 282 752;
  • 53) 0.000 842 173 130 318 282 752 × 2 = 0 + 0.001 684 346 260 636 565 504;
  • 54) 0.001 684 346 260 636 565 504 × 2 = 0 + 0.003 368 692 521 273 131 008;
  • 55) 0.003 368 692 521 273 131 008 × 2 = 0 + 0.006 737 385 042 546 262 016;
  • 56) 0.006 737 385 042 546 262 016 × 2 = 0 + 0.013 474 770 085 092 524 032;
  • 57) 0.013 474 770 085 092 524 032 × 2 = 0 + 0.026 949 540 170 185 048 064;
  • 58) 0.026 949 540 170 185 048 064 × 2 = 0 + 0.053 899 080 340 370 096 128;
  • 59) 0.053 899 080 340 370 096 128 × 2 = 0 + 0.107 798 160 680 740 192 256;
  • 60) 0.107 798 160 680 740 192 256 × 2 = 0 + 0.215 596 321 361 480 384 512;
  • 61) 0.215 596 321 361 480 384 512 × 2 = 0 + 0.431 192 642 722 960 769 024;
  • 62) 0.431 192 642 722 960 769 024 × 2 = 0 + 0.862 385 285 445 921 538 048;
  • 63) 0.862 385 285 445 921 538 048 × 2 = 1 + 0.724 770 570 891 843 076 096;
  • 64) 0.724 770 570 891 843 076 096 × 2 = 1 + 0.449 541 141 783 686 152 192;
  • 65) 0.449 541 141 783 686 152 192 × 2 = 0 + 0.899 082 283 567 372 304 384;
  • 66) 0.899 082 283 567 372 304 384 × 2 = 1 + 0.798 164 567 134 744 608 768;
  • 67) 0.798 164 567 134 744 608 768 × 2 = 1 + 0.596 329 134 269 489 217 536;
  • 68) 0.596 329 134 269 489 217 536 × 2 = 1 + 0.192 658 268 538 978 435 072;
  • 69) 0.192 658 268 538 978 435 072 × 2 = 0 + 0.385 316 537 077 956 870 144;
  • 70) 0.385 316 537 077 956 870 144 × 2 = 0 + 0.770 633 074 155 913 740 288;
  • 71) 0.770 633 074 155 913 740 288 × 2 = 1 + 0.541 266 148 311 827 480 576;
  • 72) 0.541 266 148 311 827 480 576 × 2 = 1 + 0.082 532 296 623 654 961 152;
  • 73) 0.082 532 296 623 654 961 152 × 2 = 0 + 0.165 064 593 247 309 922 304;
  • 74) 0.165 064 593 247 309 922 304 × 2 = 0 + 0.330 129 186 494 619 844 608;
  • 75) 0.330 129 186 494 619 844 608 × 2 = 0 + 0.660 258 372 989 239 689 216;
  • 76) 0.660 258 372 989 239 689 216 × 2 = 1 + 0.320 516 745 978 479 378 432;
  • 77) 0.320 516 745 978 479 378 432 × 2 = 0 + 0.641 033 491 956 958 756 864;
  • 78) 0.641 033 491 956 958 756 864 × 2 = 1 + 0.282 066 983 913 917 513 728;
  • 79) 0.282 066 983 913 917 513 728 × 2 = 0 + 0.564 133 967 827 835 027 456;
  • 80) 0.564 133 967 827 835 027 456 × 2 = 1 + 0.128 267 935 655 670 054 912;
  • 81) 0.128 267 935 655 670 054 912 × 2 = 0 + 0.256 535 871 311 340 109 824;
  • 82) 0.256 535 871 311 340 109 824 × 2 = 0 + 0.513 071 742 622 680 219 648;
  • 83) 0.513 071 742 622 680 219 648 × 2 = 1 + 0.026 143 485 245 360 439 296;
  • 84) 0.026 143 485 245 360 439 296 × 2 = 0 + 0.052 286 970 490 720 878 592;
  • 85) 0.052 286 970 490 720 878 592 × 2 = 0 + 0.104 573 940 981 441 757 184;
  • 86) 0.104 573 940 981 441 757 184 × 2 = 0 + 0.209 147 881 962 883 514 368;
  • 87) 0.209 147 881 962 883 514 368 × 2 = 0 + 0.418 295 763 925 767 028 736;
  • 88) 0.418 295 763 925 767 028 736 × 2 = 0 + 0.836 591 527 851 534 057 472;
  • 89) 0.836 591 527 851 534 057 472 × 2 = 1 + 0.673 183 055 703 068 114 944;
  • 90) 0.673 183 055 703 068 114 944 × 2 = 1 + 0.346 366 111 406 136 229 888;
  • 91) 0.346 366 111 406 136 229 888 × 2 = 0 + 0.692 732 222 812 272 459 776;
  • 92) 0.692 732 222 812 272 459 776 × 2 = 1 + 0.385 464 445 624 544 919 552;
  • 93) 0.385 464 445 624 544 919 552 × 2 = 0 + 0.770 928 891 249 089 839 104;
  • 94) 0.770 928 891 249 089 839 104 × 2 = 1 + 0.541 857 782 498 179 678 208;
  • 95) 0.541 857 782 498 179 678 208 × 2 = 1 + 0.083 715 564 996 359 356 416;
  • 96) 0.083 715 564 996 359 356 416 × 2 = 0 + 0.167 431 129 992 718 712 832;
  • 97) 0.167 431 129 992 718 712 832 × 2 = 0 + 0.334 862 259 985 437 425 664;
  • 98) 0.334 862 259 985 437 425 664 × 2 = 0 + 0.669 724 519 970 874 851 328;
  • 99) 0.669 724 519 970 874 851 328 × 2 = 1 + 0.339 449 039 941 749 702 656;
  • 100) 0.339 449 039 941 749 702 656 × 2 = 0 + 0.678 898 079 883 499 405 312;
  • 101) 0.678 898 079 883 499 405 312 × 2 = 1 + 0.357 796 159 766 998 810 624;
  • 102) 0.357 796 159 766 998 810 624 × 2 = 0 + 0.715 592 319 533 997 621 248;
  • 103) 0.715 592 319 533 997 621 248 × 2 = 1 + 0.431 184 639 067 995 242 496;
  • 104) 0.431 184 639 067 995 242 496 × 2 = 0 + 0.862 369 278 135 990 484 992;
  • 105) 0.862 369 278 135 990 484 992 × 2 = 1 + 0.724 738 556 271 980 969 984;
  • 106) 0.724 738 556 271 980 969 984 × 2 = 1 + 0.449 477 112 543 961 939 968;
  • 107) 0.449 477 112 543 961 939 968 × 2 = 0 + 0.898 954 225 087 923 879 936;
  • 108) 0.898 954 225 087 923 879 936 × 2 = 1 + 0.797 908 450 175 847 759 872;
  • 109) 0.797 908 450 175 847 759 872 × 2 = 1 + 0.595 816 900 351 695 519 744;
  • 110) 0.595 816 900 351 695 519 744 × 2 = 1 + 0.191 633 800 703 391 039 488;
  • 111) 0.191 633 800 703 391 039 488 × 2 = 0 + 0.383 267 601 406 782 078 976;
  • 112) 0.383 267 601 406 782 078 976 × 2 = 0 + 0.766 535 202 813 564 157 952;
  • 113) 0.766 535 202 813 564 157 952 × 2 = 1 + 0.533 070 405 627 128 315 904;
  • 114) 0.533 070 405 627 128 315 904 × 2 = 1 + 0.066 140 811 254 256 631 808;
  • 115) 0.066 140 811 254 256 631 808 × 2 = 0 + 0.132 281 622 508 513 263 616;

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:


0.000 000 000 000 000 000 187(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0011 0001 0101 0010 0000 1101 0110 0010 1010 1101 1100 110(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 187(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0011 0001 0101 0010 0000 1101 0110 0010 1010 1101 1100 110(2)

6. 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:


0.000 000 000 000 000 000 187(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0011 0001 0101 0010 0000 1101 0110 0010 1010 1101 1100 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 0011 0001 0101 0010 0000 1101 0110 0010 1010 1101 1100 110(2) × 20 =

1.1011 1001 1000 1010 1001 0000 0110 1011 0001 0101 0110 1110 0110(2) × 2-63


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): -63

Mantissa (not normalized):
1.1011 1001 1000 1010 1001 0000 0110 1011 0001 0101 0110 1110 0110


8. 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)


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;
  • 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;

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) =

960(10) =

011 1100 0000(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. 1011 1001 1000 1010 1001 0000 0110 1011 0001 0101 0110 1110 0110 =

1011 1001 1000 1010 1001 0000 0110 1011 0001 0101 0110 1110 0110


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 1100 0000

Mantissa (52 bits) =
1011 1001 1000 1010 1001 0000 0110 1011 0001 0101 0110 1110 0110


Decimal number 0.000 000 000 000 000 000 187 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1100 0000 - 1011 1001 1000 1010 1001 0000 0110 1011 0001 0101 0110 1110 0110

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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