-0.000 000 000 000 014 381 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 381 8(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
-0.000 000 000 000 014 381 8(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:

|-0.000 000 000 000 014 381 8| = 0.000 000 000 000 014 381 8


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: 0.000 000 000 000 014 381 8.

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) 0.000 000 000 000 014 381 8 × 2 = 0 + 0.000 000 000 000 028 763 6;
  • 2) 0.000 000 000 000 028 763 6 × 2 = 0 + 0.000 000 000 000 057 527 2;
  • 3) 0.000 000 000 000 057 527 2 × 2 = 0 + 0.000 000 000 000 115 054 4;
  • 4) 0.000 000 000 000 115 054 4 × 2 = 0 + 0.000 000 000 000 230 108 8;
  • 5) 0.000 000 000 000 230 108 8 × 2 = 0 + 0.000 000 000 000 460 217 6;
  • 6) 0.000 000 000 000 460 217 6 × 2 = 0 + 0.000 000 000 000 920 435 2;
  • 7) 0.000 000 000 000 920 435 2 × 2 = 0 + 0.000 000 000 001 840 870 4;
  • 8) 0.000 000 000 001 840 870 4 × 2 = 0 + 0.000 000 000 003 681 740 8;
  • 9) 0.000 000 000 003 681 740 8 × 2 = 0 + 0.000 000 000 007 363 481 6;
  • 10) 0.000 000 000 007 363 481 6 × 2 = 0 + 0.000 000 000 014 726 963 2;
  • 11) 0.000 000 000 014 726 963 2 × 2 = 0 + 0.000 000 000 029 453 926 4;
  • 12) 0.000 000 000 029 453 926 4 × 2 = 0 + 0.000 000 000 058 907 852 8;
  • 13) 0.000 000 000 058 907 852 8 × 2 = 0 + 0.000 000 000 117 815 705 6;
  • 14) 0.000 000 000 117 815 705 6 × 2 = 0 + 0.000 000 000 235 631 411 2;
  • 15) 0.000 000 000 235 631 411 2 × 2 = 0 + 0.000 000 000 471 262 822 4;
  • 16) 0.000 000 000 471 262 822 4 × 2 = 0 + 0.000 000 000 942 525 644 8;
  • 17) 0.000 000 000 942 525 644 8 × 2 = 0 + 0.000 000 001 885 051 289 6;
  • 18) 0.000 000 001 885 051 289 6 × 2 = 0 + 0.000 000 003 770 102 579 2;
  • 19) 0.000 000 003 770 102 579 2 × 2 = 0 + 0.000 000 007 540 205 158 4;
  • 20) 0.000 000 007 540 205 158 4 × 2 = 0 + 0.000 000 015 080 410 316 8;
  • 21) 0.000 000 015 080 410 316 8 × 2 = 0 + 0.000 000 030 160 820 633 6;
  • 22) 0.000 000 030 160 820 633 6 × 2 = 0 + 0.000 000 060 321 641 267 2;
  • 23) 0.000 000 060 321 641 267 2 × 2 = 0 + 0.000 000 120 643 282 534 4;
  • 24) 0.000 000 120 643 282 534 4 × 2 = 0 + 0.000 000 241 286 565 068 8;
  • 25) 0.000 000 241 286 565 068 8 × 2 = 0 + 0.000 000 482 573 130 137 6;
  • 26) 0.000 000 482 573 130 137 6 × 2 = 0 + 0.000 000 965 146 260 275 2;
  • 27) 0.000 000 965 146 260 275 2 × 2 = 0 + 0.000 001 930 292 520 550 4;
  • 28) 0.000 001 930 292 520 550 4 × 2 = 0 + 0.000 003 860 585 041 100 8;
  • 29) 0.000 003 860 585 041 100 8 × 2 = 0 + 0.000 007 721 170 082 201 6;
  • 30) 0.000 007 721 170 082 201 6 × 2 = 0 + 0.000 015 442 340 164 403 2;
  • 31) 0.000 015 442 340 164 403 2 × 2 = 0 + 0.000 030 884 680 328 806 4;
  • 32) 0.000 030 884 680 328 806 4 × 2 = 0 + 0.000 061 769 360 657 612 8;
  • 33) 0.000 061 769 360 657 612 8 × 2 = 0 + 0.000 123 538 721 315 225 6;
  • 34) 0.000 123 538 721 315 225 6 × 2 = 0 + 0.000 247 077 442 630 451 2;
  • 35) 0.000 247 077 442 630 451 2 × 2 = 0 + 0.000 494 154 885 260 902 4;
  • 36) 0.000 494 154 885 260 902 4 × 2 = 0 + 0.000 988 309 770 521 804 8;
  • 37) 0.000 988 309 770 521 804 8 × 2 = 0 + 0.001 976 619 541 043 609 6;
  • 38) 0.001 976 619 541 043 609 6 × 2 = 0 + 0.003 953 239 082 087 219 2;
  • 39) 0.003 953 239 082 087 219 2 × 2 = 0 + 0.007 906 478 164 174 438 4;
  • 40) 0.007 906 478 164 174 438 4 × 2 = 0 + 0.015 812 956 328 348 876 8;
  • 41) 0.015 812 956 328 348 876 8 × 2 = 0 + 0.031 625 912 656 697 753 6;
  • 42) 0.031 625 912 656 697 753 6 × 2 = 0 + 0.063 251 825 313 395 507 2;
  • 43) 0.063 251 825 313 395 507 2 × 2 = 0 + 0.126 503 650 626 791 014 4;
  • 44) 0.126 503 650 626 791 014 4 × 2 = 0 + 0.253 007 301 253 582 028 8;
  • 45) 0.253 007 301 253 582 028 8 × 2 = 0 + 0.506 014 602 507 164 057 6;
  • 46) 0.506 014 602 507 164 057 6 × 2 = 1 + 0.012 029 205 014 328 115 2;
  • 47) 0.012 029 205 014 328 115 2 × 2 = 0 + 0.024 058 410 028 656 230 4;
  • 48) 0.024 058 410 028 656 230 4 × 2 = 0 + 0.048 116 820 057 312 460 8;
  • 49) 0.048 116 820 057 312 460 8 × 2 = 0 + 0.096 233 640 114 624 921 6;
  • 50) 0.096 233 640 114 624 921 6 × 2 = 0 + 0.192 467 280 229 249 843 2;
  • 51) 0.192 467 280 229 249 843 2 × 2 = 0 + 0.384 934 560 458 499 686 4;
  • 52) 0.384 934 560 458 499 686 4 × 2 = 0 + 0.769 869 120 916 999 372 8;
  • 53) 0.769 869 120 916 999 372 8 × 2 = 1 + 0.539 738 241 833 998 745 6;
  • 54) 0.539 738 241 833 998 745 6 × 2 = 1 + 0.079 476 483 667 997 491 2;
  • 55) 0.079 476 483 667 997 491 2 × 2 = 0 + 0.158 952 967 335 994 982 4;
  • 56) 0.158 952 967 335 994 982 4 × 2 = 0 + 0.317 905 934 671 989 964 8;
  • 57) 0.317 905 934 671 989 964 8 × 2 = 0 + 0.635 811 869 343 979 929 6;
  • 58) 0.635 811 869 343 979 929 6 × 2 = 1 + 0.271 623 738 687 959 859 2;
  • 59) 0.271 623 738 687 959 859 2 × 2 = 0 + 0.543 247 477 375 919 718 4;
  • 60) 0.543 247 477 375 919 718 4 × 2 = 1 + 0.086 494 954 751 839 436 8;
  • 61) 0.086 494 954 751 839 436 8 × 2 = 0 + 0.172 989 909 503 678 873 6;
  • 62) 0.172 989 909 503 678 873 6 × 2 = 0 + 0.345 979 819 007 357 747 2;
  • 63) 0.345 979 819 007 357 747 2 × 2 = 0 + 0.691 959 638 014 715 494 4;
  • 64) 0.691 959 638 014 715 494 4 × 2 = 1 + 0.383 919 276 029 430 988 8;
  • 65) 0.383 919 276 029 430 988 8 × 2 = 0 + 0.767 838 552 058 861 977 6;
  • 66) 0.767 838 552 058 861 977 6 × 2 = 1 + 0.535 677 104 117 723 955 2;
  • 67) 0.535 677 104 117 723 955 2 × 2 = 1 + 0.071 354 208 235 447 910 4;
  • 68) 0.071 354 208 235 447 910 4 × 2 = 0 + 0.142 708 416 470 895 820 8;
  • 69) 0.142 708 416 470 895 820 8 × 2 = 0 + 0.285 416 832 941 791 641 6;
  • 70) 0.285 416 832 941 791 641 6 × 2 = 0 + 0.570 833 665 883 583 283 2;
  • 71) 0.570 833 665 883 583 283 2 × 2 = 1 + 0.141 667 331 767 166 566 4;
  • 72) 0.141 667 331 767 166 566 4 × 2 = 0 + 0.283 334 663 534 333 132 8;
  • 73) 0.283 334 663 534 333 132 8 × 2 = 0 + 0.566 669 327 068 666 265 6;
  • 74) 0.566 669 327 068 666 265 6 × 2 = 1 + 0.133 338 654 137 332 531 2;
  • 75) 0.133 338 654 137 332 531 2 × 2 = 0 + 0.266 677 308 274 665 062 4;
  • 76) 0.266 677 308 274 665 062 4 × 2 = 0 + 0.533 354 616 549 330 124 8;
  • 77) 0.533 354 616 549 330 124 8 × 2 = 1 + 0.066 709 233 098 660 249 6;
  • 78) 0.066 709 233 098 660 249 6 × 2 = 0 + 0.133 418 466 197 320 499 2;
  • 79) 0.133 418 466 197 320 499 2 × 2 = 0 + 0.266 836 932 394 640 998 4;
  • 80) 0.266 836 932 394 640 998 4 × 2 = 0 + 0.533 673 864 789 281 996 8;
  • 81) 0.533 673 864 789 281 996 8 × 2 = 1 + 0.067 347 729 578 563 993 6;
  • 82) 0.067 347 729 578 563 993 6 × 2 = 0 + 0.134 695 459 157 127 987 2;
  • 83) 0.134 695 459 157 127 987 2 × 2 = 0 + 0.269 390 918 314 255 974 4;
  • 84) 0.269 390 918 314 255 974 4 × 2 = 0 + 0.538 781 836 628 511 948 8;
  • 85) 0.538 781 836 628 511 948 8 × 2 = 1 + 0.077 563 673 257 023 897 6;
  • 86) 0.077 563 673 257 023 897 6 × 2 = 0 + 0.155 127 346 514 047 795 2;
  • 87) 0.155 127 346 514 047 795 2 × 2 = 0 + 0.310 254 693 028 095 590 4;
  • 88) 0.310 254 693 028 095 590 4 × 2 = 0 + 0.620 509 386 056 191 180 8;
  • 89) 0.620 509 386 056 191 180 8 × 2 = 1 + 0.241 018 772 112 382 361 6;
  • 90) 0.241 018 772 112 382 361 6 × 2 = 0 + 0.482 037 544 224 764 723 2;
  • 91) 0.482 037 544 224 764 723 2 × 2 = 0 + 0.964 075 088 449 529 446 4;
  • 92) 0.964 075 088 449 529 446 4 × 2 = 1 + 0.928 150 176 899 058 892 8;
  • 93) 0.928 150 176 899 058 892 8 × 2 = 1 + 0.856 300 353 798 117 785 6;
  • 94) 0.856 300 353 798 117 785 6 × 2 = 1 + 0.712 600 707 596 235 571 2;
  • 95) 0.712 600 707 596 235 571 2 × 2 = 1 + 0.425 201 415 192 471 142 4;
  • 96) 0.425 201 415 192 471 142 4 × 2 = 0 + 0.850 402 830 384 942 284 8;
  • 97) 0.850 402 830 384 942 284 8 × 2 = 1 + 0.700 805 660 769 884 569 6;
  • 98) 0.700 805 660 769 884 569 6 × 2 = 1 + 0.401 611 321 539 769 139 2;

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:


0.000 000 000 000 014 381 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 0101 0001 0110 0010 0100 1000 1000 1000 1001 1110 11(2)

6. Positive number before normalization:

0.000 000 000 000 014 381 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 0101 0001 0110 0010 0100 1000 1000 1000 1001 1110 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:


0.000 000 000 000 014 381 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 0101 0001 0110 0010 0100 1000 1000 1000 1001 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 0101 0001 0110 0010 0100 1000 1000 1000 1001 1110 11(2) × 20 =

1.0000 0011 0001 0100 0101 1000 1001 0010 0010 0010 0010 0111 1011(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 0011 0001 0100 0101 1000 1001 0010 0010 0010 0010 0111 1011


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 0011 0001 0100 0101 1000 1001 0010 0010 0010 0010 0111 1011 =

0000 0011 0001 0100 0101 1000 1001 0010 0010 0010 0010 0111 1011


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 0011 0001 0100 0101 1000 1001 0010 0010 0010 0010 0111 1011


Decimal number -0.000 000 000 000 014 381 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0011 0001 0100 0101 1000 1001 0010 0010 0010 0010 0111 1011

» Convert decimal -0.000 000 000 000 014 388 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