-0.000 000 000 010 068 461 037 74 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 010 068 461 037 74(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
-0.000 000 000 010 068 461 037 74(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:

|-0.000 000 000 010 068 461 037 74| = 0.000 000 000 010 068 461 037 74


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: 0.000 000 000 010 068 461 037 74.

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) 0.000 000 000 010 068 461 037 74 × 2 = 0 + 0.000 000 000 020 136 922 075 48;
  • 2) 0.000 000 000 020 136 922 075 48 × 2 = 0 + 0.000 000 000 040 273 844 150 96;
  • 3) 0.000 000 000 040 273 844 150 96 × 2 = 0 + 0.000 000 000 080 547 688 301 92;
  • 4) 0.000 000 000 080 547 688 301 92 × 2 = 0 + 0.000 000 000 161 095 376 603 84;
  • 5) 0.000 000 000 161 095 376 603 84 × 2 = 0 + 0.000 000 000 322 190 753 207 68;
  • 6) 0.000 000 000 322 190 753 207 68 × 2 = 0 + 0.000 000 000 644 381 506 415 36;
  • 7) 0.000 000 000 644 381 506 415 36 × 2 = 0 + 0.000 000 001 288 763 012 830 72;
  • 8) 0.000 000 001 288 763 012 830 72 × 2 = 0 + 0.000 000 002 577 526 025 661 44;
  • 9) 0.000 000 002 577 526 025 661 44 × 2 = 0 + 0.000 000 005 155 052 051 322 88;
  • 10) 0.000 000 005 155 052 051 322 88 × 2 = 0 + 0.000 000 010 310 104 102 645 76;
  • 11) 0.000 000 010 310 104 102 645 76 × 2 = 0 + 0.000 000 020 620 208 205 291 52;
  • 12) 0.000 000 020 620 208 205 291 52 × 2 = 0 + 0.000 000 041 240 416 410 583 04;
  • 13) 0.000 000 041 240 416 410 583 04 × 2 = 0 + 0.000 000 082 480 832 821 166 08;
  • 14) 0.000 000 082 480 832 821 166 08 × 2 = 0 + 0.000 000 164 961 665 642 332 16;
  • 15) 0.000 000 164 961 665 642 332 16 × 2 = 0 + 0.000 000 329 923 331 284 664 32;
  • 16) 0.000 000 329 923 331 284 664 32 × 2 = 0 + 0.000 000 659 846 662 569 328 64;
  • 17) 0.000 000 659 846 662 569 328 64 × 2 = 0 + 0.000 001 319 693 325 138 657 28;
  • 18) 0.000 001 319 693 325 138 657 28 × 2 = 0 + 0.000 002 639 386 650 277 314 56;
  • 19) 0.000 002 639 386 650 277 314 56 × 2 = 0 + 0.000 005 278 773 300 554 629 12;
  • 20) 0.000 005 278 773 300 554 629 12 × 2 = 0 + 0.000 010 557 546 601 109 258 24;
  • 21) 0.000 010 557 546 601 109 258 24 × 2 = 0 + 0.000 021 115 093 202 218 516 48;
  • 22) 0.000 021 115 093 202 218 516 48 × 2 = 0 + 0.000 042 230 186 404 437 032 96;
  • 23) 0.000 042 230 186 404 437 032 96 × 2 = 0 + 0.000 084 460 372 808 874 065 92;
  • 24) 0.000 084 460 372 808 874 065 92 × 2 = 0 + 0.000 168 920 745 617 748 131 84;
  • 25) 0.000 168 920 745 617 748 131 84 × 2 = 0 + 0.000 337 841 491 235 496 263 68;
  • 26) 0.000 337 841 491 235 496 263 68 × 2 = 0 + 0.000 675 682 982 470 992 527 36;
  • 27) 0.000 675 682 982 470 992 527 36 × 2 = 0 + 0.001 351 365 964 941 985 054 72;
  • 28) 0.001 351 365 964 941 985 054 72 × 2 = 0 + 0.002 702 731 929 883 970 109 44;
  • 29) 0.002 702 731 929 883 970 109 44 × 2 = 0 + 0.005 405 463 859 767 940 218 88;
  • 30) 0.005 405 463 859 767 940 218 88 × 2 = 0 + 0.010 810 927 719 535 880 437 76;
  • 31) 0.010 810 927 719 535 880 437 76 × 2 = 0 + 0.021 621 855 439 071 760 875 52;
  • 32) 0.021 621 855 439 071 760 875 52 × 2 = 0 + 0.043 243 710 878 143 521 751 04;
  • 33) 0.043 243 710 878 143 521 751 04 × 2 = 0 + 0.086 487 421 756 287 043 502 08;
  • 34) 0.086 487 421 756 287 043 502 08 × 2 = 0 + 0.172 974 843 512 574 087 004 16;
  • 35) 0.172 974 843 512 574 087 004 16 × 2 = 0 + 0.345 949 687 025 148 174 008 32;
  • 36) 0.345 949 687 025 148 174 008 32 × 2 = 0 + 0.691 899 374 050 296 348 016 64;
  • 37) 0.691 899 374 050 296 348 016 64 × 2 = 1 + 0.383 798 748 100 592 696 033 28;
  • 38) 0.383 798 748 100 592 696 033 28 × 2 = 0 + 0.767 597 496 201 185 392 066 56;
  • 39) 0.767 597 496 201 185 392 066 56 × 2 = 1 + 0.535 194 992 402 370 784 133 12;
  • 40) 0.535 194 992 402 370 784 133 12 × 2 = 1 + 0.070 389 984 804 741 568 266 24;
  • 41) 0.070 389 984 804 741 568 266 24 × 2 = 0 + 0.140 779 969 609 483 136 532 48;
  • 42) 0.140 779 969 609 483 136 532 48 × 2 = 0 + 0.281 559 939 218 966 273 064 96;
  • 43) 0.281 559 939 218 966 273 064 96 × 2 = 0 + 0.563 119 878 437 932 546 129 92;
  • 44) 0.563 119 878 437 932 546 129 92 × 2 = 1 + 0.126 239 756 875 865 092 259 84;
  • 45) 0.126 239 756 875 865 092 259 84 × 2 = 0 + 0.252 479 513 751 730 184 519 68;
  • 46) 0.252 479 513 751 730 184 519 68 × 2 = 0 + 0.504 959 027 503 460 369 039 36;
  • 47) 0.504 959 027 503 460 369 039 36 × 2 = 1 + 0.009 918 055 006 920 738 078 72;
  • 48) 0.009 918 055 006 920 738 078 72 × 2 = 0 + 0.019 836 110 013 841 476 157 44;
  • 49) 0.019 836 110 013 841 476 157 44 × 2 = 0 + 0.039 672 220 027 682 952 314 88;
  • 50) 0.039 672 220 027 682 952 314 88 × 2 = 0 + 0.079 344 440 055 365 904 629 76;
  • 51) 0.079 344 440 055 365 904 629 76 × 2 = 0 + 0.158 688 880 110 731 809 259 52;
  • 52) 0.158 688 880 110 731 809 259 52 × 2 = 0 + 0.317 377 760 221 463 618 519 04;
  • 53) 0.317 377 760 221 463 618 519 04 × 2 = 0 + 0.634 755 520 442 927 237 038 08;
  • 54) 0.634 755 520 442 927 237 038 08 × 2 = 1 + 0.269 511 040 885 854 474 076 16;
  • 55) 0.269 511 040 885 854 474 076 16 × 2 = 0 + 0.539 022 081 771 708 948 152 32;
  • 56) 0.539 022 081 771 708 948 152 32 × 2 = 1 + 0.078 044 163 543 417 896 304 64;
  • 57) 0.078 044 163 543 417 896 304 64 × 2 = 0 + 0.156 088 327 086 835 792 609 28;
  • 58) 0.156 088 327 086 835 792 609 28 × 2 = 0 + 0.312 176 654 173 671 585 218 56;
  • 59) 0.312 176 654 173 671 585 218 56 × 2 = 0 + 0.624 353 308 347 343 170 437 12;
  • 60) 0.624 353 308 347 343 170 437 12 × 2 = 1 + 0.248 706 616 694 686 340 874 24;
  • 61) 0.248 706 616 694 686 340 874 24 × 2 = 0 + 0.497 413 233 389 372 681 748 48;
  • 62) 0.497 413 233 389 372 681 748 48 × 2 = 0 + 0.994 826 466 778 745 363 496 96;
  • 63) 0.994 826 466 778 745 363 496 96 × 2 = 1 + 0.989 652 933 557 490 726 993 92;
  • 64) 0.989 652 933 557 490 726 993 92 × 2 = 1 + 0.979 305 867 114 981 453 987 84;
  • 65) 0.979 305 867 114 981 453 987 84 × 2 = 1 + 0.958 611 734 229 962 907 975 68;
  • 66) 0.958 611 734 229 962 907 975 68 × 2 = 1 + 0.917 223 468 459 925 815 951 36;
  • 67) 0.917 223 468 459 925 815 951 36 × 2 = 1 + 0.834 446 936 919 851 631 902 72;
  • 68) 0.834 446 936 919 851 631 902 72 × 2 = 1 + 0.668 893 873 839 703 263 805 44;
  • 69) 0.668 893 873 839 703 263 805 44 × 2 = 1 + 0.337 787 747 679 406 527 610 88;
  • 70) 0.337 787 747 679 406 527 610 88 × 2 = 0 + 0.675 575 495 358 813 055 221 76;
  • 71) 0.675 575 495 358 813 055 221 76 × 2 = 1 + 0.351 150 990 717 626 110 443 52;
  • 72) 0.351 150 990 717 626 110 443 52 × 2 = 0 + 0.702 301 981 435 252 220 887 04;
  • 73) 0.702 301 981 435 252 220 887 04 × 2 = 1 + 0.404 603 962 870 504 441 774 08;
  • 74) 0.404 603 962 870 504 441 774 08 × 2 = 0 + 0.809 207 925 741 008 883 548 16;
  • 75) 0.809 207 925 741 008 883 548 16 × 2 = 1 + 0.618 415 851 482 017 767 096 32;
  • 76) 0.618 415 851 482 017 767 096 32 × 2 = 1 + 0.236 831 702 964 035 534 192 64;
  • 77) 0.236 831 702 964 035 534 192 64 × 2 = 0 + 0.473 663 405 928 071 068 385 28;
  • 78) 0.473 663 405 928 071 068 385 28 × 2 = 0 + 0.947 326 811 856 142 136 770 56;
  • 79) 0.947 326 811 856 142 136 770 56 × 2 = 1 + 0.894 653 623 712 284 273 541 12;
  • 80) 0.894 653 623 712 284 273 541 12 × 2 = 1 + 0.789 307 247 424 568 547 082 24;
  • 81) 0.789 307 247 424 568 547 082 24 × 2 = 1 + 0.578 614 494 849 137 094 164 48;
  • 82) 0.578 614 494 849 137 094 164 48 × 2 = 1 + 0.157 228 989 698 274 188 328 96;
  • 83) 0.157 228 989 698 274 188 328 96 × 2 = 0 + 0.314 457 979 396 548 376 657 92;
  • 84) 0.314 457 979 396 548 376 657 92 × 2 = 0 + 0.628 915 958 793 096 753 315 84;
  • 85) 0.628 915 958 793 096 753 315 84 × 2 = 1 + 0.257 831 917 586 193 506 631 68;
  • 86) 0.257 831 917 586 193 506 631 68 × 2 = 0 + 0.515 663 835 172 387 013 263 36;
  • 87) 0.515 663 835 172 387 013 263 36 × 2 = 1 + 0.031 327 670 344 774 026 526 72;
  • 88) 0.031 327 670 344 774 026 526 72 × 2 = 0 + 0.062 655 340 689 548 053 053 44;
  • 89) 0.062 655 340 689 548 053 053 44 × 2 = 0 + 0.125 310 681 379 096 106 106 88;

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:


0.000 000 000 010 068 461 037 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1010 1011 0011 1100 1010 0(2)

6. Positive number before normalization:

0.000 000 000 010 068 461 037 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1010 1011 0011 1100 1010 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 37 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 010 068 461 037 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1010 1011 0011 1100 1010 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 1010 1011 0011 1100 1010 0(2) × 20 =

1.0110 0010 0100 0000 1010 0010 0111 1111 0101 0110 0111 1001 0100(2) × 2-37


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): -37

Mantissa (not normalized):
1.0110 0010 0100 0000 1010 0010 0111 1111 0101 0110 0111 1001 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-37 + 2(11-1) - 1 =

(-37 + 1 023)(10) =

986(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;
  • 986 ÷ 2 = 493 + 0;
  • 493 ÷ 2 = 246 + 1;
  • 246 ÷ 2 = 123 + 0;
  • 123 ÷ 2 = 61 + 1;
  • 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) =

986(10) =

011 1101 1010(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. 0110 0010 0100 0000 1010 0010 0111 1111 0101 0110 0111 1001 0100 =

0110 0010 0100 0000 1010 0010 0111 1111 0101 0110 0111 1001 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 1010

Mantissa (52 bits) =
0110 0010 0100 0000 1010 0010 0111 1111 0101 0110 0111 1001 0100


Decimal number -0.000 000 000 010 068 461 037 74 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1010 - 0110 0010 0100 0000 1010 0010 0111 1111 0101 0110 0111 1001 0100

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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