-0.000 000 000 000 028 422 93 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 028 422 93(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
-0.000 000 000 000 028 422 93(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:

|-0.000 000 000 000 028 422 93| = 0.000 000 000 000 028 422 93


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: 0.000 000 000 000 028 422 93.

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) 0.000 000 000 000 028 422 93 × 2 = 0 + 0.000 000 000 000 056 845 86;
  • 2) 0.000 000 000 000 056 845 86 × 2 = 0 + 0.000 000 000 000 113 691 72;
  • 3) 0.000 000 000 000 113 691 72 × 2 = 0 + 0.000 000 000 000 227 383 44;
  • 4) 0.000 000 000 000 227 383 44 × 2 = 0 + 0.000 000 000 000 454 766 88;
  • 5) 0.000 000 000 000 454 766 88 × 2 = 0 + 0.000 000 000 000 909 533 76;
  • 6) 0.000 000 000 000 909 533 76 × 2 = 0 + 0.000 000 000 001 819 067 52;
  • 7) 0.000 000 000 001 819 067 52 × 2 = 0 + 0.000 000 000 003 638 135 04;
  • 8) 0.000 000 000 003 638 135 04 × 2 = 0 + 0.000 000 000 007 276 270 08;
  • 9) 0.000 000 000 007 276 270 08 × 2 = 0 + 0.000 000 000 014 552 540 16;
  • 10) 0.000 000 000 014 552 540 16 × 2 = 0 + 0.000 000 000 029 105 080 32;
  • 11) 0.000 000 000 029 105 080 32 × 2 = 0 + 0.000 000 000 058 210 160 64;
  • 12) 0.000 000 000 058 210 160 64 × 2 = 0 + 0.000 000 000 116 420 321 28;
  • 13) 0.000 000 000 116 420 321 28 × 2 = 0 + 0.000 000 000 232 840 642 56;
  • 14) 0.000 000 000 232 840 642 56 × 2 = 0 + 0.000 000 000 465 681 285 12;
  • 15) 0.000 000 000 465 681 285 12 × 2 = 0 + 0.000 000 000 931 362 570 24;
  • 16) 0.000 000 000 931 362 570 24 × 2 = 0 + 0.000 000 001 862 725 140 48;
  • 17) 0.000 000 001 862 725 140 48 × 2 = 0 + 0.000 000 003 725 450 280 96;
  • 18) 0.000 000 003 725 450 280 96 × 2 = 0 + 0.000 000 007 450 900 561 92;
  • 19) 0.000 000 007 450 900 561 92 × 2 = 0 + 0.000 000 014 901 801 123 84;
  • 20) 0.000 000 014 901 801 123 84 × 2 = 0 + 0.000 000 029 803 602 247 68;
  • 21) 0.000 000 029 803 602 247 68 × 2 = 0 + 0.000 000 059 607 204 495 36;
  • 22) 0.000 000 059 607 204 495 36 × 2 = 0 + 0.000 000 119 214 408 990 72;
  • 23) 0.000 000 119 214 408 990 72 × 2 = 0 + 0.000 000 238 428 817 981 44;
  • 24) 0.000 000 238 428 817 981 44 × 2 = 0 + 0.000 000 476 857 635 962 88;
  • 25) 0.000 000 476 857 635 962 88 × 2 = 0 + 0.000 000 953 715 271 925 76;
  • 26) 0.000 000 953 715 271 925 76 × 2 = 0 + 0.000 001 907 430 543 851 52;
  • 27) 0.000 001 907 430 543 851 52 × 2 = 0 + 0.000 003 814 861 087 703 04;
  • 28) 0.000 003 814 861 087 703 04 × 2 = 0 + 0.000 007 629 722 175 406 08;
  • 29) 0.000 007 629 722 175 406 08 × 2 = 0 + 0.000 015 259 444 350 812 16;
  • 30) 0.000 015 259 444 350 812 16 × 2 = 0 + 0.000 030 518 888 701 624 32;
  • 31) 0.000 030 518 888 701 624 32 × 2 = 0 + 0.000 061 037 777 403 248 64;
  • 32) 0.000 061 037 777 403 248 64 × 2 = 0 + 0.000 122 075 554 806 497 28;
  • 33) 0.000 122 075 554 806 497 28 × 2 = 0 + 0.000 244 151 109 612 994 56;
  • 34) 0.000 244 151 109 612 994 56 × 2 = 0 + 0.000 488 302 219 225 989 12;
  • 35) 0.000 488 302 219 225 989 12 × 2 = 0 + 0.000 976 604 438 451 978 24;
  • 36) 0.000 976 604 438 451 978 24 × 2 = 0 + 0.001 953 208 876 903 956 48;
  • 37) 0.001 953 208 876 903 956 48 × 2 = 0 + 0.003 906 417 753 807 912 96;
  • 38) 0.003 906 417 753 807 912 96 × 2 = 0 + 0.007 812 835 507 615 825 92;
  • 39) 0.007 812 835 507 615 825 92 × 2 = 0 + 0.015 625 671 015 231 651 84;
  • 40) 0.015 625 671 015 231 651 84 × 2 = 0 + 0.031 251 342 030 463 303 68;
  • 41) 0.031 251 342 030 463 303 68 × 2 = 0 + 0.062 502 684 060 926 607 36;
  • 42) 0.062 502 684 060 926 607 36 × 2 = 0 + 0.125 005 368 121 853 214 72;
  • 43) 0.125 005 368 121 853 214 72 × 2 = 0 + 0.250 010 736 243 706 429 44;
  • 44) 0.250 010 736 243 706 429 44 × 2 = 0 + 0.500 021 472 487 412 858 88;
  • 45) 0.500 021 472 487 412 858 88 × 2 = 1 + 0.000 042 944 974 825 717 76;
  • 46) 0.000 042 944 974 825 717 76 × 2 = 0 + 0.000 085 889 949 651 435 52;
  • 47) 0.000 085 889 949 651 435 52 × 2 = 0 + 0.000 171 779 899 302 871 04;
  • 48) 0.000 171 779 899 302 871 04 × 2 = 0 + 0.000 343 559 798 605 742 08;
  • 49) 0.000 343 559 798 605 742 08 × 2 = 0 + 0.000 687 119 597 211 484 16;
  • 50) 0.000 687 119 597 211 484 16 × 2 = 0 + 0.001 374 239 194 422 968 32;
  • 51) 0.001 374 239 194 422 968 32 × 2 = 0 + 0.002 748 478 388 845 936 64;
  • 52) 0.002 748 478 388 845 936 64 × 2 = 0 + 0.005 496 956 777 691 873 28;
  • 53) 0.005 496 956 777 691 873 28 × 2 = 0 + 0.010 993 913 555 383 746 56;
  • 54) 0.010 993 913 555 383 746 56 × 2 = 0 + 0.021 987 827 110 767 493 12;
  • 55) 0.021 987 827 110 767 493 12 × 2 = 0 + 0.043 975 654 221 534 986 24;
  • 56) 0.043 975 654 221 534 986 24 × 2 = 0 + 0.087 951 308 443 069 972 48;
  • 57) 0.087 951 308 443 069 972 48 × 2 = 0 + 0.175 902 616 886 139 944 96;
  • 58) 0.175 902 616 886 139 944 96 × 2 = 0 + 0.351 805 233 772 279 889 92;
  • 59) 0.351 805 233 772 279 889 92 × 2 = 0 + 0.703 610 467 544 559 779 84;
  • 60) 0.703 610 467 544 559 779 84 × 2 = 1 + 0.407 220 935 089 119 559 68;
  • 61) 0.407 220 935 089 119 559 68 × 2 = 0 + 0.814 441 870 178 239 119 36;
  • 62) 0.814 441 870 178 239 119 36 × 2 = 1 + 0.628 883 740 356 478 238 72;
  • 63) 0.628 883 740 356 478 238 72 × 2 = 1 + 0.257 767 480 712 956 477 44;
  • 64) 0.257 767 480 712 956 477 44 × 2 = 0 + 0.515 534 961 425 912 954 88;
  • 65) 0.515 534 961 425 912 954 88 × 2 = 1 + 0.031 069 922 851 825 909 76;
  • 66) 0.031 069 922 851 825 909 76 × 2 = 0 + 0.062 139 845 703 651 819 52;
  • 67) 0.062 139 845 703 651 819 52 × 2 = 0 + 0.124 279 691 407 303 639 04;
  • 68) 0.124 279 691 407 303 639 04 × 2 = 0 + 0.248 559 382 814 607 278 08;
  • 69) 0.248 559 382 814 607 278 08 × 2 = 0 + 0.497 118 765 629 214 556 16;
  • 70) 0.497 118 765 629 214 556 16 × 2 = 0 + 0.994 237 531 258 429 112 32;
  • 71) 0.994 237 531 258 429 112 32 × 2 = 1 + 0.988 475 062 516 858 224 64;
  • 72) 0.988 475 062 516 858 224 64 × 2 = 1 + 0.976 950 125 033 716 449 28;
  • 73) 0.976 950 125 033 716 449 28 × 2 = 1 + 0.953 900 250 067 432 898 56;
  • 74) 0.953 900 250 067 432 898 56 × 2 = 1 + 0.907 800 500 134 865 797 12;
  • 75) 0.907 800 500 134 865 797 12 × 2 = 1 + 0.815 601 000 269 731 594 24;
  • 76) 0.815 601 000 269 731 594 24 × 2 = 1 + 0.631 202 000 539 463 188 48;
  • 77) 0.631 202 000 539 463 188 48 × 2 = 1 + 0.262 404 001 078 926 376 96;
  • 78) 0.262 404 001 078 926 376 96 × 2 = 0 + 0.524 808 002 157 852 753 92;
  • 79) 0.524 808 002 157 852 753 92 × 2 = 1 + 0.049 616 004 315 705 507 84;
  • 80) 0.049 616 004 315 705 507 84 × 2 = 0 + 0.099 232 008 631 411 015 68;
  • 81) 0.099 232 008 631 411 015 68 × 2 = 0 + 0.198 464 017 262 822 031 36;
  • 82) 0.198 464 017 262 822 031 36 × 2 = 0 + 0.396 928 034 525 644 062 72;
  • 83) 0.396 928 034 525 644 062 72 × 2 = 0 + 0.793 856 069 051 288 125 44;
  • 84) 0.793 856 069 051 288 125 44 × 2 = 1 + 0.587 712 138 102 576 250 88;
  • 85) 0.587 712 138 102 576 250 88 × 2 = 1 + 0.175 424 276 205 152 501 76;
  • 86) 0.175 424 276 205 152 501 76 × 2 = 0 + 0.350 848 552 410 305 003 52;
  • 87) 0.350 848 552 410 305 003 52 × 2 = 0 + 0.701 697 104 820 610 007 04;
  • 88) 0.701 697 104 820 610 007 04 × 2 = 1 + 0.403 394 209 641 220 014 08;
  • 89) 0.403 394 209 641 220 014 08 × 2 = 0 + 0.806 788 419 282 440 028 16;
  • 90) 0.806 788 419 282 440 028 16 × 2 = 1 + 0.613 576 838 564 880 056 32;
  • 91) 0.613 576 838 564 880 056 32 × 2 = 1 + 0.227 153 677 129 760 112 64;
  • 92) 0.227 153 677 129 760 112 64 × 2 = 0 + 0.454 307 354 259 520 225 28;
  • 93) 0.454 307 354 259 520 225 28 × 2 = 0 + 0.908 614 708 519 040 450 56;
  • 94) 0.908 614 708 519 040 450 56 × 2 = 1 + 0.817 229 417 038 080 901 12;
  • 95) 0.817 229 417 038 080 901 12 × 2 = 1 + 0.634 458 834 076 161 802 24;
  • 96) 0.634 458 834 076 161 802 24 × 2 = 1 + 0.268 917 668 152 323 604 48;
  • 97) 0.268 917 668 152 323 604 48 × 2 = 0 + 0.537 835 336 304 647 208 96;

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:


0.000 000 000 000 028 422 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0001 0110 1000 0011 1111 1010 0001 1001 0110 0111 0(2)

6. Positive number before normalization:

0.000 000 000 000 028 422 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0001 0110 1000 0011 1111 1010 0001 1001 0110 0111 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 45 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 028 422 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0001 0110 1000 0011 1111 1010 0001 1001 0110 0111 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0001 0110 1000 0011 1111 1010 0001 1001 0110 0111 0(2) × 20 =

1.0000 0000 0000 0010 1101 0000 0111 1111 0100 0011 0010 1100 1110(2) × 2-45


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): -45

Mantissa (not normalized):
1.0000 0000 0000 0010 1101 0000 0111 1111 0100 0011 0010 1100 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-45 + 2(11-1) - 1 =

(-45 + 1 023)(10) =

978(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;
  • 978 ÷ 2 = 489 + 0;
  • 489 ÷ 2 = 244 + 1;
  • 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) =

978(10) =

011 1101 0010(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 0000 0000 0010 1101 0000 0111 1111 0100 0011 0010 1100 1110 =

0000 0000 0000 0010 1101 0000 0111 1111 0100 0011 0010 1100 1110


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 0010

Mantissa (52 bits) =
0000 0000 0000 0010 1101 0000 0111 1111 0100 0011 0010 1100 1110


Decimal number -0.000 000 000 000 028 422 93 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0010 - 0000 0000 0000 0010 1101 0000 0111 1111 0100 0011 0010 1100 1110

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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