-0.000 000 000 000 028 422 72 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 028 422 72(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 72(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 72| = 0.000 000 000 000 028 422 72


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 72.

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 72 × 2 = 0 + 0.000 000 000 000 056 845 44;
  • 2) 0.000 000 000 000 056 845 44 × 2 = 0 + 0.000 000 000 000 113 690 88;
  • 3) 0.000 000 000 000 113 690 88 × 2 = 0 + 0.000 000 000 000 227 381 76;
  • 4) 0.000 000 000 000 227 381 76 × 2 = 0 + 0.000 000 000 000 454 763 52;
  • 5) 0.000 000 000 000 454 763 52 × 2 = 0 + 0.000 000 000 000 909 527 04;
  • 6) 0.000 000 000 000 909 527 04 × 2 = 0 + 0.000 000 000 001 819 054 08;
  • 7) 0.000 000 000 001 819 054 08 × 2 = 0 + 0.000 000 000 003 638 108 16;
  • 8) 0.000 000 000 003 638 108 16 × 2 = 0 + 0.000 000 000 007 276 216 32;
  • 9) 0.000 000 000 007 276 216 32 × 2 = 0 + 0.000 000 000 014 552 432 64;
  • 10) 0.000 000 000 014 552 432 64 × 2 = 0 + 0.000 000 000 029 104 865 28;
  • 11) 0.000 000 000 029 104 865 28 × 2 = 0 + 0.000 000 000 058 209 730 56;
  • 12) 0.000 000 000 058 209 730 56 × 2 = 0 + 0.000 000 000 116 419 461 12;
  • 13) 0.000 000 000 116 419 461 12 × 2 = 0 + 0.000 000 000 232 838 922 24;
  • 14) 0.000 000 000 232 838 922 24 × 2 = 0 + 0.000 000 000 465 677 844 48;
  • 15) 0.000 000 000 465 677 844 48 × 2 = 0 + 0.000 000 000 931 355 688 96;
  • 16) 0.000 000 000 931 355 688 96 × 2 = 0 + 0.000 000 001 862 711 377 92;
  • 17) 0.000 000 001 862 711 377 92 × 2 = 0 + 0.000 000 003 725 422 755 84;
  • 18) 0.000 000 003 725 422 755 84 × 2 = 0 + 0.000 000 007 450 845 511 68;
  • 19) 0.000 000 007 450 845 511 68 × 2 = 0 + 0.000 000 014 901 691 023 36;
  • 20) 0.000 000 014 901 691 023 36 × 2 = 0 + 0.000 000 029 803 382 046 72;
  • 21) 0.000 000 029 803 382 046 72 × 2 = 0 + 0.000 000 059 606 764 093 44;
  • 22) 0.000 000 059 606 764 093 44 × 2 = 0 + 0.000 000 119 213 528 186 88;
  • 23) 0.000 000 119 213 528 186 88 × 2 = 0 + 0.000 000 238 427 056 373 76;
  • 24) 0.000 000 238 427 056 373 76 × 2 = 0 + 0.000 000 476 854 112 747 52;
  • 25) 0.000 000 476 854 112 747 52 × 2 = 0 + 0.000 000 953 708 225 495 04;
  • 26) 0.000 000 953 708 225 495 04 × 2 = 0 + 0.000 001 907 416 450 990 08;
  • 27) 0.000 001 907 416 450 990 08 × 2 = 0 + 0.000 003 814 832 901 980 16;
  • 28) 0.000 003 814 832 901 980 16 × 2 = 0 + 0.000 007 629 665 803 960 32;
  • 29) 0.000 007 629 665 803 960 32 × 2 = 0 + 0.000 015 259 331 607 920 64;
  • 30) 0.000 015 259 331 607 920 64 × 2 = 0 + 0.000 030 518 663 215 841 28;
  • 31) 0.000 030 518 663 215 841 28 × 2 = 0 + 0.000 061 037 326 431 682 56;
  • 32) 0.000 061 037 326 431 682 56 × 2 = 0 + 0.000 122 074 652 863 365 12;
  • 33) 0.000 122 074 652 863 365 12 × 2 = 0 + 0.000 244 149 305 726 730 24;
  • 34) 0.000 244 149 305 726 730 24 × 2 = 0 + 0.000 488 298 611 453 460 48;
  • 35) 0.000 488 298 611 453 460 48 × 2 = 0 + 0.000 976 597 222 906 920 96;
  • 36) 0.000 976 597 222 906 920 96 × 2 = 0 + 0.001 953 194 445 813 841 92;
  • 37) 0.001 953 194 445 813 841 92 × 2 = 0 + 0.003 906 388 891 627 683 84;
  • 38) 0.003 906 388 891 627 683 84 × 2 = 0 + 0.007 812 777 783 255 367 68;
  • 39) 0.007 812 777 783 255 367 68 × 2 = 0 + 0.015 625 555 566 510 735 36;
  • 40) 0.015 625 555 566 510 735 36 × 2 = 0 + 0.031 251 111 133 021 470 72;
  • 41) 0.031 251 111 133 021 470 72 × 2 = 0 + 0.062 502 222 266 042 941 44;
  • 42) 0.062 502 222 266 042 941 44 × 2 = 0 + 0.125 004 444 532 085 882 88;
  • 43) 0.125 004 444 532 085 882 88 × 2 = 0 + 0.250 008 889 064 171 765 76;
  • 44) 0.250 008 889 064 171 765 76 × 2 = 0 + 0.500 017 778 128 343 531 52;
  • 45) 0.500 017 778 128 343 531 52 × 2 = 1 + 0.000 035 556 256 687 063 04;
  • 46) 0.000 035 556 256 687 063 04 × 2 = 0 + 0.000 071 112 513 374 126 08;
  • 47) 0.000 071 112 513 374 126 08 × 2 = 0 + 0.000 142 225 026 748 252 16;
  • 48) 0.000 142 225 026 748 252 16 × 2 = 0 + 0.000 284 450 053 496 504 32;
  • 49) 0.000 284 450 053 496 504 32 × 2 = 0 + 0.000 568 900 106 993 008 64;
  • 50) 0.000 568 900 106 993 008 64 × 2 = 0 + 0.001 137 800 213 986 017 28;
  • 51) 0.001 137 800 213 986 017 28 × 2 = 0 + 0.002 275 600 427 972 034 56;
  • 52) 0.002 275 600 427 972 034 56 × 2 = 0 + 0.004 551 200 855 944 069 12;
  • 53) 0.004 551 200 855 944 069 12 × 2 = 0 + 0.009 102 401 711 888 138 24;
  • 54) 0.009 102 401 711 888 138 24 × 2 = 0 + 0.018 204 803 423 776 276 48;
  • 55) 0.018 204 803 423 776 276 48 × 2 = 0 + 0.036 409 606 847 552 552 96;
  • 56) 0.036 409 606 847 552 552 96 × 2 = 0 + 0.072 819 213 695 105 105 92;
  • 57) 0.072 819 213 695 105 105 92 × 2 = 0 + 0.145 638 427 390 210 211 84;
  • 58) 0.145 638 427 390 210 211 84 × 2 = 0 + 0.291 276 854 780 420 423 68;
  • 59) 0.291 276 854 780 420 423 68 × 2 = 0 + 0.582 553 709 560 840 847 36;
  • 60) 0.582 553 709 560 840 847 36 × 2 = 1 + 0.165 107 419 121 681 694 72;
  • 61) 0.165 107 419 121 681 694 72 × 2 = 0 + 0.330 214 838 243 363 389 44;
  • 62) 0.330 214 838 243 363 389 44 × 2 = 0 + 0.660 429 676 486 726 778 88;
  • 63) 0.660 429 676 486 726 778 88 × 2 = 1 + 0.320 859 352 973 453 557 76;
  • 64) 0.320 859 352 973 453 557 76 × 2 = 0 + 0.641 718 705 946 907 115 52;
  • 65) 0.641 718 705 946 907 115 52 × 2 = 1 + 0.283 437 411 893 814 231 04;
  • 66) 0.283 437 411 893 814 231 04 × 2 = 0 + 0.566 874 823 787 628 462 08;
  • 67) 0.566 874 823 787 628 462 08 × 2 = 1 + 0.133 749 647 575 256 924 16;
  • 68) 0.133 749 647 575 256 924 16 × 2 = 0 + 0.267 499 295 150 513 848 32;
  • 69) 0.267 499 295 150 513 848 32 × 2 = 0 + 0.534 998 590 301 027 696 64;
  • 70) 0.534 998 590 301 027 696 64 × 2 = 1 + 0.069 997 180 602 055 393 28;
  • 71) 0.069 997 180 602 055 393 28 × 2 = 0 + 0.139 994 361 204 110 786 56;
  • 72) 0.139 994 361 204 110 786 56 × 2 = 0 + 0.279 988 722 408 221 573 12;
  • 73) 0.279 988 722 408 221 573 12 × 2 = 0 + 0.559 977 444 816 443 146 24;
  • 74) 0.559 977 444 816 443 146 24 × 2 = 1 + 0.119 954 889 632 886 292 48;
  • 75) 0.119 954 889 632 886 292 48 × 2 = 0 + 0.239 909 779 265 772 584 96;
  • 76) 0.239 909 779 265 772 584 96 × 2 = 0 + 0.479 819 558 531 545 169 92;
  • 77) 0.479 819 558 531 545 169 92 × 2 = 0 + 0.959 639 117 063 090 339 84;
  • 78) 0.959 639 117 063 090 339 84 × 2 = 1 + 0.919 278 234 126 180 679 68;
  • 79) 0.919 278 234 126 180 679 68 × 2 = 1 + 0.838 556 468 252 361 359 36;
  • 80) 0.838 556 468 252 361 359 36 × 2 = 1 + 0.677 112 936 504 722 718 72;
  • 81) 0.677 112 936 504 722 718 72 × 2 = 1 + 0.354 225 873 009 445 437 44;
  • 82) 0.354 225 873 009 445 437 44 × 2 = 0 + 0.708 451 746 018 890 874 88;
  • 83) 0.708 451 746 018 890 874 88 × 2 = 1 + 0.416 903 492 037 781 749 76;
  • 84) 0.416 903 492 037 781 749 76 × 2 = 0 + 0.833 806 984 075 563 499 52;
  • 85) 0.833 806 984 075 563 499 52 × 2 = 1 + 0.667 613 968 151 126 999 04;
  • 86) 0.667 613 968 151 126 999 04 × 2 = 1 + 0.335 227 936 302 253 998 08;
  • 87) 0.335 227 936 302 253 998 08 × 2 = 0 + 0.670 455 872 604 507 996 16;
  • 88) 0.670 455 872 604 507 996 16 × 2 = 1 + 0.340 911 745 209 015 992 32;
  • 89) 0.340 911 745 209 015 992 32 × 2 = 0 + 0.681 823 490 418 031 984 64;
  • 90) 0.681 823 490 418 031 984 64 × 2 = 1 + 0.363 646 980 836 063 969 28;
  • 91) 0.363 646 980 836 063 969 28 × 2 = 0 + 0.727 293 961 672 127 938 56;
  • 92) 0.727 293 961 672 127 938 56 × 2 = 1 + 0.454 587 923 344 255 877 12;
  • 93) 0.454 587 923 344 255 877 12 × 2 = 0 + 0.909 175 846 688 511 754 24;
  • 94) 0.909 175 846 688 511 754 24 × 2 = 1 + 0.818 351 693 377 023 508 48;
  • 95) 0.818 351 693 377 023 508 48 × 2 = 1 + 0.636 703 386 754 047 016 96;
  • 96) 0.636 703 386 754 047 016 96 × 2 = 1 + 0.273 406 773 508 094 033 92;
  • 97) 0.273 406 773 508 094 033 92 × 2 = 0 + 0.546 813 547 016 188 067 84;

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 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0001 0010 1010 0100 0100 0111 1010 1101 0101 0111 0(2)

6. Positive number before normalization:

0.000 000 000 000 028 422 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0001 0010 1010 0100 0100 0111 1010 1101 0101 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 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0001 0010 1010 0100 0100 0111 1010 1101 0101 0111 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0001 0010 1010 0100 0100 0111 1010 1101 0101 0111 0(2) × 20 =

1.0000 0000 0000 0010 0101 0100 1000 1000 1111 0101 1010 1010 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 0101 0100 1000 1000 1111 0101 1010 1010 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 0101 0100 1000 1000 1111 0101 1010 1010 1110 =

0000 0000 0000 0010 0101 0100 1000 1000 1111 0101 1010 1010 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 0101 0100 1000 1000 1111 0101 1010 1010 1110


Decimal number -0.000 000 000 000 028 422 72 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0010 - 0000 0000 0000 0010 0101 0100 1000 1000 1111 0101 1010 1010 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