0.000 000 000 000 000 000 277 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 000 277(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
0.000 000 000 000 000 000 277(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: 0.000 000 000 000 000 000 277.

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) 0.000 000 000 000 000 000 277 × 2 = 0 + 0.000 000 000 000 000 000 554;
  • 2) 0.000 000 000 000 000 000 554 × 2 = 0 + 0.000 000 000 000 000 001 108;
  • 3) 0.000 000 000 000 000 001 108 × 2 = 0 + 0.000 000 000 000 000 002 216;
  • 4) 0.000 000 000 000 000 002 216 × 2 = 0 + 0.000 000 000 000 000 004 432;
  • 5) 0.000 000 000 000 000 004 432 × 2 = 0 + 0.000 000 000 000 000 008 864;
  • 6) 0.000 000 000 000 000 008 864 × 2 = 0 + 0.000 000 000 000 000 017 728;
  • 7) 0.000 000 000 000 000 017 728 × 2 = 0 + 0.000 000 000 000 000 035 456;
  • 8) 0.000 000 000 000 000 035 456 × 2 = 0 + 0.000 000 000 000 000 070 912;
  • 9) 0.000 000 000 000 000 070 912 × 2 = 0 + 0.000 000 000 000 000 141 824;
  • 10) 0.000 000 000 000 000 141 824 × 2 = 0 + 0.000 000 000 000 000 283 648;
  • 11) 0.000 000 000 000 000 283 648 × 2 = 0 + 0.000 000 000 000 000 567 296;
  • 12) 0.000 000 000 000 000 567 296 × 2 = 0 + 0.000 000 000 000 001 134 592;
  • 13) 0.000 000 000 000 001 134 592 × 2 = 0 + 0.000 000 000 000 002 269 184;
  • 14) 0.000 000 000 000 002 269 184 × 2 = 0 + 0.000 000 000 000 004 538 368;
  • 15) 0.000 000 000 000 004 538 368 × 2 = 0 + 0.000 000 000 000 009 076 736;
  • 16) 0.000 000 000 000 009 076 736 × 2 = 0 + 0.000 000 000 000 018 153 472;
  • 17) 0.000 000 000 000 018 153 472 × 2 = 0 + 0.000 000 000 000 036 306 944;
  • 18) 0.000 000 000 000 036 306 944 × 2 = 0 + 0.000 000 000 000 072 613 888;
  • 19) 0.000 000 000 000 072 613 888 × 2 = 0 + 0.000 000 000 000 145 227 776;
  • 20) 0.000 000 000 000 145 227 776 × 2 = 0 + 0.000 000 000 000 290 455 552;
  • 21) 0.000 000 000 000 290 455 552 × 2 = 0 + 0.000 000 000 000 580 911 104;
  • 22) 0.000 000 000 000 580 911 104 × 2 = 0 + 0.000 000 000 001 161 822 208;
  • 23) 0.000 000 000 001 161 822 208 × 2 = 0 + 0.000 000 000 002 323 644 416;
  • 24) 0.000 000 000 002 323 644 416 × 2 = 0 + 0.000 000 000 004 647 288 832;
  • 25) 0.000 000 000 004 647 288 832 × 2 = 0 + 0.000 000 000 009 294 577 664;
  • 26) 0.000 000 000 009 294 577 664 × 2 = 0 + 0.000 000 000 018 589 155 328;
  • 27) 0.000 000 000 018 589 155 328 × 2 = 0 + 0.000 000 000 037 178 310 656;
  • 28) 0.000 000 000 037 178 310 656 × 2 = 0 + 0.000 000 000 074 356 621 312;
  • 29) 0.000 000 000 074 356 621 312 × 2 = 0 + 0.000 000 000 148 713 242 624;
  • 30) 0.000 000 000 148 713 242 624 × 2 = 0 + 0.000 000 000 297 426 485 248;
  • 31) 0.000 000 000 297 426 485 248 × 2 = 0 + 0.000 000 000 594 852 970 496;
  • 32) 0.000 000 000 594 852 970 496 × 2 = 0 + 0.000 000 001 189 705 940 992;
  • 33) 0.000 000 001 189 705 940 992 × 2 = 0 + 0.000 000 002 379 411 881 984;
  • 34) 0.000 000 002 379 411 881 984 × 2 = 0 + 0.000 000 004 758 823 763 968;
  • 35) 0.000 000 004 758 823 763 968 × 2 = 0 + 0.000 000 009 517 647 527 936;
  • 36) 0.000 000 009 517 647 527 936 × 2 = 0 + 0.000 000 019 035 295 055 872;
  • 37) 0.000 000 019 035 295 055 872 × 2 = 0 + 0.000 000 038 070 590 111 744;
  • 38) 0.000 000 038 070 590 111 744 × 2 = 0 + 0.000 000 076 141 180 223 488;
  • 39) 0.000 000 076 141 180 223 488 × 2 = 0 + 0.000 000 152 282 360 446 976;
  • 40) 0.000 000 152 282 360 446 976 × 2 = 0 + 0.000 000 304 564 720 893 952;
  • 41) 0.000 000 304 564 720 893 952 × 2 = 0 + 0.000 000 609 129 441 787 904;
  • 42) 0.000 000 609 129 441 787 904 × 2 = 0 + 0.000 001 218 258 883 575 808;
  • 43) 0.000 001 218 258 883 575 808 × 2 = 0 + 0.000 002 436 517 767 151 616;
  • 44) 0.000 002 436 517 767 151 616 × 2 = 0 + 0.000 004 873 035 534 303 232;
  • 45) 0.000 004 873 035 534 303 232 × 2 = 0 + 0.000 009 746 071 068 606 464;
  • 46) 0.000 009 746 071 068 606 464 × 2 = 0 + 0.000 019 492 142 137 212 928;
  • 47) 0.000 019 492 142 137 212 928 × 2 = 0 + 0.000 038 984 284 274 425 856;
  • 48) 0.000 038 984 284 274 425 856 × 2 = 0 + 0.000 077 968 568 548 851 712;
  • 49) 0.000 077 968 568 548 851 712 × 2 = 0 + 0.000 155 937 137 097 703 424;
  • 50) 0.000 155 937 137 097 703 424 × 2 = 0 + 0.000 311 874 274 195 406 848;
  • 51) 0.000 311 874 274 195 406 848 × 2 = 0 + 0.000 623 748 548 390 813 696;
  • 52) 0.000 623 748 548 390 813 696 × 2 = 0 + 0.001 247 497 096 781 627 392;
  • 53) 0.001 247 497 096 781 627 392 × 2 = 0 + 0.002 494 994 193 563 254 784;
  • 54) 0.002 494 994 193 563 254 784 × 2 = 0 + 0.004 989 988 387 126 509 568;
  • 55) 0.004 989 988 387 126 509 568 × 2 = 0 + 0.009 979 976 774 253 019 136;
  • 56) 0.009 979 976 774 253 019 136 × 2 = 0 + 0.019 959 953 548 506 038 272;
  • 57) 0.019 959 953 548 506 038 272 × 2 = 0 + 0.039 919 907 097 012 076 544;
  • 58) 0.039 919 907 097 012 076 544 × 2 = 0 + 0.079 839 814 194 024 153 088;
  • 59) 0.079 839 814 194 024 153 088 × 2 = 0 + 0.159 679 628 388 048 306 176;
  • 60) 0.159 679 628 388 048 306 176 × 2 = 0 + 0.319 359 256 776 096 612 352;
  • 61) 0.319 359 256 776 096 612 352 × 2 = 0 + 0.638 718 513 552 193 224 704;
  • 62) 0.638 718 513 552 193 224 704 × 2 = 1 + 0.277 437 027 104 386 449 408;
  • 63) 0.277 437 027 104 386 449 408 × 2 = 0 + 0.554 874 054 208 772 898 816;
  • 64) 0.554 874 054 208 772 898 816 × 2 = 1 + 0.109 748 108 417 545 797 632;
  • 65) 0.109 748 108 417 545 797 632 × 2 = 0 + 0.219 496 216 835 091 595 264;
  • 66) 0.219 496 216 835 091 595 264 × 2 = 0 + 0.438 992 433 670 183 190 528;
  • 67) 0.438 992 433 670 183 190 528 × 2 = 0 + 0.877 984 867 340 366 381 056;
  • 68) 0.877 984 867 340 366 381 056 × 2 = 1 + 0.755 969 734 680 732 762 112;
  • 69) 0.755 969 734 680 732 762 112 × 2 = 1 + 0.511 939 469 361 465 524 224;
  • 70) 0.511 939 469 361 465 524 224 × 2 = 1 + 0.023 878 938 722 931 048 448;
  • 71) 0.023 878 938 722 931 048 448 × 2 = 0 + 0.047 757 877 445 862 096 896;
  • 72) 0.047 757 877 445 862 096 896 × 2 = 0 + 0.095 515 754 891 724 193 792;
  • 73) 0.095 515 754 891 724 193 792 × 2 = 0 + 0.191 031 509 783 448 387 584;
  • 74) 0.191 031 509 783 448 387 584 × 2 = 0 + 0.382 063 019 566 896 775 168;
  • 75) 0.382 063 019 566 896 775 168 × 2 = 0 + 0.764 126 039 133 793 550 336;
  • 76) 0.764 126 039 133 793 550 336 × 2 = 1 + 0.528 252 078 267 587 100 672;
  • 77) 0.528 252 078 267 587 100 672 × 2 = 1 + 0.056 504 156 535 174 201 344;
  • 78) 0.056 504 156 535 174 201 344 × 2 = 0 + 0.113 008 313 070 348 402 688;
  • 79) 0.113 008 313 070 348 402 688 × 2 = 0 + 0.226 016 626 140 696 805 376;
  • 80) 0.226 016 626 140 696 805 376 × 2 = 0 + 0.452 033 252 281 393 610 752;
  • 81) 0.452 033 252 281 393 610 752 × 2 = 0 + 0.904 066 504 562 787 221 504;
  • 82) 0.904 066 504 562 787 221 504 × 2 = 1 + 0.808 133 009 125 574 443 008;
  • 83) 0.808 133 009 125 574 443 008 × 2 = 1 + 0.616 266 018 251 148 886 016;
  • 84) 0.616 266 018 251 148 886 016 × 2 = 1 + 0.232 532 036 502 297 772 032;
  • 85) 0.232 532 036 502 297 772 032 × 2 = 0 + 0.465 064 073 004 595 544 064;
  • 86) 0.465 064 073 004 595 544 064 × 2 = 0 + 0.930 128 146 009 191 088 128;
  • 87) 0.930 128 146 009 191 088 128 × 2 = 1 + 0.860 256 292 018 382 176 256;
  • 88) 0.860 256 292 018 382 176 256 × 2 = 1 + 0.720 512 584 036 764 352 512;
  • 89) 0.720 512 584 036 764 352 512 × 2 = 1 + 0.441 025 168 073 528 705 024;
  • 90) 0.441 025 168 073 528 705 024 × 2 = 0 + 0.882 050 336 147 057 410 048;
  • 91) 0.882 050 336 147 057 410 048 × 2 = 1 + 0.764 100 672 294 114 820 096;
  • 92) 0.764 100 672 294 114 820 096 × 2 = 1 + 0.528 201 344 588 229 640 192;
  • 93) 0.528 201 344 588 229 640 192 × 2 = 1 + 0.056 402 689 176 459 280 384;
  • 94) 0.056 402 689 176 459 280 384 × 2 = 0 + 0.112 805 378 352 918 560 768;
  • 95) 0.112 805 378 352 918 560 768 × 2 = 0 + 0.225 610 756 705 837 121 536;
  • 96) 0.225 610 756 705 837 121 536 × 2 = 0 + 0.451 221 513 411 674 243 072;
  • 97) 0.451 221 513 411 674 243 072 × 2 = 0 + 0.902 443 026 823 348 486 144;
  • 98) 0.902 443 026 823 348 486 144 × 2 = 1 + 0.804 886 053 646 696 972 288;
  • 99) 0.804 886 053 646 696 972 288 × 2 = 1 + 0.609 772 107 293 393 944 576;
  • 100) 0.609 772 107 293 393 944 576 × 2 = 1 + 0.219 544 214 586 787 889 152;
  • 101) 0.219 544 214 586 787 889 152 × 2 = 0 + 0.439 088 429 173 575 778 304;
  • 102) 0.439 088 429 173 575 778 304 × 2 = 0 + 0.878 176 858 347 151 556 608;
  • 103) 0.878 176 858 347 151 556 608 × 2 = 1 + 0.756 353 716 694 303 113 216;
  • 104) 0.756 353 716 694 303 113 216 × 2 = 1 + 0.512 707 433 388 606 226 432;
  • 105) 0.512 707 433 388 606 226 432 × 2 = 1 + 0.025 414 866 777 212 452 864;
  • 106) 0.025 414 866 777 212 452 864 × 2 = 0 + 0.050 829 733 554 424 905 728;
  • 107) 0.050 829 733 554 424 905 728 × 2 = 0 + 0.101 659 467 108 849 811 456;
  • 108) 0.101 659 467 108 849 811 456 × 2 = 0 + 0.203 318 934 217 699 622 912;
  • 109) 0.203 318 934 217 699 622 912 × 2 = 0 + 0.406 637 868 435 399 245 824;
  • 110) 0.406 637 868 435 399 245 824 × 2 = 0 + 0.813 275 736 870 798 491 648;
  • 111) 0.813 275 736 870 798 491 648 × 2 = 1 + 0.626 551 473 741 596 983 296;
  • 112) 0.626 551 473 741 596 983 296 × 2 = 1 + 0.253 102 947 483 193 966 592;
  • 113) 0.253 102 947 483 193 966 592 × 2 = 0 + 0.506 205 894 966 387 933 184;
  • 114) 0.506 205 894 966 387 933 184 × 2 = 1 + 0.012 411 789 932 775 866 368;

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:


0.000 000 000 000 000 000 277(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0001 1100 0001 1000 0111 0011 1011 1000 0111 0011 1000 0011 01(2)

5. Positive number before normalization:

0.000 000 000 000 000 000 277(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0001 1100 0001 1000 0111 0011 1011 1000 0111 0011 1000 0011 01(2)

6. 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:


0.000 000 000 000 000 000 277(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0001 1100 0001 1000 0111 0011 1011 1000 0111 0011 1000 0011 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0001 1100 0001 1000 0111 0011 1011 1000 0111 0011 1000 0011 01(2) × 20 =

1.0100 0111 0000 0110 0001 1100 1110 1110 0001 1100 1110 0000 1101(2) × 2-62


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): -62

Mantissa (not normalized):
1.0100 0111 0000 0110 0001 1100 1110 1110 0001 1100 1110 0000 1101


8. 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)


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;
  • 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;

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) =

961(10) =

011 1100 0001(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. 0100 0111 0000 0110 0001 1100 1110 1110 0001 1100 1110 0000 1101 =

0100 0111 0000 0110 0001 1100 1110 1110 0001 1100 1110 0000 1101


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 0001

Mantissa (52 bits) =
0100 0111 0000 0110 0001 1100 1110 1110 0001 1100 1110 0000 1101


Decimal number 0.000 000 000 000 000 000 277 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1100 0001 - 0100 0111 0000 0110 0001 1100 1110 1110 0001 1100 1110 0000 1101

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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