1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 346 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 346 1(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
1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 346 1(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. First, convert to binary (in base 2) the integer part: 1.
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;
  • 1 ÷ 2 = 0 + 1;

2. 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.123 412 341 234 123 412 341 234 123 412 341 234 123 412 346 1.

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) 0.123 412 341 234 123 412 341 234 123 412 341 234 123 412 346 1 × 2 = 0 + 0.246 824 682 468 246 824 682 468 246 824 682 468 246 824 692 2;
  • 2) 0.246 824 682 468 246 824 682 468 246 824 682 468 246 824 692 2 × 2 = 0 + 0.493 649 364 936 493 649 364 936 493 649 364 936 493 649 384 4;
  • 3) 0.493 649 364 936 493 649 364 936 493 649 364 936 493 649 384 4 × 2 = 0 + 0.987 298 729 872 987 298 729 872 987 298 729 872 987 298 768 8;
  • 4) 0.987 298 729 872 987 298 729 872 987 298 729 872 987 298 768 8 × 2 = 1 + 0.974 597 459 745 974 597 459 745 974 597 459 745 974 597 537 6;
  • 5) 0.974 597 459 745 974 597 459 745 974 597 459 745 974 597 537 6 × 2 = 1 + 0.949 194 919 491 949 194 919 491 949 194 919 491 949 195 075 2;
  • 6) 0.949 194 919 491 949 194 919 491 949 194 919 491 949 195 075 2 × 2 = 1 + 0.898 389 838 983 898 389 838 983 898 389 838 983 898 390 150 4;
  • 7) 0.898 389 838 983 898 389 838 983 898 389 838 983 898 390 150 4 × 2 = 1 + 0.796 779 677 967 796 779 677 967 796 779 677 967 796 780 300 8;
  • 8) 0.796 779 677 967 796 779 677 967 796 779 677 967 796 780 300 8 × 2 = 1 + 0.593 559 355 935 593 559 355 935 593 559 355 935 593 560 601 6;
  • 9) 0.593 559 355 935 593 559 355 935 593 559 355 935 593 560 601 6 × 2 = 1 + 0.187 118 711 871 187 118 711 871 187 118 711 871 187 121 203 2;
  • 10) 0.187 118 711 871 187 118 711 871 187 118 711 871 187 121 203 2 × 2 = 0 + 0.374 237 423 742 374 237 423 742 374 237 423 742 374 242 406 4;
  • 11) 0.374 237 423 742 374 237 423 742 374 237 423 742 374 242 406 4 × 2 = 0 + 0.748 474 847 484 748 474 847 484 748 474 847 484 748 484 812 8;
  • 12) 0.748 474 847 484 748 474 847 484 748 474 847 484 748 484 812 8 × 2 = 1 + 0.496 949 694 969 496 949 694 969 496 949 694 969 496 969 625 6;
  • 13) 0.496 949 694 969 496 949 694 969 496 949 694 969 496 969 625 6 × 2 = 0 + 0.993 899 389 938 993 899 389 938 993 899 389 938 993 939 251 2;
  • 14) 0.993 899 389 938 993 899 389 938 993 899 389 938 993 939 251 2 × 2 = 1 + 0.987 798 779 877 987 798 779 877 987 798 779 877 987 878 502 4;
  • 15) 0.987 798 779 877 987 798 779 877 987 798 779 877 987 878 502 4 × 2 = 1 + 0.975 597 559 755 975 597 559 755 975 597 559 755 975 757 004 8;
  • 16) 0.975 597 559 755 975 597 559 755 975 597 559 755 975 757 004 8 × 2 = 1 + 0.951 195 119 511 951 195 119 511 951 195 119 511 951 514 009 6;
  • 17) 0.951 195 119 511 951 195 119 511 951 195 119 511 951 514 009 6 × 2 = 1 + 0.902 390 239 023 902 390 239 023 902 390 239 023 903 028 019 2;
  • 18) 0.902 390 239 023 902 390 239 023 902 390 239 023 903 028 019 2 × 2 = 1 + 0.804 780 478 047 804 780 478 047 804 780 478 047 806 056 038 4;
  • 19) 0.804 780 478 047 804 780 478 047 804 780 478 047 806 056 038 4 × 2 = 1 + 0.609 560 956 095 609 560 956 095 609 560 956 095 612 112 076 8;
  • 20) 0.609 560 956 095 609 560 956 095 609 560 956 095 612 112 076 8 × 2 = 1 + 0.219 121 912 191 219 121 912 191 219 121 912 191 224 224 153 6;
  • 21) 0.219 121 912 191 219 121 912 191 219 121 912 191 224 224 153 6 × 2 = 0 + 0.438 243 824 382 438 243 824 382 438 243 824 382 448 448 307 2;
  • 22) 0.438 243 824 382 438 243 824 382 438 243 824 382 448 448 307 2 × 2 = 0 + 0.876 487 648 764 876 487 648 764 876 487 648 764 896 896 614 4;
  • 23) 0.876 487 648 764 876 487 648 764 876 487 648 764 896 896 614 4 × 2 = 1 + 0.752 975 297 529 752 975 297 529 752 975 297 529 793 793 228 8;
  • 24) 0.752 975 297 529 752 975 297 529 752 975 297 529 793 793 228 8 × 2 = 1 + 0.505 950 595 059 505 950 595 059 505 950 595 059 587 586 457 6;
  • 25) 0.505 950 595 059 505 950 595 059 505 950 595 059 587 586 457 6 × 2 = 1 + 0.011 901 190 119 011 901 190 119 011 901 190 119 175 172 915 2;
  • 26) 0.011 901 190 119 011 901 190 119 011 901 190 119 175 172 915 2 × 2 = 0 + 0.023 802 380 238 023 802 380 238 023 802 380 238 350 345 830 4;
  • 27) 0.023 802 380 238 023 802 380 238 023 802 380 238 350 345 830 4 × 2 = 0 + 0.047 604 760 476 047 604 760 476 047 604 760 476 700 691 660 8;
  • 28) 0.047 604 760 476 047 604 760 476 047 604 760 476 700 691 660 8 × 2 = 0 + 0.095 209 520 952 095 209 520 952 095 209 520 953 401 383 321 6;
  • 29) 0.095 209 520 952 095 209 520 952 095 209 520 953 401 383 321 6 × 2 = 0 + 0.190 419 041 904 190 419 041 904 190 419 041 906 802 766 643 2;
  • 30) 0.190 419 041 904 190 419 041 904 190 419 041 906 802 766 643 2 × 2 = 0 + 0.380 838 083 808 380 838 083 808 380 838 083 813 605 533 286 4;
  • 31) 0.380 838 083 808 380 838 083 808 380 838 083 813 605 533 286 4 × 2 = 0 + 0.761 676 167 616 761 676 167 616 761 676 167 627 211 066 572 8;
  • 32) 0.761 676 167 616 761 676 167 616 761 676 167 627 211 066 572 8 × 2 = 1 + 0.523 352 335 233 523 352 335 233 523 352 335 254 422 133 145 6;
  • 33) 0.523 352 335 233 523 352 335 233 523 352 335 254 422 133 145 6 × 2 = 1 + 0.046 704 670 467 046 704 670 467 046 704 670 508 844 266 291 2;
  • 34) 0.046 704 670 467 046 704 670 467 046 704 670 508 844 266 291 2 × 2 = 0 + 0.093 409 340 934 093 409 340 934 093 409 341 017 688 532 582 4;
  • 35) 0.093 409 340 934 093 409 340 934 093 409 341 017 688 532 582 4 × 2 = 0 + 0.186 818 681 868 186 818 681 868 186 818 682 035 377 065 164 8;
  • 36) 0.186 818 681 868 186 818 681 868 186 818 682 035 377 065 164 8 × 2 = 0 + 0.373 637 363 736 373 637 363 736 373 637 364 070 754 130 329 6;
  • 37) 0.373 637 363 736 373 637 363 736 373 637 364 070 754 130 329 6 × 2 = 0 + 0.747 274 727 472 747 274 727 472 747 274 728 141 508 260 659 2;
  • 38) 0.747 274 727 472 747 274 727 472 747 274 728 141 508 260 659 2 × 2 = 1 + 0.494 549 454 945 494 549 454 945 494 549 456 283 016 521 318 4;
  • 39) 0.494 549 454 945 494 549 454 945 494 549 456 283 016 521 318 4 × 2 = 0 + 0.989 098 909 890 989 098 909 890 989 098 912 566 033 042 636 8;
  • 40) 0.989 098 909 890 989 098 909 890 989 098 912 566 033 042 636 8 × 2 = 1 + 0.978 197 819 781 978 197 819 781 978 197 825 132 066 085 273 6;
  • 41) 0.978 197 819 781 978 197 819 781 978 197 825 132 066 085 273 6 × 2 = 1 + 0.956 395 639 563 956 395 639 563 956 395 650 264 132 170 547 2;
  • 42) 0.956 395 639 563 956 395 639 563 956 395 650 264 132 170 547 2 × 2 = 1 + 0.912 791 279 127 912 791 279 127 912 791 300 528 264 341 094 4;
  • 43) 0.912 791 279 127 912 791 279 127 912 791 300 528 264 341 094 4 × 2 = 1 + 0.825 582 558 255 825 582 558 255 825 582 601 056 528 682 188 8;
  • 44) 0.825 582 558 255 825 582 558 255 825 582 601 056 528 682 188 8 × 2 = 1 + 0.651 165 116 511 651 165 116 511 651 165 202 113 057 364 377 6;
  • 45) 0.651 165 116 511 651 165 116 511 651 165 202 113 057 364 377 6 × 2 = 1 + 0.302 330 233 023 302 330 233 023 302 330 404 226 114 728 755 2;
  • 46) 0.302 330 233 023 302 330 233 023 302 330 404 226 114 728 755 2 × 2 = 0 + 0.604 660 466 046 604 660 466 046 604 660 808 452 229 457 510 4;
  • 47) 0.604 660 466 046 604 660 466 046 604 660 808 452 229 457 510 4 × 2 = 1 + 0.209 320 932 093 209 320 932 093 209 321 616 904 458 915 020 8;
  • 48) 0.209 320 932 093 209 320 932 093 209 321 616 904 458 915 020 8 × 2 = 0 + 0.418 641 864 186 418 641 864 186 418 643 233 808 917 830 041 6;
  • 49) 0.418 641 864 186 418 641 864 186 418 643 233 808 917 830 041 6 × 2 = 0 + 0.837 283 728 372 837 283 728 372 837 286 467 617 835 660 083 2;
  • 50) 0.837 283 728 372 837 283 728 372 837 286 467 617 835 660 083 2 × 2 = 1 + 0.674 567 456 745 674 567 456 745 674 572 935 235 671 320 166 4;
  • 51) 0.674 567 456 745 674 567 456 745 674 572 935 235 671 320 166 4 × 2 = 1 + 0.349 134 913 491 349 134 913 491 349 145 870 471 342 640 332 8;
  • 52) 0.349 134 913 491 349 134 913 491 349 145 870 471 342 640 332 8 × 2 = 0 + 0.698 269 826 982 698 269 826 982 698 291 740 942 685 280 665 6;
  • 53) 0.698 269 826 982 698 269 826 982 698 291 740 942 685 280 665 6 × 2 = 1 + 0.396 539 653 965 396 539 653 965 396 583 481 885 370 561 331 2;

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).


4. 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:


0.123 412 341 234 123 412 341 234 123 412 341 234 123 412 346 1(10) =

0.0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1(2)

5. Positive number before normalization:

1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 346 1(10) =

1.0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 346 1(10) =

1.0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1(2) =

1.0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1(2) × 20


7. 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 0 (a positive number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(10)


9. 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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1023(10) =

011 1111 1111(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110 1 =

0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1111

Mantissa (52 bits) =
0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110


Decimal number 1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 346 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110

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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