-0.000 000 000 010 068 461 036 74 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

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

|-0.000 000 000 010 068 461 036 74| = 0.000 000 000 010 068 461 036 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: 0.000 000 000 010 068 461 036 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) 0.000 000 000 010 068 461 036 74 × 2 = 0 + 0.000 000 000 020 136 922 073 48;
  • 2) 0.000 000 000 020 136 922 073 48 × 2 = 0 + 0.000 000 000 040 273 844 146 96;
  • 3) 0.000 000 000 040 273 844 146 96 × 2 = 0 + 0.000 000 000 080 547 688 293 92;
  • 4) 0.000 000 000 080 547 688 293 92 × 2 = 0 + 0.000 000 000 161 095 376 587 84;
  • 5) 0.000 000 000 161 095 376 587 84 × 2 = 0 + 0.000 000 000 322 190 753 175 68;
  • 6) 0.000 000 000 322 190 753 175 68 × 2 = 0 + 0.000 000 000 644 381 506 351 36;
  • 7) 0.000 000 000 644 381 506 351 36 × 2 = 0 + 0.000 000 001 288 763 012 702 72;
  • 8) 0.000 000 001 288 763 012 702 72 × 2 = 0 + 0.000 000 002 577 526 025 405 44;
  • 9) 0.000 000 002 577 526 025 405 44 × 2 = 0 + 0.000 000 005 155 052 050 810 88;
  • 10) 0.000 000 005 155 052 050 810 88 × 2 = 0 + 0.000 000 010 310 104 101 621 76;
  • 11) 0.000 000 010 310 104 101 621 76 × 2 = 0 + 0.000 000 020 620 208 203 243 52;
  • 12) 0.000 000 020 620 208 203 243 52 × 2 = 0 + 0.000 000 041 240 416 406 487 04;
  • 13) 0.000 000 041 240 416 406 487 04 × 2 = 0 + 0.000 000 082 480 832 812 974 08;
  • 14) 0.000 000 082 480 832 812 974 08 × 2 = 0 + 0.000 000 164 961 665 625 948 16;
  • 15) 0.000 000 164 961 665 625 948 16 × 2 = 0 + 0.000 000 329 923 331 251 896 32;
  • 16) 0.000 000 329 923 331 251 896 32 × 2 = 0 + 0.000 000 659 846 662 503 792 64;
  • 17) 0.000 000 659 846 662 503 792 64 × 2 = 0 + 0.000 001 319 693 325 007 585 28;
  • 18) 0.000 001 319 693 325 007 585 28 × 2 = 0 + 0.000 002 639 386 650 015 170 56;
  • 19) 0.000 002 639 386 650 015 170 56 × 2 = 0 + 0.000 005 278 773 300 030 341 12;
  • 20) 0.000 005 278 773 300 030 341 12 × 2 = 0 + 0.000 010 557 546 600 060 682 24;
  • 21) 0.000 010 557 546 600 060 682 24 × 2 = 0 + 0.000 021 115 093 200 121 364 48;
  • 22) 0.000 021 115 093 200 121 364 48 × 2 = 0 + 0.000 042 230 186 400 242 728 96;
  • 23) 0.000 042 230 186 400 242 728 96 × 2 = 0 + 0.000 084 460 372 800 485 457 92;
  • 24) 0.000 084 460 372 800 485 457 92 × 2 = 0 + 0.000 168 920 745 600 970 915 84;
  • 25) 0.000 168 920 745 600 970 915 84 × 2 = 0 + 0.000 337 841 491 201 941 831 68;
  • 26) 0.000 337 841 491 201 941 831 68 × 2 = 0 + 0.000 675 682 982 403 883 663 36;
  • 27) 0.000 675 682 982 403 883 663 36 × 2 = 0 + 0.001 351 365 964 807 767 326 72;
  • 28) 0.001 351 365 964 807 767 326 72 × 2 = 0 + 0.002 702 731 929 615 534 653 44;
  • 29) 0.002 702 731 929 615 534 653 44 × 2 = 0 + 0.005 405 463 859 231 069 306 88;
  • 30) 0.005 405 463 859 231 069 306 88 × 2 = 0 + 0.010 810 927 718 462 138 613 76;
  • 31) 0.010 810 927 718 462 138 613 76 × 2 = 0 + 0.021 621 855 436 924 277 227 52;
  • 32) 0.021 621 855 436 924 277 227 52 × 2 = 0 + 0.043 243 710 873 848 554 455 04;
  • 33) 0.043 243 710 873 848 554 455 04 × 2 = 0 + 0.086 487 421 747 697 108 910 08;
  • 34) 0.086 487 421 747 697 108 910 08 × 2 = 0 + 0.172 974 843 495 394 217 820 16;
  • 35) 0.172 974 843 495 394 217 820 16 × 2 = 0 + 0.345 949 686 990 788 435 640 32;
  • 36) 0.345 949 686 990 788 435 640 32 × 2 = 0 + 0.691 899 373 981 576 871 280 64;
  • 37) 0.691 899 373 981 576 871 280 64 × 2 = 1 + 0.383 798 747 963 153 742 561 28;
  • 38) 0.383 798 747 963 153 742 561 28 × 2 = 0 + 0.767 597 495 926 307 485 122 56;
  • 39) 0.767 597 495 926 307 485 122 56 × 2 = 1 + 0.535 194 991 852 614 970 245 12;
  • 40) 0.535 194 991 852 614 970 245 12 × 2 = 1 + 0.070 389 983 705 229 940 490 24;
  • 41) 0.070 389 983 705 229 940 490 24 × 2 = 0 + 0.140 779 967 410 459 880 980 48;
  • 42) 0.140 779 967 410 459 880 980 48 × 2 = 0 + 0.281 559 934 820 919 761 960 96;
  • 43) 0.281 559 934 820 919 761 960 96 × 2 = 0 + 0.563 119 869 641 839 523 921 92;
  • 44) 0.563 119 869 641 839 523 921 92 × 2 = 1 + 0.126 239 739 283 679 047 843 84;
  • 45) 0.126 239 739 283 679 047 843 84 × 2 = 0 + 0.252 479 478 567 358 095 687 68;
  • 46) 0.252 479 478 567 358 095 687 68 × 2 = 0 + 0.504 958 957 134 716 191 375 36;
  • 47) 0.504 958 957 134 716 191 375 36 × 2 = 1 + 0.009 917 914 269 432 382 750 72;
  • 48) 0.009 917 914 269 432 382 750 72 × 2 = 0 + 0.019 835 828 538 864 765 501 44;
  • 49) 0.019 835 828 538 864 765 501 44 × 2 = 0 + 0.039 671 657 077 729 531 002 88;
  • 50) 0.039 671 657 077 729 531 002 88 × 2 = 0 + 0.079 343 314 155 459 062 005 76;
  • 51) 0.079 343 314 155 459 062 005 76 × 2 = 0 + 0.158 686 628 310 918 124 011 52;
  • 52) 0.158 686 628 310 918 124 011 52 × 2 = 0 + 0.317 373 256 621 836 248 023 04;
  • 53) 0.317 373 256 621 836 248 023 04 × 2 = 0 + 0.634 746 513 243 672 496 046 08;
  • 54) 0.634 746 513 243 672 496 046 08 × 2 = 1 + 0.269 493 026 487 344 992 092 16;
  • 55) 0.269 493 026 487 344 992 092 16 × 2 = 0 + 0.538 986 052 974 689 984 184 32;
  • 56) 0.538 986 052 974 689 984 184 32 × 2 = 1 + 0.077 972 105 949 379 968 368 64;
  • 57) 0.077 972 105 949 379 968 368 64 × 2 = 0 + 0.155 944 211 898 759 936 737 28;
  • 58) 0.155 944 211 898 759 936 737 28 × 2 = 0 + 0.311 888 423 797 519 873 474 56;
  • 59) 0.311 888 423 797 519 873 474 56 × 2 = 0 + 0.623 776 847 595 039 746 949 12;
  • 60) 0.623 776 847 595 039 746 949 12 × 2 = 1 + 0.247 553 695 190 079 493 898 24;
  • 61) 0.247 553 695 190 079 493 898 24 × 2 = 0 + 0.495 107 390 380 158 987 796 48;
  • 62) 0.495 107 390 380 158 987 796 48 × 2 = 0 + 0.990 214 780 760 317 975 592 96;
  • 63) 0.990 214 780 760 317 975 592 96 × 2 = 1 + 0.980 429 561 520 635 951 185 92;
  • 64) 0.980 429 561 520 635 951 185 92 × 2 = 1 + 0.960 859 123 041 271 902 371 84;
  • 65) 0.960 859 123 041 271 902 371 84 × 2 = 1 + 0.921 718 246 082 543 804 743 68;
  • 66) 0.921 718 246 082 543 804 743 68 × 2 = 1 + 0.843 436 492 165 087 609 487 36;
  • 67) 0.843 436 492 165 087 609 487 36 × 2 = 1 + 0.686 872 984 330 175 218 974 72;
  • 68) 0.686 872 984 330 175 218 974 72 × 2 = 1 + 0.373 745 968 660 350 437 949 44;
  • 69) 0.373 745 968 660 350 437 949 44 × 2 = 0 + 0.747 491 937 320 700 875 898 88;
  • 70) 0.747 491 937 320 700 875 898 88 × 2 = 1 + 0.494 983 874 641 401 751 797 76;
  • 71) 0.494 983 874 641 401 751 797 76 × 2 = 0 + 0.989 967 749 282 803 503 595 52;
  • 72) 0.989 967 749 282 803 503 595 52 × 2 = 1 + 0.979 935 498 565 607 007 191 04;
  • 73) 0.979 935 498 565 607 007 191 04 × 2 = 1 + 0.959 870 997 131 214 014 382 08;
  • 74) 0.959 870 997 131 214 014 382 08 × 2 = 1 + 0.919 741 994 262 428 028 764 16;
  • 75) 0.919 741 994 262 428 028 764 16 × 2 = 1 + 0.839 483 988 524 856 057 528 32;
  • 76) 0.839 483 988 524 856 057 528 32 × 2 = 1 + 0.678 967 977 049 712 115 056 64;
  • 77) 0.678 967 977 049 712 115 056 64 × 2 = 1 + 0.357 935 954 099 424 230 113 28;
  • 78) 0.357 935 954 099 424 230 113 28 × 2 = 0 + 0.715 871 908 198 848 460 226 56;
  • 79) 0.715 871 908 198 848 460 226 56 × 2 = 1 + 0.431 743 816 397 696 920 453 12;
  • 80) 0.431 743 816 397 696 920 453 12 × 2 = 0 + 0.863 487 632 795 393 840 906 24;
  • 81) 0.863 487 632 795 393 840 906 24 × 2 = 1 + 0.726 975 265 590 787 681 812 48;
  • 82) 0.726 975 265 590 787 681 812 48 × 2 = 1 + 0.453 950 531 181 575 363 624 96;
  • 83) 0.453 950 531 181 575 363 624 96 × 2 = 0 + 0.907 901 062 363 150 727 249 92;
  • 84) 0.907 901 062 363 150 727 249 92 × 2 = 1 + 0.815 802 124 726 301 454 499 84;
  • 85) 0.815 802 124 726 301 454 499 84 × 2 = 1 + 0.631 604 249 452 602 908 999 68;
  • 86) 0.631 604 249 452 602 908 999 68 × 2 = 1 + 0.263 208 498 905 205 817 999 36;
  • 87) 0.263 208 498 905 205 817 999 36 × 2 = 0 + 0.526 416 997 810 411 635 998 72;
  • 88) 0.526 416 997 810 411 635 998 72 × 2 = 1 + 0.052 833 995 620 823 271 997 44;
  • 89) 0.052 833 995 620 823 271 997 44 × 2 = 0 + 0.105 667 991 241 646 543 994 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 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 0101 1111 1010 1101 1101 0(2)

6. Positive number before normalization:

0.000 000 000 010 068 461 036 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 0101 1111 1010 1101 1101 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 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 0101 1111 1010 1101 1101 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0011 1111 0101 1111 1010 1101 1101 0(2) × 20 =

1.0110 0010 0100 0000 1010 0010 0111 1110 1011 1111 0101 1011 1010(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 1011 1111 0101 1011 1010


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 1011 1111 0101 1011 1010 =

0110 0010 0100 0000 1010 0010 0111 1110 1011 1111 0101 1011 1010


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 1011 1111 0101 1011 1010


Decimal number -0.000 000 000 010 068 461 036 74 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1010 - 0110 0010 0100 0000 1010 0010 0111 1110 1011 1111 0101 1011 1010

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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