-0.000 000 000 000 176 557 108 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 108 6(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
-0.000 000 000 000 176 557 108 6(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:

|-0.000 000 000 000 176 557 108 6| = 0.000 000 000 000 176 557 108 6


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: 0.000 000 000 000 176 557 108 6.

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) 0.000 000 000 000 176 557 108 6 × 2 = 0 + 0.000 000 000 000 353 114 217 2;
  • 2) 0.000 000 000 000 353 114 217 2 × 2 = 0 + 0.000 000 000 000 706 228 434 4;
  • 3) 0.000 000 000 000 706 228 434 4 × 2 = 0 + 0.000 000 000 001 412 456 868 8;
  • 4) 0.000 000 000 001 412 456 868 8 × 2 = 0 + 0.000 000 000 002 824 913 737 6;
  • 5) 0.000 000 000 002 824 913 737 6 × 2 = 0 + 0.000 000 000 005 649 827 475 2;
  • 6) 0.000 000 000 005 649 827 475 2 × 2 = 0 + 0.000 000 000 011 299 654 950 4;
  • 7) 0.000 000 000 011 299 654 950 4 × 2 = 0 + 0.000 000 000 022 599 309 900 8;
  • 8) 0.000 000 000 022 599 309 900 8 × 2 = 0 + 0.000 000 000 045 198 619 801 6;
  • 9) 0.000 000 000 045 198 619 801 6 × 2 = 0 + 0.000 000 000 090 397 239 603 2;
  • 10) 0.000 000 000 090 397 239 603 2 × 2 = 0 + 0.000 000 000 180 794 479 206 4;
  • 11) 0.000 000 000 180 794 479 206 4 × 2 = 0 + 0.000 000 000 361 588 958 412 8;
  • 12) 0.000 000 000 361 588 958 412 8 × 2 = 0 + 0.000 000 000 723 177 916 825 6;
  • 13) 0.000 000 000 723 177 916 825 6 × 2 = 0 + 0.000 000 001 446 355 833 651 2;
  • 14) 0.000 000 001 446 355 833 651 2 × 2 = 0 + 0.000 000 002 892 711 667 302 4;
  • 15) 0.000 000 002 892 711 667 302 4 × 2 = 0 + 0.000 000 005 785 423 334 604 8;
  • 16) 0.000 000 005 785 423 334 604 8 × 2 = 0 + 0.000 000 011 570 846 669 209 6;
  • 17) 0.000 000 011 570 846 669 209 6 × 2 = 0 + 0.000 000 023 141 693 338 419 2;
  • 18) 0.000 000 023 141 693 338 419 2 × 2 = 0 + 0.000 000 046 283 386 676 838 4;
  • 19) 0.000 000 046 283 386 676 838 4 × 2 = 0 + 0.000 000 092 566 773 353 676 8;
  • 20) 0.000 000 092 566 773 353 676 8 × 2 = 0 + 0.000 000 185 133 546 707 353 6;
  • 21) 0.000 000 185 133 546 707 353 6 × 2 = 0 + 0.000 000 370 267 093 414 707 2;
  • 22) 0.000 000 370 267 093 414 707 2 × 2 = 0 + 0.000 000 740 534 186 829 414 4;
  • 23) 0.000 000 740 534 186 829 414 4 × 2 = 0 + 0.000 001 481 068 373 658 828 8;
  • 24) 0.000 001 481 068 373 658 828 8 × 2 = 0 + 0.000 002 962 136 747 317 657 6;
  • 25) 0.000 002 962 136 747 317 657 6 × 2 = 0 + 0.000 005 924 273 494 635 315 2;
  • 26) 0.000 005 924 273 494 635 315 2 × 2 = 0 + 0.000 011 848 546 989 270 630 4;
  • 27) 0.000 011 848 546 989 270 630 4 × 2 = 0 + 0.000 023 697 093 978 541 260 8;
  • 28) 0.000 023 697 093 978 541 260 8 × 2 = 0 + 0.000 047 394 187 957 082 521 6;
  • 29) 0.000 047 394 187 957 082 521 6 × 2 = 0 + 0.000 094 788 375 914 165 043 2;
  • 30) 0.000 094 788 375 914 165 043 2 × 2 = 0 + 0.000 189 576 751 828 330 086 4;
  • 31) 0.000 189 576 751 828 330 086 4 × 2 = 0 + 0.000 379 153 503 656 660 172 8;
  • 32) 0.000 379 153 503 656 660 172 8 × 2 = 0 + 0.000 758 307 007 313 320 345 6;
  • 33) 0.000 758 307 007 313 320 345 6 × 2 = 0 + 0.001 516 614 014 626 640 691 2;
  • 34) 0.001 516 614 014 626 640 691 2 × 2 = 0 + 0.003 033 228 029 253 281 382 4;
  • 35) 0.003 033 228 029 253 281 382 4 × 2 = 0 + 0.006 066 456 058 506 562 764 8;
  • 36) 0.006 066 456 058 506 562 764 8 × 2 = 0 + 0.012 132 912 117 013 125 529 6;
  • 37) 0.012 132 912 117 013 125 529 6 × 2 = 0 + 0.024 265 824 234 026 251 059 2;
  • 38) 0.024 265 824 234 026 251 059 2 × 2 = 0 + 0.048 531 648 468 052 502 118 4;
  • 39) 0.048 531 648 468 052 502 118 4 × 2 = 0 + 0.097 063 296 936 105 004 236 8;
  • 40) 0.097 063 296 936 105 004 236 8 × 2 = 0 + 0.194 126 593 872 210 008 473 6;
  • 41) 0.194 126 593 872 210 008 473 6 × 2 = 0 + 0.388 253 187 744 420 016 947 2;
  • 42) 0.388 253 187 744 420 016 947 2 × 2 = 0 + 0.776 506 375 488 840 033 894 4;
  • 43) 0.776 506 375 488 840 033 894 4 × 2 = 1 + 0.553 012 750 977 680 067 788 8;
  • 44) 0.553 012 750 977 680 067 788 8 × 2 = 1 + 0.106 025 501 955 360 135 577 6;
  • 45) 0.106 025 501 955 360 135 577 6 × 2 = 0 + 0.212 051 003 910 720 271 155 2;
  • 46) 0.212 051 003 910 720 271 155 2 × 2 = 0 + 0.424 102 007 821 440 542 310 4;
  • 47) 0.424 102 007 821 440 542 310 4 × 2 = 0 + 0.848 204 015 642 881 084 620 8;
  • 48) 0.848 204 015 642 881 084 620 8 × 2 = 1 + 0.696 408 031 285 762 169 241 6;
  • 49) 0.696 408 031 285 762 169 241 6 × 2 = 1 + 0.392 816 062 571 524 338 483 2;
  • 50) 0.392 816 062 571 524 338 483 2 × 2 = 0 + 0.785 632 125 143 048 676 966 4;
  • 51) 0.785 632 125 143 048 676 966 4 × 2 = 1 + 0.571 264 250 286 097 353 932 8;
  • 52) 0.571 264 250 286 097 353 932 8 × 2 = 1 + 0.142 528 500 572 194 707 865 6;
  • 53) 0.142 528 500 572 194 707 865 6 × 2 = 0 + 0.285 057 001 144 389 415 731 2;
  • 54) 0.285 057 001 144 389 415 731 2 × 2 = 0 + 0.570 114 002 288 778 831 462 4;
  • 55) 0.570 114 002 288 778 831 462 4 × 2 = 1 + 0.140 228 004 577 557 662 924 8;
  • 56) 0.140 228 004 577 557 662 924 8 × 2 = 0 + 0.280 456 009 155 115 325 849 6;
  • 57) 0.280 456 009 155 115 325 849 6 × 2 = 0 + 0.560 912 018 310 230 651 699 2;
  • 58) 0.560 912 018 310 230 651 699 2 × 2 = 1 + 0.121 824 036 620 461 303 398 4;
  • 59) 0.121 824 036 620 461 303 398 4 × 2 = 0 + 0.243 648 073 240 922 606 796 8;
  • 60) 0.243 648 073 240 922 606 796 8 × 2 = 0 + 0.487 296 146 481 845 213 593 6;
  • 61) 0.487 296 146 481 845 213 593 6 × 2 = 0 + 0.974 592 292 963 690 427 187 2;
  • 62) 0.974 592 292 963 690 427 187 2 × 2 = 1 + 0.949 184 585 927 380 854 374 4;
  • 63) 0.949 184 585 927 380 854 374 4 × 2 = 1 + 0.898 369 171 854 761 708 748 8;
  • 64) 0.898 369 171 854 761 708 748 8 × 2 = 1 + 0.796 738 343 709 523 417 497 6;
  • 65) 0.796 738 343 709 523 417 497 6 × 2 = 1 + 0.593 476 687 419 046 834 995 2;
  • 66) 0.593 476 687 419 046 834 995 2 × 2 = 1 + 0.186 953 374 838 093 669 990 4;
  • 67) 0.186 953 374 838 093 669 990 4 × 2 = 0 + 0.373 906 749 676 187 339 980 8;
  • 68) 0.373 906 749 676 187 339 980 8 × 2 = 0 + 0.747 813 499 352 374 679 961 6;
  • 69) 0.747 813 499 352 374 679 961 6 × 2 = 1 + 0.495 626 998 704 749 359 923 2;
  • 70) 0.495 626 998 704 749 359 923 2 × 2 = 0 + 0.991 253 997 409 498 719 846 4;
  • 71) 0.991 253 997 409 498 719 846 4 × 2 = 1 + 0.982 507 994 818 997 439 692 8;
  • 72) 0.982 507 994 818 997 439 692 8 × 2 = 1 + 0.965 015 989 637 994 879 385 6;
  • 73) 0.965 015 989 637 994 879 385 6 × 2 = 1 + 0.930 031 979 275 989 758 771 2;
  • 74) 0.930 031 979 275 989 758 771 2 × 2 = 1 + 0.860 063 958 551 979 517 542 4;
  • 75) 0.860 063 958 551 979 517 542 4 × 2 = 1 + 0.720 127 917 103 959 035 084 8;
  • 76) 0.720 127 917 103 959 035 084 8 × 2 = 1 + 0.440 255 834 207 918 070 169 6;
  • 77) 0.440 255 834 207 918 070 169 6 × 2 = 0 + 0.880 511 668 415 836 140 339 2;
  • 78) 0.880 511 668 415 836 140 339 2 × 2 = 1 + 0.761 023 336 831 672 280 678 4;
  • 79) 0.761 023 336 831 672 280 678 4 × 2 = 1 + 0.522 046 673 663 344 561 356 8;
  • 80) 0.522 046 673 663 344 561 356 8 × 2 = 1 + 0.044 093 347 326 689 122 713 6;
  • 81) 0.044 093 347 326 689 122 713 6 × 2 = 0 + 0.088 186 694 653 378 245 427 2;
  • 82) 0.088 186 694 653 378 245 427 2 × 2 = 0 + 0.176 373 389 306 756 490 854 4;
  • 83) 0.176 373 389 306 756 490 854 4 × 2 = 0 + 0.352 746 778 613 512 981 708 8;
  • 84) 0.352 746 778 613 512 981 708 8 × 2 = 0 + 0.705 493 557 227 025 963 417 6;
  • 85) 0.705 493 557 227 025 963 417 6 × 2 = 1 + 0.410 987 114 454 051 926 835 2;
  • 86) 0.410 987 114 454 051 926 835 2 × 2 = 0 + 0.821 974 228 908 103 853 670 4;
  • 87) 0.821 974 228 908 103 853 670 4 × 2 = 1 + 0.643 948 457 816 207 707 340 8;
  • 88) 0.643 948 457 816 207 707 340 8 × 2 = 1 + 0.287 896 915 632 415 414 681 6;
  • 89) 0.287 896 915 632 415 414 681 6 × 2 = 0 + 0.575 793 831 264 830 829 363 2;
  • 90) 0.575 793 831 264 830 829 363 2 × 2 = 1 + 0.151 587 662 529 661 658 726 4;
  • 91) 0.151 587 662 529 661 658 726 4 × 2 = 0 + 0.303 175 325 059 323 317 452 8;
  • 92) 0.303 175 325 059 323 317 452 8 × 2 = 0 + 0.606 350 650 118 646 634 905 6;
  • 93) 0.606 350 650 118 646 634 905 6 × 2 = 1 + 0.212 701 300 237 293 269 811 2;
  • 94) 0.212 701 300 237 293 269 811 2 × 2 = 0 + 0.425 402 600 474 586 539 622 4;
  • 95) 0.425 402 600 474 586 539 622 4 × 2 = 0 + 0.850 805 200 949 173 079 244 8;

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:


0.000 000 000 000 176 557 108 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0111 1100 1011 1111 0111 0000 1011 0100 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 108 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0111 1100 1011 1111 0111 0000 1011 0100 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 557 108 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0111 1100 1011 1111 0111 0000 1011 0100 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0111 1100 1011 1111 0111 0000 1011 0100 100(2) × 20 =

1.1000 1101 1001 0010 0011 1110 0101 1111 1011 1000 0101 1010 0100(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 1001 0010 0011 1110 0101 1111 1011 1000 0101 1010 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 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;

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) =

980(10) =

011 1101 0100(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. 1000 1101 1001 0010 0011 1110 0101 1111 1011 1000 0101 1010 0100 =

1000 1101 1001 0010 0011 1110 0101 1111 1011 1000 0101 1010 0100


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 1101 0100

Mantissa (52 bits) =
1000 1101 1001 0010 0011 1110 0101 1111 1011 1000 0101 1010 0100


Decimal number -0.000 000 000 000 176 557 108 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0011 1110 0101 1111 1011 1000 0101 1010 0100

» Convert decimal -0.000 000 000 000 176 557 109 4 to 64 bit double precision IEEE 754 binary floating point representation standard

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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