-0.000 000 000 000 014 356 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 356 9(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
-0.000 000 000 000 014 356 9(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:

|-0.000 000 000 000 014 356 9| = 0.000 000 000 000 014 356 9


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: 0.000 000 000 000 014 356 9.

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) 0.000 000 000 000 014 356 9 × 2 = 0 + 0.000 000 000 000 028 713 8;
  • 2) 0.000 000 000 000 028 713 8 × 2 = 0 + 0.000 000 000 000 057 427 6;
  • 3) 0.000 000 000 000 057 427 6 × 2 = 0 + 0.000 000 000 000 114 855 2;
  • 4) 0.000 000 000 000 114 855 2 × 2 = 0 + 0.000 000 000 000 229 710 4;
  • 5) 0.000 000 000 000 229 710 4 × 2 = 0 + 0.000 000 000 000 459 420 8;
  • 6) 0.000 000 000 000 459 420 8 × 2 = 0 + 0.000 000 000 000 918 841 6;
  • 7) 0.000 000 000 000 918 841 6 × 2 = 0 + 0.000 000 000 001 837 683 2;
  • 8) 0.000 000 000 001 837 683 2 × 2 = 0 + 0.000 000 000 003 675 366 4;
  • 9) 0.000 000 000 003 675 366 4 × 2 = 0 + 0.000 000 000 007 350 732 8;
  • 10) 0.000 000 000 007 350 732 8 × 2 = 0 + 0.000 000 000 014 701 465 6;
  • 11) 0.000 000 000 014 701 465 6 × 2 = 0 + 0.000 000 000 029 402 931 2;
  • 12) 0.000 000 000 029 402 931 2 × 2 = 0 + 0.000 000 000 058 805 862 4;
  • 13) 0.000 000 000 058 805 862 4 × 2 = 0 + 0.000 000 000 117 611 724 8;
  • 14) 0.000 000 000 117 611 724 8 × 2 = 0 + 0.000 000 000 235 223 449 6;
  • 15) 0.000 000 000 235 223 449 6 × 2 = 0 + 0.000 000 000 470 446 899 2;
  • 16) 0.000 000 000 470 446 899 2 × 2 = 0 + 0.000 000 000 940 893 798 4;
  • 17) 0.000 000 000 940 893 798 4 × 2 = 0 + 0.000 000 001 881 787 596 8;
  • 18) 0.000 000 001 881 787 596 8 × 2 = 0 + 0.000 000 003 763 575 193 6;
  • 19) 0.000 000 003 763 575 193 6 × 2 = 0 + 0.000 000 007 527 150 387 2;
  • 20) 0.000 000 007 527 150 387 2 × 2 = 0 + 0.000 000 015 054 300 774 4;
  • 21) 0.000 000 015 054 300 774 4 × 2 = 0 + 0.000 000 030 108 601 548 8;
  • 22) 0.000 000 030 108 601 548 8 × 2 = 0 + 0.000 000 060 217 203 097 6;
  • 23) 0.000 000 060 217 203 097 6 × 2 = 0 + 0.000 000 120 434 406 195 2;
  • 24) 0.000 000 120 434 406 195 2 × 2 = 0 + 0.000 000 240 868 812 390 4;
  • 25) 0.000 000 240 868 812 390 4 × 2 = 0 + 0.000 000 481 737 624 780 8;
  • 26) 0.000 000 481 737 624 780 8 × 2 = 0 + 0.000 000 963 475 249 561 6;
  • 27) 0.000 000 963 475 249 561 6 × 2 = 0 + 0.000 001 926 950 499 123 2;
  • 28) 0.000 001 926 950 499 123 2 × 2 = 0 + 0.000 003 853 900 998 246 4;
  • 29) 0.000 003 853 900 998 246 4 × 2 = 0 + 0.000 007 707 801 996 492 8;
  • 30) 0.000 007 707 801 996 492 8 × 2 = 0 + 0.000 015 415 603 992 985 6;
  • 31) 0.000 015 415 603 992 985 6 × 2 = 0 + 0.000 030 831 207 985 971 2;
  • 32) 0.000 030 831 207 985 971 2 × 2 = 0 + 0.000 061 662 415 971 942 4;
  • 33) 0.000 061 662 415 971 942 4 × 2 = 0 + 0.000 123 324 831 943 884 8;
  • 34) 0.000 123 324 831 943 884 8 × 2 = 0 + 0.000 246 649 663 887 769 6;
  • 35) 0.000 246 649 663 887 769 6 × 2 = 0 + 0.000 493 299 327 775 539 2;
  • 36) 0.000 493 299 327 775 539 2 × 2 = 0 + 0.000 986 598 655 551 078 4;
  • 37) 0.000 986 598 655 551 078 4 × 2 = 0 + 0.001 973 197 311 102 156 8;
  • 38) 0.001 973 197 311 102 156 8 × 2 = 0 + 0.003 946 394 622 204 313 6;
  • 39) 0.003 946 394 622 204 313 6 × 2 = 0 + 0.007 892 789 244 408 627 2;
  • 40) 0.007 892 789 244 408 627 2 × 2 = 0 + 0.015 785 578 488 817 254 4;
  • 41) 0.015 785 578 488 817 254 4 × 2 = 0 + 0.031 571 156 977 634 508 8;
  • 42) 0.031 571 156 977 634 508 8 × 2 = 0 + 0.063 142 313 955 269 017 6;
  • 43) 0.063 142 313 955 269 017 6 × 2 = 0 + 0.126 284 627 910 538 035 2;
  • 44) 0.126 284 627 910 538 035 2 × 2 = 0 + 0.252 569 255 821 076 070 4;
  • 45) 0.252 569 255 821 076 070 4 × 2 = 0 + 0.505 138 511 642 152 140 8;
  • 46) 0.505 138 511 642 152 140 8 × 2 = 1 + 0.010 277 023 284 304 281 6;
  • 47) 0.010 277 023 284 304 281 6 × 2 = 0 + 0.020 554 046 568 608 563 2;
  • 48) 0.020 554 046 568 608 563 2 × 2 = 0 + 0.041 108 093 137 217 126 4;
  • 49) 0.041 108 093 137 217 126 4 × 2 = 0 + 0.082 216 186 274 434 252 8;
  • 50) 0.082 216 186 274 434 252 8 × 2 = 0 + 0.164 432 372 548 868 505 6;
  • 51) 0.164 432 372 548 868 505 6 × 2 = 0 + 0.328 864 745 097 737 011 2;
  • 52) 0.328 864 745 097 737 011 2 × 2 = 0 + 0.657 729 490 195 474 022 4;
  • 53) 0.657 729 490 195 474 022 4 × 2 = 1 + 0.315 458 980 390 948 044 8;
  • 54) 0.315 458 980 390 948 044 8 × 2 = 0 + 0.630 917 960 781 896 089 6;
  • 55) 0.630 917 960 781 896 089 6 × 2 = 1 + 0.261 835 921 563 792 179 2;
  • 56) 0.261 835 921 563 792 179 2 × 2 = 0 + 0.523 671 843 127 584 358 4;
  • 57) 0.523 671 843 127 584 358 4 × 2 = 1 + 0.047 343 686 255 168 716 8;
  • 58) 0.047 343 686 255 168 716 8 × 2 = 0 + 0.094 687 372 510 337 433 6;
  • 59) 0.094 687 372 510 337 433 6 × 2 = 0 + 0.189 374 745 020 674 867 2;
  • 60) 0.189 374 745 020 674 867 2 × 2 = 0 + 0.378 749 490 041 349 734 4;
  • 61) 0.378 749 490 041 349 734 4 × 2 = 0 + 0.757 498 980 082 699 468 8;
  • 62) 0.757 498 980 082 699 468 8 × 2 = 1 + 0.514 997 960 165 398 937 6;
  • 63) 0.514 997 960 165 398 937 6 × 2 = 1 + 0.029 995 920 330 797 875 2;
  • 64) 0.029 995 920 330 797 875 2 × 2 = 0 + 0.059 991 840 661 595 750 4;
  • 65) 0.059 991 840 661 595 750 4 × 2 = 0 + 0.119 983 681 323 191 500 8;
  • 66) 0.119 983 681 323 191 500 8 × 2 = 0 + 0.239 967 362 646 383 001 6;
  • 67) 0.239 967 362 646 383 001 6 × 2 = 0 + 0.479 934 725 292 766 003 2;
  • 68) 0.479 934 725 292 766 003 2 × 2 = 0 + 0.959 869 450 585 532 006 4;
  • 69) 0.959 869 450 585 532 006 4 × 2 = 1 + 0.919 738 901 171 064 012 8;
  • 70) 0.919 738 901 171 064 012 8 × 2 = 1 + 0.839 477 802 342 128 025 6;
  • 71) 0.839 477 802 342 128 025 6 × 2 = 1 + 0.678 955 604 684 256 051 2;
  • 72) 0.678 955 604 684 256 051 2 × 2 = 1 + 0.357 911 209 368 512 102 4;
  • 73) 0.357 911 209 368 512 102 4 × 2 = 0 + 0.715 822 418 737 024 204 8;
  • 74) 0.715 822 418 737 024 204 8 × 2 = 1 + 0.431 644 837 474 048 409 6;
  • 75) 0.431 644 837 474 048 409 6 × 2 = 0 + 0.863 289 674 948 096 819 2;
  • 76) 0.863 289 674 948 096 819 2 × 2 = 1 + 0.726 579 349 896 193 638 4;
  • 77) 0.726 579 349 896 193 638 4 × 2 = 1 + 0.453 158 699 792 387 276 8;
  • 78) 0.453 158 699 792 387 276 8 × 2 = 0 + 0.906 317 399 584 774 553 6;
  • 79) 0.906 317 399 584 774 553 6 × 2 = 1 + 0.812 634 799 169 549 107 2;
  • 80) 0.812 634 799 169 549 107 2 × 2 = 1 + 0.625 269 598 339 098 214 4;
  • 81) 0.625 269 598 339 098 214 4 × 2 = 1 + 0.250 539 196 678 196 428 8;
  • 82) 0.250 539 196 678 196 428 8 × 2 = 0 + 0.501 078 393 356 392 857 6;
  • 83) 0.501 078 393 356 392 857 6 × 2 = 1 + 0.002 156 786 712 785 715 2;
  • 84) 0.002 156 786 712 785 715 2 × 2 = 0 + 0.004 313 573 425 571 430 4;
  • 85) 0.004 313 573 425 571 430 4 × 2 = 0 + 0.008 627 146 851 142 860 8;
  • 86) 0.008 627 146 851 142 860 8 × 2 = 0 + 0.017 254 293 702 285 721 6;
  • 87) 0.017 254 293 702 285 721 6 × 2 = 0 + 0.034 508 587 404 571 443 2;
  • 88) 0.034 508 587 404 571 443 2 × 2 = 0 + 0.069 017 174 809 142 886 4;
  • 89) 0.069 017 174 809 142 886 4 × 2 = 0 + 0.138 034 349 618 285 772 8;
  • 90) 0.138 034 349 618 285 772 8 × 2 = 0 + 0.276 068 699 236 571 545 6;
  • 91) 0.276 068 699 236 571 545 6 × 2 = 0 + 0.552 137 398 473 143 091 2;
  • 92) 0.552 137 398 473 143 091 2 × 2 = 1 + 0.104 274 796 946 286 182 4;
  • 93) 0.104 274 796 946 286 182 4 × 2 = 0 + 0.208 549 593 892 572 364 8;
  • 94) 0.208 549 593 892 572 364 8 × 2 = 0 + 0.417 099 187 785 144 729 6;
  • 95) 0.417 099 187 785 144 729 6 × 2 = 0 + 0.834 198 375 570 289 459 2;
  • 96) 0.834 198 375 570 289 459 2 × 2 = 1 + 0.668 396 751 140 578 918 4;
  • 97) 0.668 396 751 140 578 918 4 × 2 = 1 + 0.336 793 502 281 157 836 8;
  • 98) 0.336 793 502 281 157 836 8 × 2 = 0 + 0.673 587 004 562 315 673 6;

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:


0.000 000 000 000 014 356 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1000 0110 0000 1111 0101 1011 1010 0000 0001 0001 10(2)

6. Positive number before normalization:

0.000 000 000 000 014 356 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1000 0110 0000 1111 0101 1011 1010 0000 0001 0001 10(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:


0.000 000 000 000 014 356 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1000 0110 0000 1111 0101 1011 1010 0000 0001 0001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1000 0110 0000 1111 0101 1011 1010 0000 0001 0001 10(2) × 20 =

1.0000 0010 1010 0001 1000 0011 1101 0110 1110 1000 0000 0100 0110(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 1010 0001 1000 0011 1101 0110 1110 1000 0000 0100 0110


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 1010 0001 1000 0011 1101 0110 1110 1000 0000 0100 0110 =

0000 0010 1010 0001 1000 0011 1101 0110 1110 1000 0000 0100 0110


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 1010 0001 1000 0011 1101 0110 1110 1000 0000 0100 0110


Decimal number -0.000 000 000 000 014 356 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1010 0001 1000 0011 1101 0110 1110 1000 0000 0100 0110

» Convert decimal -0.000 000 000 000 014 349 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