Convert the Number 0.000 000 000 000 755 840 011 110 107 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number. Detailed Explanations
Number 0.000 000 000 000 755 840 011 110 107(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)
The first steps we'll go through to make the conversion:
Convert to binary (to base 2) the integer part of the number.
Convert to binary the fractional part of the number.
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: 0.000 000 000 000 755 840 011 110 107.
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) 0.000 000 000 000 755 840 011 110 107 × 2 = 0 + 0.000 000 000 001 511 680 022 220 214;
- 2) 0.000 000 000 001 511 680 022 220 214 × 2 = 0 + 0.000 000 000 003 023 360 044 440 428;
- 3) 0.000 000 000 003 023 360 044 440 428 × 2 = 0 + 0.000 000 000 006 046 720 088 880 856;
- 4) 0.000 000 000 006 046 720 088 880 856 × 2 = 0 + 0.000 000 000 012 093 440 177 761 712;
- 5) 0.000 000 000 012 093 440 177 761 712 × 2 = 0 + 0.000 000 000 024 186 880 355 523 424;
- 6) 0.000 000 000 024 186 880 355 523 424 × 2 = 0 + 0.000 000 000 048 373 760 711 046 848;
- 7) 0.000 000 000 048 373 760 711 046 848 × 2 = 0 + 0.000 000 000 096 747 521 422 093 696;
- 8) 0.000 000 000 096 747 521 422 093 696 × 2 = 0 + 0.000 000 000 193 495 042 844 187 392;
- 9) 0.000 000 000 193 495 042 844 187 392 × 2 = 0 + 0.000 000 000 386 990 085 688 374 784;
- 10) 0.000 000 000 386 990 085 688 374 784 × 2 = 0 + 0.000 000 000 773 980 171 376 749 568;
- 11) 0.000 000 000 773 980 171 376 749 568 × 2 = 0 + 0.000 000 001 547 960 342 753 499 136;
- 12) 0.000 000 001 547 960 342 753 499 136 × 2 = 0 + 0.000 000 003 095 920 685 506 998 272;
- 13) 0.000 000 003 095 920 685 506 998 272 × 2 = 0 + 0.000 000 006 191 841 371 013 996 544;
- 14) 0.000 000 006 191 841 371 013 996 544 × 2 = 0 + 0.000 000 012 383 682 742 027 993 088;
- 15) 0.000 000 012 383 682 742 027 993 088 × 2 = 0 + 0.000 000 024 767 365 484 055 986 176;
- 16) 0.000 000 024 767 365 484 055 986 176 × 2 = 0 + 0.000 000 049 534 730 968 111 972 352;
- 17) 0.000 000 049 534 730 968 111 972 352 × 2 = 0 + 0.000 000 099 069 461 936 223 944 704;
- 18) 0.000 000 099 069 461 936 223 944 704 × 2 = 0 + 0.000 000 198 138 923 872 447 889 408;
- 19) 0.000 000 198 138 923 872 447 889 408 × 2 = 0 + 0.000 000 396 277 847 744 895 778 816;
- 20) 0.000 000 396 277 847 744 895 778 816 × 2 = 0 + 0.000 000 792 555 695 489 791 557 632;
- 21) 0.000 000 792 555 695 489 791 557 632 × 2 = 0 + 0.000 001 585 111 390 979 583 115 264;
- 22) 0.000 001 585 111 390 979 583 115 264 × 2 = 0 + 0.000 003 170 222 781 959 166 230 528;
- 23) 0.000 003 170 222 781 959 166 230 528 × 2 = 0 + 0.000 006 340 445 563 918 332 461 056;
- 24) 0.000 006 340 445 563 918 332 461 056 × 2 = 0 + 0.000 012 680 891 127 836 664 922 112;
- 25) 0.000 012 680 891 127 836 664 922 112 × 2 = 0 + 0.000 025 361 782 255 673 329 844 224;
- 26) 0.000 025 361 782 255 673 329 844 224 × 2 = 0 + 0.000 050 723 564 511 346 659 688 448;
- 27) 0.000 050 723 564 511 346 659 688 448 × 2 = 0 + 0.000 101 447 129 022 693 319 376 896;
- 28) 0.000 101 447 129 022 693 319 376 896 × 2 = 0 + 0.000 202 894 258 045 386 638 753 792;
- 29) 0.000 202 894 258 045 386 638 753 792 × 2 = 0 + 0.000 405 788 516 090 773 277 507 584;
- 30) 0.000 405 788 516 090 773 277 507 584 × 2 = 0 + 0.000 811 577 032 181 546 555 015 168;
- 31) 0.000 811 577 032 181 546 555 015 168 × 2 = 0 + 0.001 623 154 064 363 093 110 030 336;
- 32) 0.001 623 154 064 363 093 110 030 336 × 2 = 0 + 0.003 246 308 128 726 186 220 060 672;
- 33) 0.003 246 308 128 726 186 220 060 672 × 2 = 0 + 0.006 492 616 257 452 372 440 121 344;
- 34) 0.006 492 616 257 452 372 440 121 344 × 2 = 0 + 0.012 985 232 514 904 744 880 242 688;
- 35) 0.012 985 232 514 904 744 880 242 688 × 2 = 0 + 0.025 970 465 029 809 489 760 485 376;
- 36) 0.025 970 465 029 809 489 760 485 376 × 2 = 0 + 0.051 940 930 059 618 979 520 970 752;
- 37) 0.051 940 930 059 618 979 520 970 752 × 2 = 0 + 0.103 881 860 119 237 959 041 941 504;
- 38) 0.103 881 860 119 237 959 041 941 504 × 2 = 0 + 0.207 763 720 238 475 918 083 883 008;
- 39) 0.207 763 720 238 475 918 083 883 008 × 2 = 0 + 0.415 527 440 476 951 836 167 766 016;
- 40) 0.415 527 440 476 951 836 167 766 016 × 2 = 0 + 0.831 054 880 953 903 672 335 532 032;
- 41) 0.831 054 880 953 903 672 335 532 032 × 2 = 1 + 0.662 109 761 907 807 344 671 064 064;
- 42) 0.662 109 761 907 807 344 671 064 064 × 2 = 1 + 0.324 219 523 815 614 689 342 128 128;
- 43) 0.324 219 523 815 614 689 342 128 128 × 2 = 0 + 0.648 439 047 631 229 378 684 256 256;
- 44) 0.648 439 047 631 229 378 684 256 256 × 2 = 1 + 0.296 878 095 262 458 757 368 512 512;
- 45) 0.296 878 095 262 458 757 368 512 512 × 2 = 0 + 0.593 756 190 524 917 514 737 025 024;
- 46) 0.593 756 190 524 917 514 737 025 024 × 2 = 1 + 0.187 512 381 049 835 029 474 050 048;
- 47) 0.187 512 381 049 835 029 474 050 048 × 2 = 0 + 0.375 024 762 099 670 058 948 100 096;
- 48) 0.375 024 762 099 670 058 948 100 096 × 2 = 0 + 0.750 049 524 199 340 117 896 200 192;
- 49) 0.750 049 524 199 340 117 896 200 192 × 2 = 1 + 0.500 099 048 398 680 235 792 400 384;
- 50) 0.500 099 048 398 680 235 792 400 384 × 2 = 1 + 0.000 198 096 797 360 471 584 800 768;
- 51) 0.000 198 096 797 360 471 584 800 768 × 2 = 0 + 0.000 396 193 594 720 943 169 601 536;
- 52) 0.000 396 193 594 720 943 169 601 536 × 2 = 0 + 0.000 792 387 189 441 886 339 203 072;
- 53) 0.000 792 387 189 441 886 339 203 072 × 2 = 0 + 0.001 584 774 378 883 772 678 406 144;
- 54) 0.001 584 774 378 883 772 678 406 144 × 2 = 0 + 0.003 169 548 757 767 545 356 812 288;
- 55) 0.003 169 548 757 767 545 356 812 288 × 2 = 0 + 0.006 339 097 515 535 090 713 624 576;
- 56) 0.006 339 097 515 535 090 713 624 576 × 2 = 0 + 0.012 678 195 031 070 181 427 249 152;
- 57) 0.012 678 195 031 070 181 427 249 152 × 2 = 0 + 0.025 356 390 062 140 362 854 498 304;
- 58) 0.025 356 390 062 140 362 854 498 304 × 2 = 0 + 0.050 712 780 124 280 725 708 996 608;
- 59) 0.050 712 780 124 280 725 708 996 608 × 2 = 0 + 0.101 425 560 248 561 451 417 993 216;
- 60) 0.101 425 560 248 561 451 417 993 216 × 2 = 0 + 0.202 851 120 497 122 902 835 986 432;
- 61) 0.202 851 120 497 122 902 835 986 432 × 2 = 0 + 0.405 702 240 994 245 805 671 972 864;
- 62) 0.405 702 240 994 245 805 671 972 864 × 2 = 0 + 0.811 404 481 988 491 611 343 945 728;
- 63) 0.811 404 481 988 491 611 343 945 728 × 2 = 1 + 0.622 808 963 976 983 222 687 891 456;
- 64) 0.622 808 963 976 983 222 687 891 456 × 2 = 1 + 0.245 617 927 953 966 445 375 782 912;
- 65) 0.245 617 927 953 966 445 375 782 912 × 2 = 0 + 0.491 235 855 907 932 890 751 565 824;
- 66) 0.491 235 855 907 932 890 751 565 824 × 2 = 0 + 0.982 471 711 815 865 781 503 131 648;
- 67) 0.982 471 711 815 865 781 503 131 648 × 2 = 1 + 0.964 943 423 631 731 563 006 263 296;
- 68) 0.964 943 423 631 731 563 006 263 296 × 2 = 1 + 0.929 886 847 263 463 126 012 526 592;
- 69) 0.929 886 847 263 463 126 012 526 592 × 2 = 1 + 0.859 773 694 526 926 252 025 053 184;
- 70) 0.859 773 694 526 926 252 025 053 184 × 2 = 1 + 0.719 547 389 053 852 504 050 106 368;
- 71) 0.719 547 389 053 852 504 050 106 368 × 2 = 1 + 0.439 094 778 107 705 008 100 212 736;
- 72) 0.439 094 778 107 705 008 100 212 736 × 2 = 0 + 0.878 189 556 215 410 016 200 425 472;
- 73) 0.878 189 556 215 410 016 200 425 472 × 2 = 1 + 0.756 379 112 430 820 032 400 850 944;
- 74) 0.756 379 112 430 820 032 400 850 944 × 2 = 1 + 0.512 758 224 861 640 064 801 701 888;
- 75) 0.512 758 224 861 640 064 801 701 888 × 2 = 1 + 0.025 516 449 723 280 129 603 403 776;
- 76) 0.025 516 449 723 280 129 603 403 776 × 2 = 0 + 0.051 032 899 446 560 259 206 807 552;
- 77) 0.051 032 899 446 560 259 206 807 552 × 2 = 0 + 0.102 065 798 893 120 518 413 615 104;
- 78) 0.102 065 798 893 120 518 413 615 104 × 2 = 0 + 0.204 131 597 786 241 036 827 230 208;
- 79) 0.204 131 597 786 241 036 827 230 208 × 2 = 0 + 0.408 263 195 572 482 073 654 460 416;
- 80) 0.408 263 195 572 482 073 654 460 416 × 2 = 0 + 0.816 526 391 144 964 147 308 920 832;
- 81) 0.816 526 391 144 964 147 308 920 832 × 2 = 1 + 0.633 052 782 289 928 294 617 841 664;
- 82) 0.633 052 782 289 928 294 617 841 664 × 2 = 1 + 0.266 105 564 579 856 589 235 683 328;
- 83) 0.266 105 564 579 856 589 235 683 328 × 2 = 0 + 0.532 211 129 159 713 178 471 366 656;
- 84) 0.532 211 129 159 713 178 471 366 656 × 2 = 1 + 0.064 422 258 319 426 356 942 733 312;
- 85) 0.064 422 258 319 426 356 942 733 312 × 2 = 0 + 0.128 844 516 638 852 713 885 466 624;
- 86) 0.128 844 516 638 852 713 885 466 624 × 2 = 0 + 0.257 689 033 277 705 427 770 933 248;
- 87) 0.257 689 033 277 705 427 770 933 248 × 2 = 0 + 0.515 378 066 555 410 855 541 866 496;
- 88) 0.515 378 066 555 410 855 541 866 496 × 2 = 1 + 0.030 756 133 110 821 711 083 732 992;
- 89) 0.030 756 133 110 821 711 083 732 992 × 2 = 0 + 0.061 512 266 221 643 422 167 465 984;
- 90) 0.061 512 266 221 643 422 167 465 984 × 2 = 0 + 0.123 024 532 443 286 844 334 931 968;
- 91) 0.123 024 532 443 286 844 334 931 968 × 2 = 0 + 0.246 049 064 886 573 688 669 863 936;
- 92) 0.246 049 064 886 573 688 669 863 936 × 2 = 0 + 0.492 098 129 773 147 377 339 727 872;
- 93) 0.492 098 129 773 147 377 339 727 872 × 2 = 0 + 0.984 196 259 546 294 754 679 455 744;
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...)
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:
0.000 000 000 000 755 840 011 110 107(10) =
0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1101 0100 1100 0000 0000 0011 0011 1110 1110 0000 1101 0001 0000 0(2)
5. Positive number before normalization:
0.000 000 000 000 755 840 011 110 107(10) =
0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1101 0100 1100 0000 0000 0011 0011 1110 1110 0000 1101 0001 0000 0(2)
The last steps we'll go through to make the conversion:
Normalize the binary representation of the number.
Adjust the exponent.
Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.
Normalize the mantissa.
6. Normalize the binary representation of the number.
Shift the decimal mark 41 positions to the right, so that only one non zero digit remains to the left of it:
0.000 000 000 000 755 840 011 110 107(10) =
0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1101 0100 1100 0000 0000 0011 0011 1110 1110 0000 1101 0001 0000 0(2) =
0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1101 0100 1100 0000 0000 0011 0011 1110 1110 0000 1101 0001 0000 0(2) × 20 =
1.1010 1001 1000 0000 0000 0110 0111 1101 1100 0001 1010 0010 0000(2) × 2-41
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): -41
Mantissa (not normalized):
1.1010 1001 1000 0000 0000 0110 0111 1101 1100 0001 1010 0010 0000
8. Adjust the exponent.
Use the 11 bit excess/bias notation:
Exponent (adjusted) =
Exponent (unadjusted) + 2(11-1) - 1 =
-41 + 2(11-1) - 1 =
(-41 + 1 023)(10) =
982(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;
- 982 ÷ 2 = 491 + 0;
- 491 ÷ 2 = 245 + 1;
- 245 ÷ 2 = 122 + 1;
- 122 ÷ 2 = 61 + 0;
- 61 ÷ 2 = 30 + 1;
- 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) =
982(10) =
011 1101 0110(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. 1010 1001 1000 0000 0000 0110 0111 1101 1100 0001 1010 0010 0000 =
1010 1001 1000 0000 0000 0110 0111 1101 1100 0001 1010 0010 0000
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 1101 0110
Mantissa (52 bits) =
1010 1001 1000 0000 0000 0110 0111 1101 1100 0001 1010 0010 0000
The base ten decimal number 0.000 000 000 000 755 840 011 110 107 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1101 0110 - 1010 1001 1000 0000 0000 0110 0111 1101 1100 0001 1010 0010 0000
Sign (1 bit):
Exponent (11 bits):
0
621
611
601
591
580
571
560
551
541
530
52
Mantissa (52 bits):
1
510
501
490
481
470
460
451
441
430
420
410
400
390
380
370
360
350
340
330
320
311
301
290
280
271
261
251
241
231
220
211
201
191
180
170
160
150
140
131
121
110
101
90
80
70
61
50
40
30
20
10
0
Convert to 64 bit double precision IEEE 754 binary floating point representation standard
A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)
The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation
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
Available Base Conversions Between Decimal and Binary Systems
Conversions Between Decimal System Numbers (Written in Base Ten) and Binary System Numbers (Base Two and Computer Representation):
1. Integer -> Binary
2. Decimal -> Binary
3. Binary -> Integer
4. Binary -> Decimal