-0.000 000 000 010 068 461 036 99 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 010 068 461 036 99(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
-0.000 000 000 010 068 461 036 99(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:

|-0.000 000 000 010 068 461 036 99| = 0.000 000 000 010 068 461 036 99


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: 0.000 000 000 010 068 461 036 99.

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) 0.000 000 000 010 068 461 036 99 × 2 = 0 + 0.000 000 000 020 136 922 073 98;
  • 2) 0.000 000 000 020 136 922 073 98 × 2 = 0 + 0.000 000 000 040 273 844 147 96;
  • 3) 0.000 000 000 040 273 844 147 96 × 2 = 0 + 0.000 000 000 080 547 688 295 92;
  • 4) 0.000 000 000 080 547 688 295 92 × 2 = 0 + 0.000 000 000 161 095 376 591 84;
  • 5) 0.000 000 000 161 095 376 591 84 × 2 = 0 + 0.000 000 000 322 190 753 183 68;
  • 6) 0.000 000 000 322 190 753 183 68 × 2 = 0 + 0.000 000 000 644 381 506 367 36;
  • 7) 0.000 000 000 644 381 506 367 36 × 2 = 0 + 0.000 000 001 288 763 012 734 72;
  • 8) 0.000 000 001 288 763 012 734 72 × 2 = 0 + 0.000 000 002 577 526 025 469 44;
  • 9) 0.000 000 002 577 526 025 469 44 × 2 = 0 + 0.000 000 005 155 052 050 938 88;
  • 10) 0.000 000 005 155 052 050 938 88 × 2 = 0 + 0.000 000 010 310 104 101 877 76;
  • 11) 0.000 000 010 310 104 101 877 76 × 2 = 0 + 0.000 000 020 620 208 203 755 52;
  • 12) 0.000 000 020 620 208 203 755 52 × 2 = 0 + 0.000 000 041 240 416 407 511 04;
  • 13) 0.000 000 041 240 416 407 511 04 × 2 = 0 + 0.000 000 082 480 832 815 022 08;
  • 14) 0.000 000 082 480 832 815 022 08 × 2 = 0 + 0.000 000 164 961 665 630 044 16;
  • 15) 0.000 000 164 961 665 630 044 16 × 2 = 0 + 0.000 000 329 923 331 260 088 32;
  • 16) 0.000 000 329 923 331 260 088 32 × 2 = 0 + 0.000 000 659 846 662 520 176 64;
  • 17) 0.000 000 659 846 662 520 176 64 × 2 = 0 + 0.000 001 319 693 325 040 353 28;
  • 18) 0.000 001 319 693 325 040 353 28 × 2 = 0 + 0.000 002 639 386 650 080 706 56;
  • 19) 0.000 002 639 386 650 080 706 56 × 2 = 0 + 0.000 005 278 773 300 161 413 12;
  • 20) 0.000 005 278 773 300 161 413 12 × 2 = 0 + 0.000 010 557 546 600 322 826 24;
  • 21) 0.000 010 557 546 600 322 826 24 × 2 = 0 + 0.000 021 115 093 200 645 652 48;
  • 22) 0.000 021 115 093 200 645 652 48 × 2 = 0 + 0.000 042 230 186 401 291 304 96;
  • 23) 0.000 042 230 186 401 291 304 96 × 2 = 0 + 0.000 084 460 372 802 582 609 92;
  • 24) 0.000 084 460 372 802 582 609 92 × 2 = 0 + 0.000 168 920 745 605 165 219 84;
  • 25) 0.000 168 920 745 605 165 219 84 × 2 = 0 + 0.000 337 841 491 210 330 439 68;
  • 26) 0.000 337 841 491 210 330 439 68 × 2 = 0 + 0.000 675 682 982 420 660 879 36;
  • 27) 0.000 675 682 982 420 660 879 36 × 2 = 0 + 0.001 351 365 964 841 321 758 72;
  • 28) 0.001 351 365 964 841 321 758 72 × 2 = 0 + 0.002 702 731 929 682 643 517 44;
  • 29) 0.002 702 731 929 682 643 517 44 × 2 = 0 + 0.005 405 463 859 365 287 034 88;
  • 30) 0.005 405 463 859 365 287 034 88 × 2 = 0 + 0.010 810 927 718 730 574 069 76;
  • 31) 0.010 810 927 718 730 574 069 76 × 2 = 0 + 0.021 621 855 437 461 148 139 52;
  • 32) 0.021 621 855 437 461 148 139 52 × 2 = 0 + 0.043 243 710 874 922 296 279 04;
  • 33) 0.043 243 710 874 922 296 279 04 × 2 = 0 + 0.086 487 421 749 844 592 558 08;
  • 34) 0.086 487 421 749 844 592 558 08 × 2 = 0 + 0.172 974 843 499 689 185 116 16;
  • 35) 0.172 974 843 499 689 185 116 16 × 2 = 0 + 0.345 949 686 999 378 370 232 32;
  • 36) 0.345 949 686 999 378 370 232 32 × 2 = 0 + 0.691 899 373 998 756 740 464 64;
  • 37) 0.691 899 373 998 756 740 464 64 × 2 = 1 + 0.383 798 747 997 513 480 929 28;
  • 38) 0.383 798 747 997 513 480 929 28 × 2 = 0 + 0.767 597 495 995 026 961 858 56;
  • 39) 0.767 597 495 995 026 961 858 56 × 2 = 1 + 0.535 194 991 990 053 923 717 12;
  • 40) 0.535 194 991 990 053 923 717 12 × 2 = 1 + 0.070 389 983 980 107 847 434 24;
  • 41) 0.070 389 983 980 107 847 434 24 × 2 = 0 + 0.140 779 967 960 215 694 868 48;
  • 42) 0.140 779 967 960 215 694 868 48 × 2 = 0 + 0.281 559 935 920 431 389 736 96;
  • 43) 0.281 559 935 920 431 389 736 96 × 2 = 0 + 0.563 119 871 840 862 779 473 92;
  • 44) 0.563 119 871 840 862 779 473 92 × 2 = 1 + 0.126 239 743 681 725 558 947 84;
  • 45) 0.126 239 743 681 725 558 947 84 × 2 = 0 + 0.252 479 487 363 451 117 895 68;
  • 46) 0.252 479 487 363 451 117 895 68 × 2 = 0 + 0.504 958 974 726 902 235 791 36;
  • 47) 0.504 958 974 726 902 235 791 36 × 2 = 1 + 0.009 917 949 453 804 471 582 72;
  • 48) 0.009 917 949 453 804 471 582 72 × 2 = 0 + 0.019 835 898 907 608 943 165 44;
  • 49) 0.019 835 898 907 608 943 165 44 × 2 = 0 + 0.039 671 797 815 217 886 330 88;
  • 50) 0.039 671 797 815 217 886 330 88 × 2 = 0 + 0.079 343 595 630 435 772 661 76;
  • 51) 0.079 343 595 630 435 772 661 76 × 2 = 0 + 0.158 687 191 260 871 545 323 52;
  • 52) 0.158 687 191 260 871 545 323 52 × 2 = 0 + 0.317 374 382 521 743 090 647 04;
  • 53) 0.317 374 382 521 743 090 647 04 × 2 = 0 + 0.634 748 765 043 486 181 294 08;
  • 54) 0.634 748 765 043 486 181 294 08 × 2 = 1 + 0.269 497 530 086 972 362 588 16;
  • 55) 0.269 497 530 086 972 362 588 16 × 2 = 0 + 0.538 995 060 173 944 725 176 32;
  • 56) 0.538 995 060 173 944 725 176 32 × 2 = 1 + 0.077 990 120 347 889 450 352 64;
  • 57) 0.077 990 120 347 889 450 352 64 × 2 = 0 + 0.155 980 240 695 778 900 705 28;
  • 58) 0.155 980 240 695 778 900 705 28 × 2 = 0 + 0.311 960 481 391 557 801 410 56;
  • 59) 0.311 960 481 391 557 801 410 56 × 2 = 0 + 0.623 920 962 783 115 602 821 12;
  • 60) 0.623 920 962 783 115 602 821 12 × 2 = 1 + 0.247 841 925 566 231 205 642 24;
  • 61) 0.247 841 925 566 231 205 642 24 × 2 = 0 + 0.495 683 851 132 462 411 284 48;
  • 62) 0.495 683 851 132 462 411 284 48 × 2 = 0 + 0.991 367 702 264 924 822 568 96;
  • 63) 0.991 367 702 264 924 822 568 96 × 2 = 1 + 0.982 735 404 529 849 645 137 92;
  • 64) 0.982 735 404 529 849 645 137 92 × 2 = 1 + 0.965 470 809 059 699 290 275 84;
  • 65) 0.965 470 809 059 699 290 275 84 × 2 = 1 + 0.930 941 618 119 398 580 551 68;
  • 66) 0.930 941 618 119 398 580 551 68 × 2 = 1 + 0.861 883 236 238 797 161 103 36;
  • 67) 0.861 883 236 238 797 161 103 36 × 2 = 1 + 0.723 766 472 477 594 322 206 72;
  • 68) 0.723 766 472 477 594 322 206 72 × 2 = 1 + 0.447 532 944 955 188 644 413 44;
  • 69) 0.447 532 944 955 188 644 413 44 × 2 = 0 + 0.895 065 889 910 377 288 826 88;
  • 70) 0.895 065 889 910 377 288 826 88 × 2 = 1 + 0.790 131 779 820 754 577 653 76;
  • 71) 0.790 131 779 820 754 577 653 76 × 2 = 1 + 0.580 263 559 641 509 155 307 52;
  • 72) 0.580 263 559 641 509 155 307 52 × 2 = 1 + 0.160 527 119 283 018 310 615 04;
  • 73) 0.160 527 119 283 018 310 615 04 × 2 = 0 + 0.321 054 238 566 036 621 230 08;
  • 74) 0.321 054 238 566 036 621 230 08 × 2 = 0 + 0.642 108 477 132 073 242 460 16;
  • 75) 0.642 108 477 132 073 242 460 16 × 2 = 1 + 0.284 216 954 264 146 484 920 32;
  • 76) 0.284 216 954 264 146 484 920 32 × 2 = 0 + 0.568 433 908 528 292 969 840 64;
  • 77) 0.568 433 908 528 292 969 840 64 × 2 = 1 + 0.136 867 817 056 585 939 681 28;
  • 78) 0.136 867 817 056 585 939 681 28 × 2 = 0 + 0.273 735 634 113 171 879 362 56;
  • 79) 0.273 735 634 113 171 879 362 56 × 2 = 0 + 0.547 471 268 226 343 758 725 12;
  • 80) 0.547 471 268 226 343 758 725 12 × 2 = 1 + 0.094 942 536 452 687 517 450 24;
  • 81) 0.094 942 536 452 687 517 450 24 × 2 = 0 + 0.189 885 072 905 375 034 900 48;
  • 82) 0.189 885 072 905 375 034 900 48 × 2 = 0 + 0.379 770 145 810 750 069 800 96;
  • 83) 0.379 770 145 810 750 069 800 96 × 2 = 0 + 0.759 540 291 621 500 139 601 92;
  • 84) 0.759 540 291 621 500 139 601 92 × 2 = 1 + 0.519 080 583 243 000 279 203 84;
  • 85) 0.519 080 583 243 000 279 203 84 × 2 = 1 + 0.038 161 166 486 000 558 407 68;
  • 86) 0.038 161 166 486 000 558 407 68 × 2 = 0 + 0.076 322 332 972 001 116 815 36;
  • 87) 0.076 322 332 972 001 116 815 36 × 2 = 0 + 0.152 644 665 944 002 233 630 72;
  • 88) 0.152 644 665 944 002 233 630 72 × 2 = 0 + 0.305 289 331 888 004 467 261 44;
  • 89) 0.305 289 331 888 004 467 261 44 × 2 = 0 + 0.610 578 663 776 008 934 522 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:


0.000 000 000 010 068 461 036 99(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 0111 0010 1001 0001 1000 0(2)

6. Positive number before normalization:

0.000 000 000 010 068 461 036 99(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 0111 0010 1001 0001 1000 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:


0.000 000 000 010 068 461 036 99(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 0111 0010 1001 0001 1000 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 0111 0010 1001 0001 1000 0(2) × 20 =

1.0110 0010 0100 0000 1010 0010 0111 1110 1110 0101 0010 0011 0000(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 1110 1110 0101 0010 0011 0000


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 1110 1110 0101 0010 0011 0000 =

0110 0010 0100 0000 1010 0010 0111 1110 1110 0101 0010 0011 0000


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 1110 1110 0101 0010 0011 0000


Decimal number -0.000 000 000 010 068 461 036 99 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1010 - 0110 0010 0100 0000 1010 0010 0111 1110 1110 0101 0010 0011 0000

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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