-0.000 000 000 000 014 362 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 362 1(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
-0.000 000 000 000 014 362 1(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:

|-0.000 000 000 000 014 362 1| = 0.000 000 000 000 014 362 1


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: 0.000 000 000 000 014 362 1.

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) 0.000 000 000 000 014 362 1 × 2 = 0 + 0.000 000 000 000 028 724 2;
  • 2) 0.000 000 000 000 028 724 2 × 2 = 0 + 0.000 000 000 000 057 448 4;
  • 3) 0.000 000 000 000 057 448 4 × 2 = 0 + 0.000 000 000 000 114 896 8;
  • 4) 0.000 000 000 000 114 896 8 × 2 = 0 + 0.000 000 000 000 229 793 6;
  • 5) 0.000 000 000 000 229 793 6 × 2 = 0 + 0.000 000 000 000 459 587 2;
  • 6) 0.000 000 000 000 459 587 2 × 2 = 0 + 0.000 000 000 000 919 174 4;
  • 7) 0.000 000 000 000 919 174 4 × 2 = 0 + 0.000 000 000 001 838 348 8;
  • 8) 0.000 000 000 001 838 348 8 × 2 = 0 + 0.000 000 000 003 676 697 6;
  • 9) 0.000 000 000 003 676 697 6 × 2 = 0 + 0.000 000 000 007 353 395 2;
  • 10) 0.000 000 000 007 353 395 2 × 2 = 0 + 0.000 000 000 014 706 790 4;
  • 11) 0.000 000 000 014 706 790 4 × 2 = 0 + 0.000 000 000 029 413 580 8;
  • 12) 0.000 000 000 029 413 580 8 × 2 = 0 + 0.000 000 000 058 827 161 6;
  • 13) 0.000 000 000 058 827 161 6 × 2 = 0 + 0.000 000 000 117 654 323 2;
  • 14) 0.000 000 000 117 654 323 2 × 2 = 0 + 0.000 000 000 235 308 646 4;
  • 15) 0.000 000 000 235 308 646 4 × 2 = 0 + 0.000 000 000 470 617 292 8;
  • 16) 0.000 000 000 470 617 292 8 × 2 = 0 + 0.000 000 000 941 234 585 6;
  • 17) 0.000 000 000 941 234 585 6 × 2 = 0 + 0.000 000 001 882 469 171 2;
  • 18) 0.000 000 001 882 469 171 2 × 2 = 0 + 0.000 000 003 764 938 342 4;
  • 19) 0.000 000 003 764 938 342 4 × 2 = 0 + 0.000 000 007 529 876 684 8;
  • 20) 0.000 000 007 529 876 684 8 × 2 = 0 + 0.000 000 015 059 753 369 6;
  • 21) 0.000 000 015 059 753 369 6 × 2 = 0 + 0.000 000 030 119 506 739 2;
  • 22) 0.000 000 030 119 506 739 2 × 2 = 0 + 0.000 000 060 239 013 478 4;
  • 23) 0.000 000 060 239 013 478 4 × 2 = 0 + 0.000 000 120 478 026 956 8;
  • 24) 0.000 000 120 478 026 956 8 × 2 = 0 + 0.000 000 240 956 053 913 6;
  • 25) 0.000 000 240 956 053 913 6 × 2 = 0 + 0.000 000 481 912 107 827 2;
  • 26) 0.000 000 481 912 107 827 2 × 2 = 0 + 0.000 000 963 824 215 654 4;
  • 27) 0.000 000 963 824 215 654 4 × 2 = 0 + 0.000 001 927 648 431 308 8;
  • 28) 0.000 001 927 648 431 308 8 × 2 = 0 + 0.000 003 855 296 862 617 6;
  • 29) 0.000 003 855 296 862 617 6 × 2 = 0 + 0.000 007 710 593 725 235 2;
  • 30) 0.000 007 710 593 725 235 2 × 2 = 0 + 0.000 015 421 187 450 470 4;
  • 31) 0.000 015 421 187 450 470 4 × 2 = 0 + 0.000 030 842 374 900 940 8;
  • 32) 0.000 030 842 374 900 940 8 × 2 = 0 + 0.000 061 684 749 801 881 6;
  • 33) 0.000 061 684 749 801 881 6 × 2 = 0 + 0.000 123 369 499 603 763 2;
  • 34) 0.000 123 369 499 603 763 2 × 2 = 0 + 0.000 246 738 999 207 526 4;
  • 35) 0.000 246 738 999 207 526 4 × 2 = 0 + 0.000 493 477 998 415 052 8;
  • 36) 0.000 493 477 998 415 052 8 × 2 = 0 + 0.000 986 955 996 830 105 6;
  • 37) 0.000 986 955 996 830 105 6 × 2 = 0 + 0.001 973 911 993 660 211 2;
  • 38) 0.001 973 911 993 660 211 2 × 2 = 0 + 0.003 947 823 987 320 422 4;
  • 39) 0.003 947 823 987 320 422 4 × 2 = 0 + 0.007 895 647 974 640 844 8;
  • 40) 0.007 895 647 974 640 844 8 × 2 = 0 + 0.015 791 295 949 281 689 6;
  • 41) 0.015 791 295 949 281 689 6 × 2 = 0 + 0.031 582 591 898 563 379 2;
  • 42) 0.031 582 591 898 563 379 2 × 2 = 0 + 0.063 165 183 797 126 758 4;
  • 43) 0.063 165 183 797 126 758 4 × 2 = 0 + 0.126 330 367 594 253 516 8;
  • 44) 0.126 330 367 594 253 516 8 × 2 = 0 + 0.252 660 735 188 507 033 6;
  • 45) 0.252 660 735 188 507 033 6 × 2 = 0 + 0.505 321 470 377 014 067 2;
  • 46) 0.505 321 470 377 014 067 2 × 2 = 1 + 0.010 642 940 754 028 134 4;
  • 47) 0.010 642 940 754 028 134 4 × 2 = 0 + 0.021 285 881 508 056 268 8;
  • 48) 0.021 285 881 508 056 268 8 × 2 = 0 + 0.042 571 763 016 112 537 6;
  • 49) 0.042 571 763 016 112 537 6 × 2 = 0 + 0.085 143 526 032 225 075 2;
  • 50) 0.085 143 526 032 225 075 2 × 2 = 0 + 0.170 287 052 064 450 150 4;
  • 51) 0.170 287 052 064 450 150 4 × 2 = 0 + 0.340 574 104 128 900 300 8;
  • 52) 0.340 574 104 128 900 300 8 × 2 = 0 + 0.681 148 208 257 800 601 6;
  • 53) 0.681 148 208 257 800 601 6 × 2 = 1 + 0.362 296 416 515 601 203 2;
  • 54) 0.362 296 416 515 601 203 2 × 2 = 0 + 0.724 592 833 031 202 406 4;
  • 55) 0.724 592 833 031 202 406 4 × 2 = 1 + 0.449 185 666 062 404 812 8;
  • 56) 0.449 185 666 062 404 812 8 × 2 = 0 + 0.898 371 332 124 809 625 6;
  • 57) 0.898 371 332 124 809 625 6 × 2 = 1 + 0.796 742 664 249 619 251 2;
  • 58) 0.796 742 664 249 619 251 2 × 2 = 1 + 0.593 485 328 499 238 502 4;
  • 59) 0.593 485 328 499 238 502 4 × 2 = 1 + 0.186 970 656 998 477 004 8;
  • 60) 0.186 970 656 998 477 004 8 × 2 = 0 + 0.373 941 313 996 954 009 6;
  • 61) 0.373 941 313 996 954 009 6 × 2 = 0 + 0.747 882 627 993 908 019 2;
  • 62) 0.747 882 627 993 908 019 2 × 2 = 1 + 0.495 765 255 987 816 038 4;
  • 63) 0.495 765 255 987 816 038 4 × 2 = 0 + 0.991 530 511 975 632 076 8;
  • 64) 0.991 530 511 975 632 076 8 × 2 = 1 + 0.983 061 023 951 264 153 6;
  • 65) 0.983 061 023 951 264 153 6 × 2 = 1 + 0.966 122 047 902 528 307 2;
  • 66) 0.966 122 047 902 528 307 2 × 2 = 1 + 0.932 244 095 805 056 614 4;
  • 67) 0.932 244 095 805 056 614 4 × 2 = 1 + 0.864 488 191 610 113 228 8;
  • 68) 0.864 488 191 610 113 228 8 × 2 = 1 + 0.728 976 383 220 226 457 6;
  • 69) 0.728 976 383 220 226 457 6 × 2 = 1 + 0.457 952 766 440 452 915 2;
  • 70) 0.457 952 766 440 452 915 2 × 2 = 0 + 0.915 905 532 880 905 830 4;
  • 71) 0.915 905 532 880 905 830 4 × 2 = 1 + 0.831 811 065 761 811 660 8;
  • 72) 0.831 811 065 761 811 660 8 × 2 = 1 + 0.663 622 131 523 623 321 6;
  • 73) 0.663 622 131 523 623 321 6 × 2 = 1 + 0.327 244 263 047 246 643 2;
  • 74) 0.327 244 263 047 246 643 2 × 2 = 0 + 0.654 488 526 094 493 286 4;
  • 75) 0.654 488 526 094 493 286 4 × 2 = 1 + 0.308 977 052 188 986 572 8;
  • 76) 0.308 977 052 188 986 572 8 × 2 = 0 + 0.617 954 104 377 973 145 6;
  • 77) 0.617 954 104 377 973 145 6 × 2 = 1 + 0.235 908 208 755 946 291 2;
  • 78) 0.235 908 208 755 946 291 2 × 2 = 0 + 0.471 816 417 511 892 582 4;
  • 79) 0.471 816 417 511 892 582 4 × 2 = 0 + 0.943 632 835 023 785 164 8;
  • 80) 0.943 632 835 023 785 164 8 × 2 = 1 + 0.887 265 670 047 570 329 6;
  • 81) 0.887 265 670 047 570 329 6 × 2 = 1 + 0.774 531 340 095 140 659 2;
  • 82) 0.774 531 340 095 140 659 2 × 2 = 1 + 0.549 062 680 190 281 318 4;
  • 83) 0.549 062 680 190 281 318 4 × 2 = 1 + 0.098 125 360 380 562 636 8;
  • 84) 0.098 125 360 380 562 636 8 × 2 = 0 + 0.196 250 720 761 125 273 6;
  • 85) 0.196 250 720 761 125 273 6 × 2 = 0 + 0.392 501 441 522 250 547 2;
  • 86) 0.392 501 441 522 250 547 2 × 2 = 0 + 0.785 002 883 044 501 094 4;
  • 87) 0.785 002 883 044 501 094 4 × 2 = 1 + 0.570 005 766 089 002 188 8;
  • 88) 0.570 005 766 089 002 188 8 × 2 = 1 + 0.140 011 532 178 004 377 6;
  • 89) 0.140 011 532 178 004 377 6 × 2 = 0 + 0.280 023 064 356 008 755 2;
  • 90) 0.280 023 064 356 008 755 2 × 2 = 0 + 0.560 046 128 712 017 510 4;
  • 91) 0.560 046 128 712 017 510 4 × 2 = 1 + 0.120 092 257 424 035 020 8;
  • 92) 0.120 092 257 424 035 020 8 × 2 = 0 + 0.240 184 514 848 070 041 6;
  • 93) 0.240 184 514 848 070 041 6 × 2 = 0 + 0.480 369 029 696 140 083 2;
  • 94) 0.480 369 029 696 140 083 2 × 2 = 0 + 0.960 738 059 392 280 166 4;
  • 95) 0.960 738 059 392 280 166 4 × 2 = 1 + 0.921 476 118 784 560 332 8;
  • 96) 0.921 476 118 784 560 332 8 × 2 = 1 + 0.842 952 237 569 120 665 6;
  • 97) 0.842 952 237 569 120 665 6 × 2 = 1 + 0.685 904 475 138 241 331 2;
  • 98) 0.685 904 475 138 241 331 2 × 2 = 1 + 0.371 808 950 276 482 662 4;

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:


0.000 000 000 000 014 362 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1110 0101 1111 1011 1010 1001 1110 0011 0010 0011 11(2)

6. Positive number before normalization:

0.000 000 000 000 014 362 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1110 0101 1111 1011 1010 1001 1110 0011 0010 0011 11(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:


0.000 000 000 000 014 362 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1110 0101 1111 1011 1010 1001 1110 0011 0010 0011 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1110 0101 1111 1011 1010 1001 1110 0011 0010 0011 11(2) × 20 =

1.0000 0010 1011 1001 0111 1110 1110 1010 0111 1000 1100 1000 1111(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 1011 1001 0111 1110 1110 1010 0111 1000 1100 1000 1111


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 1011 1001 0111 1110 1110 1010 0111 1000 1100 1000 1111 =

0000 0010 1011 1001 0111 1110 1110 1010 0111 1000 1100 1000 1111


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 1011 1001 0111 1110 1110 1010 0111 1000 1100 1000 1111


Decimal number -0.000 000 000 000 014 362 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1011 1001 0111 1110 1110 1010 0111 1000 1100 1000 1111

» Convert decimal -0.000 000 000 000 014 353 9 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