-0.000 000 000 000 014 350 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 350 7(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
-0.000 000 000 000 014 350 7(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:

|-0.000 000 000 000 014 350 7| = 0.000 000 000 000 014 350 7


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: 0.000 000 000 000 014 350 7.

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) 0.000 000 000 000 014 350 7 × 2 = 0 + 0.000 000 000 000 028 701 4;
  • 2) 0.000 000 000 000 028 701 4 × 2 = 0 + 0.000 000 000 000 057 402 8;
  • 3) 0.000 000 000 000 057 402 8 × 2 = 0 + 0.000 000 000 000 114 805 6;
  • 4) 0.000 000 000 000 114 805 6 × 2 = 0 + 0.000 000 000 000 229 611 2;
  • 5) 0.000 000 000 000 229 611 2 × 2 = 0 + 0.000 000 000 000 459 222 4;
  • 6) 0.000 000 000 000 459 222 4 × 2 = 0 + 0.000 000 000 000 918 444 8;
  • 7) 0.000 000 000 000 918 444 8 × 2 = 0 + 0.000 000 000 001 836 889 6;
  • 8) 0.000 000 000 001 836 889 6 × 2 = 0 + 0.000 000 000 003 673 779 2;
  • 9) 0.000 000 000 003 673 779 2 × 2 = 0 + 0.000 000 000 007 347 558 4;
  • 10) 0.000 000 000 007 347 558 4 × 2 = 0 + 0.000 000 000 014 695 116 8;
  • 11) 0.000 000 000 014 695 116 8 × 2 = 0 + 0.000 000 000 029 390 233 6;
  • 12) 0.000 000 000 029 390 233 6 × 2 = 0 + 0.000 000 000 058 780 467 2;
  • 13) 0.000 000 000 058 780 467 2 × 2 = 0 + 0.000 000 000 117 560 934 4;
  • 14) 0.000 000 000 117 560 934 4 × 2 = 0 + 0.000 000 000 235 121 868 8;
  • 15) 0.000 000 000 235 121 868 8 × 2 = 0 + 0.000 000 000 470 243 737 6;
  • 16) 0.000 000 000 470 243 737 6 × 2 = 0 + 0.000 000 000 940 487 475 2;
  • 17) 0.000 000 000 940 487 475 2 × 2 = 0 + 0.000 000 001 880 974 950 4;
  • 18) 0.000 000 001 880 974 950 4 × 2 = 0 + 0.000 000 003 761 949 900 8;
  • 19) 0.000 000 003 761 949 900 8 × 2 = 0 + 0.000 000 007 523 899 801 6;
  • 20) 0.000 000 007 523 899 801 6 × 2 = 0 + 0.000 000 015 047 799 603 2;
  • 21) 0.000 000 015 047 799 603 2 × 2 = 0 + 0.000 000 030 095 599 206 4;
  • 22) 0.000 000 030 095 599 206 4 × 2 = 0 + 0.000 000 060 191 198 412 8;
  • 23) 0.000 000 060 191 198 412 8 × 2 = 0 + 0.000 000 120 382 396 825 6;
  • 24) 0.000 000 120 382 396 825 6 × 2 = 0 + 0.000 000 240 764 793 651 2;
  • 25) 0.000 000 240 764 793 651 2 × 2 = 0 + 0.000 000 481 529 587 302 4;
  • 26) 0.000 000 481 529 587 302 4 × 2 = 0 + 0.000 000 963 059 174 604 8;
  • 27) 0.000 000 963 059 174 604 8 × 2 = 0 + 0.000 001 926 118 349 209 6;
  • 28) 0.000 001 926 118 349 209 6 × 2 = 0 + 0.000 003 852 236 698 419 2;
  • 29) 0.000 003 852 236 698 419 2 × 2 = 0 + 0.000 007 704 473 396 838 4;
  • 30) 0.000 007 704 473 396 838 4 × 2 = 0 + 0.000 015 408 946 793 676 8;
  • 31) 0.000 015 408 946 793 676 8 × 2 = 0 + 0.000 030 817 893 587 353 6;
  • 32) 0.000 030 817 893 587 353 6 × 2 = 0 + 0.000 061 635 787 174 707 2;
  • 33) 0.000 061 635 787 174 707 2 × 2 = 0 + 0.000 123 271 574 349 414 4;
  • 34) 0.000 123 271 574 349 414 4 × 2 = 0 + 0.000 246 543 148 698 828 8;
  • 35) 0.000 246 543 148 698 828 8 × 2 = 0 + 0.000 493 086 297 397 657 6;
  • 36) 0.000 493 086 297 397 657 6 × 2 = 0 + 0.000 986 172 594 795 315 2;
  • 37) 0.000 986 172 594 795 315 2 × 2 = 0 + 0.001 972 345 189 590 630 4;
  • 38) 0.001 972 345 189 590 630 4 × 2 = 0 + 0.003 944 690 379 181 260 8;
  • 39) 0.003 944 690 379 181 260 8 × 2 = 0 + 0.007 889 380 758 362 521 6;
  • 40) 0.007 889 380 758 362 521 6 × 2 = 0 + 0.015 778 761 516 725 043 2;
  • 41) 0.015 778 761 516 725 043 2 × 2 = 0 + 0.031 557 523 033 450 086 4;
  • 42) 0.031 557 523 033 450 086 4 × 2 = 0 + 0.063 115 046 066 900 172 8;
  • 43) 0.063 115 046 066 900 172 8 × 2 = 0 + 0.126 230 092 133 800 345 6;
  • 44) 0.126 230 092 133 800 345 6 × 2 = 0 + 0.252 460 184 267 600 691 2;
  • 45) 0.252 460 184 267 600 691 2 × 2 = 0 + 0.504 920 368 535 201 382 4;
  • 46) 0.504 920 368 535 201 382 4 × 2 = 1 + 0.009 840 737 070 402 764 8;
  • 47) 0.009 840 737 070 402 764 8 × 2 = 0 + 0.019 681 474 140 805 529 6;
  • 48) 0.019 681 474 140 805 529 6 × 2 = 0 + 0.039 362 948 281 611 059 2;
  • 49) 0.039 362 948 281 611 059 2 × 2 = 0 + 0.078 725 896 563 222 118 4;
  • 50) 0.078 725 896 563 222 118 4 × 2 = 0 + 0.157 451 793 126 444 236 8;
  • 51) 0.157 451 793 126 444 236 8 × 2 = 0 + 0.314 903 586 252 888 473 6;
  • 52) 0.314 903 586 252 888 473 6 × 2 = 0 + 0.629 807 172 505 776 947 2;
  • 53) 0.629 807 172 505 776 947 2 × 2 = 1 + 0.259 614 345 011 553 894 4;
  • 54) 0.259 614 345 011 553 894 4 × 2 = 0 + 0.519 228 690 023 107 788 8;
  • 55) 0.519 228 690 023 107 788 8 × 2 = 1 + 0.038 457 380 046 215 577 6;
  • 56) 0.038 457 380 046 215 577 6 × 2 = 0 + 0.076 914 760 092 431 155 2;
  • 57) 0.076 914 760 092 431 155 2 × 2 = 0 + 0.153 829 520 184 862 310 4;
  • 58) 0.153 829 520 184 862 310 4 × 2 = 0 + 0.307 659 040 369 724 620 8;
  • 59) 0.307 659 040 369 724 620 8 × 2 = 0 + 0.615 318 080 739 449 241 6;
  • 60) 0.615 318 080 739 449 241 6 × 2 = 1 + 0.230 636 161 478 898 483 2;
  • 61) 0.230 636 161 478 898 483 2 × 2 = 0 + 0.461 272 322 957 796 966 4;
  • 62) 0.461 272 322 957 796 966 4 × 2 = 0 + 0.922 544 645 915 593 932 8;
  • 63) 0.922 544 645 915 593 932 8 × 2 = 1 + 0.845 089 291 831 187 865 6;
  • 64) 0.845 089 291 831 187 865 6 × 2 = 1 + 0.690 178 583 662 375 731 2;
  • 65) 0.690 178 583 662 375 731 2 × 2 = 1 + 0.380 357 167 324 751 462 4;
  • 66) 0.380 357 167 324 751 462 4 × 2 = 0 + 0.760 714 334 649 502 924 8;
  • 67) 0.760 714 334 649 502 924 8 × 2 = 1 + 0.521 428 669 299 005 849 6;
  • 68) 0.521 428 669 299 005 849 6 × 2 = 1 + 0.042 857 338 598 011 699 2;
  • 69) 0.042 857 338 598 011 699 2 × 2 = 0 + 0.085 714 677 196 023 398 4;
  • 70) 0.085 714 677 196 023 398 4 × 2 = 0 + 0.171 429 354 392 046 796 8;
  • 71) 0.171 429 354 392 046 796 8 × 2 = 0 + 0.342 858 708 784 093 593 6;
  • 72) 0.342 858 708 784 093 593 6 × 2 = 0 + 0.685 717 417 568 187 187 2;
  • 73) 0.685 717 417 568 187 187 2 × 2 = 1 + 0.371 434 835 136 374 374 4;
  • 74) 0.371 434 835 136 374 374 4 × 2 = 0 + 0.742 869 670 272 748 748 8;
  • 75) 0.742 869 670 272 748 748 8 × 2 = 1 + 0.485 739 340 545 497 497 6;
  • 76) 0.485 739 340 545 497 497 6 × 2 = 0 + 0.971 478 681 090 994 995 2;
  • 77) 0.971 478 681 090 994 995 2 × 2 = 1 + 0.942 957 362 181 989 990 4;
  • 78) 0.942 957 362 181 989 990 4 × 2 = 1 + 0.885 914 724 363 979 980 8;
  • 79) 0.885 914 724 363 979 980 8 × 2 = 1 + 0.771 829 448 727 959 961 6;
  • 80) 0.771 829 448 727 959 961 6 × 2 = 1 + 0.543 658 897 455 919 923 2;
  • 81) 0.543 658 897 455 919 923 2 × 2 = 1 + 0.087 317 794 911 839 846 4;
  • 82) 0.087 317 794 911 839 846 4 × 2 = 0 + 0.174 635 589 823 679 692 8;
  • 83) 0.174 635 589 823 679 692 8 × 2 = 0 + 0.349 271 179 647 359 385 6;
  • 84) 0.349 271 179 647 359 385 6 × 2 = 0 + 0.698 542 359 294 718 771 2;
  • 85) 0.698 542 359 294 718 771 2 × 2 = 1 + 0.397 084 718 589 437 542 4;
  • 86) 0.397 084 718 589 437 542 4 × 2 = 0 + 0.794 169 437 178 875 084 8;
  • 87) 0.794 169 437 178 875 084 8 × 2 = 1 + 0.588 338 874 357 750 169 6;
  • 88) 0.588 338 874 357 750 169 6 × 2 = 1 + 0.176 677 748 715 500 339 2;
  • 89) 0.176 677 748 715 500 339 2 × 2 = 0 + 0.353 355 497 431 000 678 4;
  • 90) 0.353 355 497 431 000 678 4 × 2 = 0 + 0.706 710 994 862 001 356 8;
  • 91) 0.706 710 994 862 001 356 8 × 2 = 1 + 0.413 421 989 724 002 713 6;
  • 92) 0.413 421 989 724 002 713 6 × 2 = 0 + 0.826 843 979 448 005 427 2;
  • 93) 0.826 843 979 448 005 427 2 × 2 = 1 + 0.653 687 958 896 010 854 4;
  • 94) 0.653 687 958 896 010 854 4 × 2 = 1 + 0.307 375 917 792 021 708 8;
  • 95) 0.307 375 917 792 021 708 8 × 2 = 0 + 0.614 751 835 584 043 417 6;
  • 96) 0.614 751 835 584 043 417 6 × 2 = 1 + 0.229 503 671 168 086 835 2;
  • 97) 0.229 503 671 168 086 835 2 × 2 = 0 + 0.459 007 342 336 173 670 4;
  • 98) 0.459 007 342 336 173 670 4 × 2 = 0 + 0.918 014 684 672 347 340 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:


0.000 000 000 000 014 350 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 0001 0011 1011 0000 1010 1111 1000 1011 0010 1101 00(2)

6. Positive number before normalization:

0.000 000 000 000 014 350 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 0001 0011 1011 0000 1010 1111 1000 1011 0010 1101 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 014 350 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 0001 0011 1011 0000 1010 1111 1000 1011 0010 1101 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 0001 0011 1011 0000 1010 1111 1000 1011 0010 1101 00(2) × 20 =

1.0000 0010 1000 0100 1110 1100 0010 1011 1110 0010 1100 1011 0100(2) × 2-46


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): -46

Mantissa (not normalized):
1.0000 0010 1000 0100 1110 1100 0010 1011 1110 0010 1100 1011 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 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) =

977(10) =

011 1101 0001(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. 0000 0010 1000 0100 1110 1100 0010 1011 1110 0010 1100 1011 0100 =

0000 0010 1000 0100 1110 1100 0010 1011 1110 0010 1100 1011 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 0001

Mantissa (52 bits) =
0000 0010 1000 0100 1110 1100 0010 1011 1110 0010 1100 1011 0100


Decimal number -0.000 000 000 000 014 350 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1000 0100 1110 1100 0010 1011 1110 0010 1100 1011 0100

» Convert decimal -0.000 000 000 000 014 343 3 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