Decimal to 64 Bit IEEE 754 Binary: Convert Number -0.000 000 000 000 028 421 76 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -0.000 000 000 000 028 421 76(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 000 028 421 76| = 0.000 000 000 000 028 421 76


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: 0.000 000 000 000 028 421 76.

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) 0.000 000 000 000 028 421 76 × 2 = 0 + 0.000 000 000 000 056 843 52;
  • 2) 0.000 000 000 000 056 843 52 × 2 = 0 + 0.000 000 000 000 113 687 04;
  • 3) 0.000 000 000 000 113 687 04 × 2 = 0 + 0.000 000 000 000 227 374 08;
  • 4) 0.000 000 000 000 227 374 08 × 2 = 0 + 0.000 000 000 000 454 748 16;
  • 5) 0.000 000 000 000 454 748 16 × 2 = 0 + 0.000 000 000 000 909 496 32;
  • 6) 0.000 000 000 000 909 496 32 × 2 = 0 + 0.000 000 000 001 818 992 64;
  • 7) 0.000 000 000 001 818 992 64 × 2 = 0 + 0.000 000 000 003 637 985 28;
  • 8) 0.000 000 000 003 637 985 28 × 2 = 0 + 0.000 000 000 007 275 970 56;
  • 9) 0.000 000 000 007 275 970 56 × 2 = 0 + 0.000 000 000 014 551 941 12;
  • 10) 0.000 000 000 014 551 941 12 × 2 = 0 + 0.000 000 000 029 103 882 24;
  • 11) 0.000 000 000 029 103 882 24 × 2 = 0 + 0.000 000 000 058 207 764 48;
  • 12) 0.000 000 000 058 207 764 48 × 2 = 0 + 0.000 000 000 116 415 528 96;
  • 13) 0.000 000 000 116 415 528 96 × 2 = 0 + 0.000 000 000 232 831 057 92;
  • 14) 0.000 000 000 232 831 057 92 × 2 = 0 + 0.000 000 000 465 662 115 84;
  • 15) 0.000 000 000 465 662 115 84 × 2 = 0 + 0.000 000 000 931 324 231 68;
  • 16) 0.000 000 000 931 324 231 68 × 2 = 0 + 0.000 000 001 862 648 463 36;
  • 17) 0.000 000 001 862 648 463 36 × 2 = 0 + 0.000 000 003 725 296 926 72;
  • 18) 0.000 000 003 725 296 926 72 × 2 = 0 + 0.000 000 007 450 593 853 44;
  • 19) 0.000 000 007 450 593 853 44 × 2 = 0 + 0.000 000 014 901 187 706 88;
  • 20) 0.000 000 014 901 187 706 88 × 2 = 0 + 0.000 000 029 802 375 413 76;
  • 21) 0.000 000 029 802 375 413 76 × 2 = 0 + 0.000 000 059 604 750 827 52;
  • 22) 0.000 000 059 604 750 827 52 × 2 = 0 + 0.000 000 119 209 501 655 04;
  • 23) 0.000 000 119 209 501 655 04 × 2 = 0 + 0.000 000 238 419 003 310 08;
  • 24) 0.000 000 238 419 003 310 08 × 2 = 0 + 0.000 000 476 838 006 620 16;
  • 25) 0.000 000 476 838 006 620 16 × 2 = 0 + 0.000 000 953 676 013 240 32;
  • 26) 0.000 000 953 676 013 240 32 × 2 = 0 + 0.000 001 907 352 026 480 64;
  • 27) 0.000 001 907 352 026 480 64 × 2 = 0 + 0.000 003 814 704 052 961 28;
  • 28) 0.000 003 814 704 052 961 28 × 2 = 0 + 0.000 007 629 408 105 922 56;
  • 29) 0.000 007 629 408 105 922 56 × 2 = 0 + 0.000 015 258 816 211 845 12;
  • 30) 0.000 015 258 816 211 845 12 × 2 = 0 + 0.000 030 517 632 423 690 24;
  • 31) 0.000 030 517 632 423 690 24 × 2 = 0 + 0.000 061 035 264 847 380 48;
  • 32) 0.000 061 035 264 847 380 48 × 2 = 0 + 0.000 122 070 529 694 760 96;
  • 33) 0.000 122 070 529 694 760 96 × 2 = 0 + 0.000 244 141 059 389 521 92;
  • 34) 0.000 244 141 059 389 521 92 × 2 = 0 + 0.000 488 282 118 779 043 84;
  • 35) 0.000 488 282 118 779 043 84 × 2 = 0 + 0.000 976 564 237 558 087 68;
  • 36) 0.000 976 564 237 558 087 68 × 2 = 0 + 0.001 953 128 475 116 175 36;
  • 37) 0.001 953 128 475 116 175 36 × 2 = 0 + 0.003 906 256 950 232 350 72;
  • 38) 0.003 906 256 950 232 350 72 × 2 = 0 + 0.007 812 513 900 464 701 44;
  • 39) 0.007 812 513 900 464 701 44 × 2 = 0 + 0.015 625 027 800 929 402 88;
  • 40) 0.015 625 027 800 929 402 88 × 2 = 0 + 0.031 250 055 601 858 805 76;
  • 41) 0.031 250 055 601 858 805 76 × 2 = 0 + 0.062 500 111 203 717 611 52;
  • 42) 0.062 500 111 203 717 611 52 × 2 = 0 + 0.125 000 222 407 435 223 04;
  • 43) 0.125 000 222 407 435 223 04 × 2 = 0 + 0.250 000 444 814 870 446 08;
  • 44) 0.250 000 444 814 870 446 08 × 2 = 0 + 0.500 000 889 629 740 892 16;
  • 45) 0.500 000 889 629 740 892 16 × 2 = 1 + 0.000 001 779 259 481 784 32;
  • 46) 0.000 001 779 259 481 784 32 × 2 = 0 + 0.000 003 558 518 963 568 64;
  • 47) 0.000 003 558 518 963 568 64 × 2 = 0 + 0.000 007 117 037 927 137 28;
  • 48) 0.000 007 117 037 927 137 28 × 2 = 0 + 0.000 014 234 075 854 274 56;
  • 49) 0.000 014 234 075 854 274 56 × 2 = 0 + 0.000 028 468 151 708 549 12;
  • 50) 0.000 028 468 151 708 549 12 × 2 = 0 + 0.000 056 936 303 417 098 24;
  • 51) 0.000 056 936 303 417 098 24 × 2 = 0 + 0.000 113 872 606 834 196 48;
  • 52) 0.000 113 872 606 834 196 48 × 2 = 0 + 0.000 227 745 213 668 392 96;
  • 53) 0.000 227 745 213 668 392 96 × 2 = 0 + 0.000 455 490 427 336 785 92;
  • 54) 0.000 455 490 427 336 785 92 × 2 = 0 + 0.000 910 980 854 673 571 84;
  • 55) 0.000 910 980 854 673 571 84 × 2 = 0 + 0.001 821 961 709 347 143 68;
  • 56) 0.001 821 961 709 347 143 68 × 2 = 0 + 0.003 643 923 418 694 287 36;
  • 57) 0.003 643 923 418 694 287 36 × 2 = 0 + 0.007 287 846 837 388 574 72;
  • 58) 0.007 287 846 837 388 574 72 × 2 = 0 + 0.014 575 693 674 777 149 44;
  • 59) 0.014 575 693 674 777 149 44 × 2 = 0 + 0.029 151 387 349 554 298 88;
  • 60) 0.029 151 387 349 554 298 88 × 2 = 0 + 0.058 302 774 699 108 597 76;
  • 61) 0.058 302 774 699 108 597 76 × 2 = 0 + 0.116 605 549 398 217 195 52;
  • 62) 0.116 605 549 398 217 195 52 × 2 = 0 + 0.233 211 098 796 434 391 04;
  • 63) 0.233 211 098 796 434 391 04 × 2 = 0 + 0.466 422 197 592 868 782 08;
  • 64) 0.466 422 197 592 868 782 08 × 2 = 0 + 0.932 844 395 185 737 564 16;
  • 65) 0.932 844 395 185 737 564 16 × 2 = 1 + 0.865 688 790 371 475 128 32;
  • 66) 0.865 688 790 371 475 128 32 × 2 = 1 + 0.731 377 580 742 950 256 64;
  • 67) 0.731 377 580 742 950 256 64 × 2 = 1 + 0.462 755 161 485 900 513 28;
  • 68) 0.462 755 161 485 900 513 28 × 2 = 0 + 0.925 510 322 971 801 026 56;
  • 69) 0.925 510 322 971 801 026 56 × 2 = 1 + 0.851 020 645 943 602 053 12;
  • 70) 0.851 020 645 943 602 053 12 × 2 = 1 + 0.702 041 291 887 204 106 24;
  • 71) 0.702 041 291 887 204 106 24 × 2 = 1 + 0.404 082 583 774 408 212 48;
  • 72) 0.404 082 583 774 408 212 48 × 2 = 0 + 0.808 165 167 548 816 424 96;
  • 73) 0.808 165 167 548 816 424 96 × 2 = 1 + 0.616 330 335 097 632 849 92;
  • 74) 0.616 330 335 097 632 849 92 × 2 = 1 + 0.232 660 670 195 265 699 84;
  • 75) 0.232 660 670 195 265 699 84 × 2 = 0 + 0.465 321 340 390 531 399 68;
  • 76) 0.465 321 340 390 531 399 68 × 2 = 0 + 0.930 642 680 781 062 799 36;
  • 77) 0.930 642 680 781 062 799 36 × 2 = 1 + 0.861 285 361 562 125 598 72;
  • 78) 0.861 285 361 562 125 598 72 × 2 = 1 + 0.722 570 723 124 251 197 44;
  • 79) 0.722 570 723 124 251 197 44 × 2 = 1 + 0.445 141 446 248 502 394 88;
  • 80) 0.445 141 446 248 502 394 88 × 2 = 0 + 0.890 282 892 497 004 789 76;
  • 81) 0.890 282 892 497 004 789 76 × 2 = 1 + 0.780 565 784 994 009 579 52;
  • 82) 0.780 565 784 994 009 579 52 × 2 = 1 + 0.561 131 569 988 019 159 04;
  • 83) 0.561 131 569 988 019 159 04 × 2 = 1 + 0.122 263 139 976 038 318 08;
  • 84) 0.122 263 139 976 038 318 08 × 2 = 0 + 0.244 526 279 952 076 636 16;
  • 85) 0.244 526 279 952 076 636 16 × 2 = 0 + 0.489 052 559 904 153 272 32;
  • 86) 0.489 052 559 904 153 272 32 × 2 = 0 + 0.978 105 119 808 306 544 64;
  • 87) 0.978 105 119 808 306 544 64 × 2 = 1 + 0.956 210 239 616 613 089 28;
  • 88) 0.956 210 239 616 613 089 28 × 2 = 1 + 0.912 420 479 233 226 178 56;
  • 89) 0.912 420 479 233 226 178 56 × 2 = 1 + 0.824 840 958 466 452 357 12;
  • 90) 0.824 840 958 466 452 357 12 × 2 = 1 + 0.649 681 916 932 904 714 24;
  • 91) 0.649 681 916 932 904 714 24 × 2 = 1 + 0.299 363 833 865 809 428 48;
  • 92) 0.299 363 833 865 809 428 48 × 2 = 0 + 0.598 727 667 731 618 856 96;
  • 93) 0.598 727 667 731 618 856 96 × 2 = 1 + 0.197 455 335 463 237 713 92;
  • 94) 0.197 455 335 463 237 713 92 × 2 = 0 + 0.394 910 670 926 475 427 84;
  • 95) 0.394 910 670 926 475 427 84 × 2 = 0 + 0.789 821 341 852 950 855 68;
  • 96) 0.789 821 341 852 950 855 68 × 2 = 1 + 0.579 642 683 705 901 711 36;
  • 97) 0.579 642 683 705 901 711 36 × 2 = 1 + 0.159 285 367 411 803 422 72;

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:


0.000 000 000 000 028 421 76(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0000 0000 1110 1110 1100 1110 1110 0011 1110 1001 1(2)

6. Positive number before normalization:

0.000 000 000 000 028 421 76(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0000 0000 1110 1110 1100 1110 1110 0011 1110 1001 1(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:


0.000 000 000 000 028 421 76(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0000 0000 1110 1110 1100 1110 1110 0011 1110 1001 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0000 0000 1110 1110 1100 1110 1110 0011 1110 1001 1(2) × 20 =

1.0000 0000 0000 0000 0001 1101 1101 1001 1101 1100 0111 1101 0011(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 0000 0001 1101 1101 1001 1101 1100 0111 1101 0011


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 0000 0001 1101 1101 1001 1101 1100 0111 1101 0011 =

0000 0000 0000 0000 0001 1101 1101 1001 1101 1100 0111 1101 0011


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 0000 0001 1101 1101 1001 1101 1100 0111 1101 0011


The base ten decimal number -0.000 000 000 000 028 421 76 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0010 - 0000 0000 0000 0000 0001 1101 1101 1001 1101 1100 0111 1101 0011

» Decimal number in base ten -0.000 000 000 000 028 422 05 converted and written as 64 bit double precision IEEE 754 binary floating point representation standard

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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