-0.000 000 000 000 014 365 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 365 4(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
-0.000 000 000 000 014 365 4(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:

|-0.000 000 000 000 014 365 4| = 0.000 000 000 000 014 365 4


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: 0.000 000 000 000 014 365 4.

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) 0.000 000 000 000 014 365 4 × 2 = 0 + 0.000 000 000 000 028 730 8;
  • 2) 0.000 000 000 000 028 730 8 × 2 = 0 + 0.000 000 000 000 057 461 6;
  • 3) 0.000 000 000 000 057 461 6 × 2 = 0 + 0.000 000 000 000 114 923 2;
  • 4) 0.000 000 000 000 114 923 2 × 2 = 0 + 0.000 000 000 000 229 846 4;
  • 5) 0.000 000 000 000 229 846 4 × 2 = 0 + 0.000 000 000 000 459 692 8;
  • 6) 0.000 000 000 000 459 692 8 × 2 = 0 + 0.000 000 000 000 919 385 6;
  • 7) 0.000 000 000 000 919 385 6 × 2 = 0 + 0.000 000 000 001 838 771 2;
  • 8) 0.000 000 000 001 838 771 2 × 2 = 0 + 0.000 000 000 003 677 542 4;
  • 9) 0.000 000 000 003 677 542 4 × 2 = 0 + 0.000 000 000 007 355 084 8;
  • 10) 0.000 000 000 007 355 084 8 × 2 = 0 + 0.000 000 000 014 710 169 6;
  • 11) 0.000 000 000 014 710 169 6 × 2 = 0 + 0.000 000 000 029 420 339 2;
  • 12) 0.000 000 000 029 420 339 2 × 2 = 0 + 0.000 000 000 058 840 678 4;
  • 13) 0.000 000 000 058 840 678 4 × 2 = 0 + 0.000 000 000 117 681 356 8;
  • 14) 0.000 000 000 117 681 356 8 × 2 = 0 + 0.000 000 000 235 362 713 6;
  • 15) 0.000 000 000 235 362 713 6 × 2 = 0 + 0.000 000 000 470 725 427 2;
  • 16) 0.000 000 000 470 725 427 2 × 2 = 0 + 0.000 000 000 941 450 854 4;
  • 17) 0.000 000 000 941 450 854 4 × 2 = 0 + 0.000 000 001 882 901 708 8;
  • 18) 0.000 000 001 882 901 708 8 × 2 = 0 + 0.000 000 003 765 803 417 6;
  • 19) 0.000 000 003 765 803 417 6 × 2 = 0 + 0.000 000 007 531 606 835 2;
  • 20) 0.000 000 007 531 606 835 2 × 2 = 0 + 0.000 000 015 063 213 670 4;
  • 21) 0.000 000 015 063 213 670 4 × 2 = 0 + 0.000 000 030 126 427 340 8;
  • 22) 0.000 000 030 126 427 340 8 × 2 = 0 + 0.000 000 060 252 854 681 6;
  • 23) 0.000 000 060 252 854 681 6 × 2 = 0 + 0.000 000 120 505 709 363 2;
  • 24) 0.000 000 120 505 709 363 2 × 2 = 0 + 0.000 000 241 011 418 726 4;
  • 25) 0.000 000 241 011 418 726 4 × 2 = 0 + 0.000 000 482 022 837 452 8;
  • 26) 0.000 000 482 022 837 452 8 × 2 = 0 + 0.000 000 964 045 674 905 6;
  • 27) 0.000 000 964 045 674 905 6 × 2 = 0 + 0.000 001 928 091 349 811 2;
  • 28) 0.000 001 928 091 349 811 2 × 2 = 0 + 0.000 003 856 182 699 622 4;
  • 29) 0.000 003 856 182 699 622 4 × 2 = 0 + 0.000 007 712 365 399 244 8;
  • 30) 0.000 007 712 365 399 244 8 × 2 = 0 + 0.000 015 424 730 798 489 6;
  • 31) 0.000 015 424 730 798 489 6 × 2 = 0 + 0.000 030 849 461 596 979 2;
  • 32) 0.000 030 849 461 596 979 2 × 2 = 0 + 0.000 061 698 923 193 958 4;
  • 33) 0.000 061 698 923 193 958 4 × 2 = 0 + 0.000 123 397 846 387 916 8;
  • 34) 0.000 123 397 846 387 916 8 × 2 = 0 + 0.000 246 795 692 775 833 6;
  • 35) 0.000 246 795 692 775 833 6 × 2 = 0 + 0.000 493 591 385 551 667 2;
  • 36) 0.000 493 591 385 551 667 2 × 2 = 0 + 0.000 987 182 771 103 334 4;
  • 37) 0.000 987 182 771 103 334 4 × 2 = 0 + 0.001 974 365 542 206 668 8;
  • 38) 0.001 974 365 542 206 668 8 × 2 = 0 + 0.003 948 731 084 413 337 6;
  • 39) 0.003 948 731 084 413 337 6 × 2 = 0 + 0.007 897 462 168 826 675 2;
  • 40) 0.007 897 462 168 826 675 2 × 2 = 0 + 0.015 794 924 337 653 350 4;
  • 41) 0.015 794 924 337 653 350 4 × 2 = 0 + 0.031 589 848 675 306 700 8;
  • 42) 0.031 589 848 675 306 700 8 × 2 = 0 + 0.063 179 697 350 613 401 6;
  • 43) 0.063 179 697 350 613 401 6 × 2 = 0 + 0.126 359 394 701 226 803 2;
  • 44) 0.126 359 394 701 226 803 2 × 2 = 0 + 0.252 718 789 402 453 606 4;
  • 45) 0.252 718 789 402 453 606 4 × 2 = 0 + 0.505 437 578 804 907 212 8;
  • 46) 0.505 437 578 804 907 212 8 × 2 = 1 + 0.010 875 157 609 814 425 6;
  • 47) 0.010 875 157 609 814 425 6 × 2 = 0 + 0.021 750 315 219 628 851 2;
  • 48) 0.021 750 315 219 628 851 2 × 2 = 0 + 0.043 500 630 439 257 702 4;
  • 49) 0.043 500 630 439 257 702 4 × 2 = 0 + 0.087 001 260 878 515 404 8;
  • 50) 0.087 001 260 878 515 404 8 × 2 = 0 + 0.174 002 521 757 030 809 6;
  • 51) 0.174 002 521 757 030 809 6 × 2 = 0 + 0.348 005 043 514 061 619 2;
  • 52) 0.348 005 043 514 061 619 2 × 2 = 0 + 0.696 010 087 028 123 238 4;
  • 53) 0.696 010 087 028 123 238 4 × 2 = 1 + 0.392 020 174 056 246 476 8;
  • 54) 0.392 020 174 056 246 476 8 × 2 = 0 + 0.784 040 348 112 492 953 6;
  • 55) 0.784 040 348 112 492 953 6 × 2 = 1 + 0.568 080 696 224 985 907 2;
  • 56) 0.568 080 696 224 985 907 2 × 2 = 1 + 0.136 161 392 449 971 814 4;
  • 57) 0.136 161 392 449 971 814 4 × 2 = 0 + 0.272 322 784 899 943 628 8;
  • 58) 0.272 322 784 899 943 628 8 × 2 = 0 + 0.544 645 569 799 887 257 6;
  • 59) 0.544 645 569 799 887 257 6 × 2 = 1 + 0.089 291 139 599 774 515 2;
  • 60) 0.089 291 139 599 774 515 2 × 2 = 0 + 0.178 582 279 199 549 030 4;
  • 61) 0.178 582 279 199 549 030 4 × 2 = 0 + 0.357 164 558 399 098 060 8;
  • 62) 0.357 164 558 399 098 060 8 × 2 = 0 + 0.714 329 116 798 196 121 6;
  • 63) 0.714 329 116 798 196 121 6 × 2 = 1 + 0.428 658 233 596 392 243 2;
  • 64) 0.428 658 233 596 392 243 2 × 2 = 0 + 0.857 316 467 192 784 486 4;
  • 65) 0.857 316 467 192 784 486 4 × 2 = 1 + 0.714 632 934 385 568 972 8;
  • 66) 0.714 632 934 385 568 972 8 × 2 = 1 + 0.429 265 868 771 137 945 6;
  • 67) 0.429 265 868 771 137 945 6 × 2 = 0 + 0.858 531 737 542 275 891 2;
  • 68) 0.858 531 737 542 275 891 2 × 2 = 1 + 0.717 063 475 084 551 782 4;
  • 69) 0.717 063 475 084 551 782 4 × 2 = 1 + 0.434 126 950 169 103 564 8;
  • 70) 0.434 126 950 169 103 564 8 × 2 = 0 + 0.868 253 900 338 207 129 6;
  • 71) 0.868 253 900 338 207 129 6 × 2 = 1 + 0.736 507 800 676 414 259 2;
  • 72) 0.736 507 800 676 414 259 2 × 2 = 1 + 0.473 015 601 352 828 518 4;
  • 73) 0.473 015 601 352 828 518 4 × 2 = 0 + 0.946 031 202 705 657 036 8;
  • 74) 0.946 031 202 705 657 036 8 × 2 = 1 + 0.892 062 405 411 314 073 6;
  • 75) 0.892 062 405 411 314 073 6 × 2 = 1 + 0.784 124 810 822 628 147 2;
  • 76) 0.784 124 810 822 628 147 2 × 2 = 1 + 0.568 249 621 645 256 294 4;
  • 77) 0.568 249 621 645 256 294 4 × 2 = 1 + 0.136 499 243 290 512 588 8;
  • 78) 0.136 499 243 290 512 588 8 × 2 = 0 + 0.272 998 486 581 025 177 6;
  • 79) 0.272 998 486 581 025 177 6 × 2 = 0 + 0.545 996 973 162 050 355 2;
  • 80) 0.545 996 973 162 050 355 2 × 2 = 1 + 0.091 993 946 324 100 710 4;
  • 81) 0.091 993 946 324 100 710 4 × 2 = 0 + 0.183 987 892 648 201 420 8;
  • 82) 0.183 987 892 648 201 420 8 × 2 = 0 + 0.367 975 785 296 402 841 6;
  • 83) 0.367 975 785 296 402 841 6 × 2 = 0 + 0.735 951 570 592 805 683 2;
  • 84) 0.735 951 570 592 805 683 2 × 2 = 1 + 0.471 903 141 185 611 366 4;
  • 85) 0.471 903 141 185 611 366 4 × 2 = 0 + 0.943 806 282 371 222 732 8;
  • 86) 0.943 806 282 371 222 732 8 × 2 = 1 + 0.887 612 564 742 445 465 6;
  • 87) 0.887 612 564 742 445 465 6 × 2 = 1 + 0.775 225 129 484 890 931 2;
  • 88) 0.775 225 129 484 890 931 2 × 2 = 1 + 0.550 450 258 969 781 862 4;
  • 89) 0.550 450 258 969 781 862 4 × 2 = 1 + 0.100 900 517 939 563 724 8;
  • 90) 0.100 900 517 939 563 724 8 × 2 = 0 + 0.201 801 035 879 127 449 6;
  • 91) 0.201 801 035 879 127 449 6 × 2 = 0 + 0.403 602 071 758 254 899 2;
  • 92) 0.403 602 071 758 254 899 2 × 2 = 0 + 0.807 204 143 516 509 798 4;
  • 93) 0.807 204 143 516 509 798 4 × 2 = 1 + 0.614 408 287 033 019 596 8;
  • 94) 0.614 408 287 033 019 596 8 × 2 = 1 + 0.228 816 574 066 039 193 6;
  • 95) 0.228 816 574 066 039 193 6 × 2 = 0 + 0.457 633 148 132 078 387 2;
  • 96) 0.457 633 148 132 078 387 2 × 2 = 0 + 0.915 266 296 264 156 774 4;
  • 97) 0.915 266 296 264 156 774 4 × 2 = 1 + 0.830 532 592 528 313 548 8;
  • 98) 0.830 532 592 528 313 548 8 × 2 = 1 + 0.661 065 185 056 627 097 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:


0.000 000 000 000 014 365 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0010 0010 1101 1011 0111 1001 0001 0111 1000 1100 11(2)

6. Positive number before normalization:

0.000 000 000 000 014 365 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0010 0010 1101 1011 0111 1001 0001 0111 1000 1100 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:


0.000 000 000 000 014 365 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0010 0010 1101 1011 0111 1001 0001 0111 1000 1100 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0010 0010 1101 1011 0111 1001 0001 0111 1000 1100 11(2) × 20 =

1.0000 0010 1100 1000 1011 0110 1101 1110 0100 0101 1110 0011 0011(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 1100 1000 1011 0110 1101 1110 0100 0101 1110 0011 0011


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 1100 1000 1011 0110 1101 1110 0100 0101 1110 0011 0011 =

0000 0010 1100 1000 1011 0110 1101 1110 0100 0101 1110 0011 0011


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 1100 1000 1011 0110 1101 1110 0100 0101 1110 0011 0011


Decimal number -0.000 000 000 000 014 365 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1100 1000 1011 0110 1101 1110 0100 0101 1110 0011 0011

» Convert decimal -0.000 000 000 000 014 357 6 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