Convert 1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 341 23 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 341 23(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (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 the binary (base 2) the fractional part: 0.123 412 341 234 123 412 341 234 123 412 341 234 123 412 341 23.

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) 0.123 412 341 234 123 412 341 234 123 412 341 234 123 412 341 23 × 2 = 0 + 0.246 824 682 468 246 824 682 468 246 824 682 468 246 824 682 46;
  • 2) 0.246 824 682 468 246 824 682 468 246 824 682 468 246 824 682 46 × 2 = 0 + 0.493 649 364 936 493 649 364 936 493 649 364 936 493 649 364 92;
  • 3) 0.493 649 364 936 493 649 364 936 493 649 364 936 493 649 364 92 × 2 = 0 + 0.987 298 729 872 987 298 729 872 987 298 729 872 987 298 729 84;
  • 4) 0.987 298 729 872 987 298 729 872 987 298 729 872 987 298 729 84 × 2 = 1 + 0.974 597 459 745 974 597 459 745 974 597 459 745 974 597 459 68;
  • 5) 0.974 597 459 745 974 597 459 745 974 597 459 745 974 597 459 68 × 2 = 1 + 0.949 194 919 491 949 194 919 491 949 194 919 491 949 194 919 36;
  • 6) 0.949 194 919 491 949 194 919 491 949 194 919 491 949 194 919 36 × 2 = 1 + 0.898 389 838 983 898 389 838 983 898 389 838 983 898 389 838 72;
  • 7) 0.898 389 838 983 898 389 838 983 898 389 838 983 898 389 838 72 × 2 = 1 + 0.796 779 677 967 796 779 677 967 796 779 677 967 796 779 677 44;
  • 8) 0.796 779 677 967 796 779 677 967 796 779 677 967 796 779 677 44 × 2 = 1 + 0.593 559 355 935 593 559 355 935 593 559 355 935 593 559 354 88;
  • 9) 0.593 559 355 935 593 559 355 935 593 559 355 935 593 559 354 88 × 2 = 1 + 0.187 118 711 871 187 118 711 871 187 118 711 871 187 118 709 76;
  • 10) 0.187 118 711 871 187 118 711 871 187 118 711 871 187 118 709 76 × 2 = 0 + 0.374 237 423 742 374 237 423 742 374 237 423 742 374 237 419 52;
  • 11) 0.374 237 423 742 374 237 423 742 374 237 423 742 374 237 419 52 × 2 = 0 + 0.748 474 847 484 748 474 847 484 748 474 847 484 748 474 839 04;
  • 12) 0.748 474 847 484 748 474 847 484 748 474 847 484 748 474 839 04 × 2 = 1 + 0.496 949 694 969 496 949 694 969 496 949 694 969 496 949 678 08;
  • 13) 0.496 949 694 969 496 949 694 969 496 949 694 969 496 949 678 08 × 2 = 0 + 0.993 899 389 938 993 899 389 938 993 899 389 938 993 899 356 16;
  • 14) 0.993 899 389 938 993 899 389 938 993 899 389 938 993 899 356 16 × 2 = 1 + 0.987 798 779 877 987 798 779 877 987 798 779 877 987 798 712 32;
  • 15) 0.987 798 779 877 987 798 779 877 987 798 779 877 987 798 712 32 × 2 = 1 + 0.975 597 559 755 975 597 559 755 975 597 559 755 975 597 424 64;
  • 16) 0.975 597 559 755 975 597 559 755 975 597 559 755 975 597 424 64 × 2 = 1 + 0.951 195 119 511 951 195 119 511 951 195 119 511 951 194 849 28;
  • 17) 0.951 195 119 511 951 195 119 511 951 195 119 511 951 194 849 28 × 2 = 1 + 0.902 390 239 023 902 390 239 023 902 390 239 023 902 389 698 56;
  • 18) 0.902 390 239 023 902 390 239 023 902 390 239 023 902 389 698 56 × 2 = 1 + 0.804 780 478 047 804 780 478 047 804 780 478 047 804 779 397 12;
  • 19) 0.804 780 478 047 804 780 478 047 804 780 478 047 804 779 397 12 × 2 = 1 + 0.609 560 956 095 609 560 956 095 609 560 956 095 609 558 794 24;
  • 20) 0.609 560 956 095 609 560 956 095 609 560 956 095 609 558 794 24 × 2 = 1 + 0.219 121 912 191 219 121 912 191 219 121 912 191 219 117 588 48;
  • 21) 0.219 121 912 191 219 121 912 191 219 121 912 191 219 117 588 48 × 2 = 0 + 0.438 243 824 382 438 243 824 382 438 243 824 382 438 235 176 96;
  • 22) 0.438 243 824 382 438 243 824 382 438 243 824 382 438 235 176 96 × 2 = 0 + 0.876 487 648 764 876 487 648 764 876 487 648 764 876 470 353 92;
  • 23) 0.876 487 648 764 876 487 648 764 876 487 648 764 876 470 353 92 × 2 = 1 + 0.752 975 297 529 752 975 297 529 752 975 297 529 752 940 707 84;
  • 24) 0.752 975 297 529 752 975 297 529 752 975 297 529 752 940 707 84 × 2 = 1 + 0.505 950 595 059 505 950 595 059 505 950 595 059 505 881 415 68;
  • 25) 0.505 950 595 059 505 950 595 059 505 950 595 059 505 881 415 68 × 2 = 1 + 0.011 901 190 119 011 901 190 119 011 901 190 119 011 762 831 36;
  • 26) 0.011 901 190 119 011 901 190 119 011 901 190 119 011 762 831 36 × 2 = 0 + 0.023 802 380 238 023 802 380 238 023 802 380 238 023 525 662 72;
  • 27) 0.023 802 380 238 023 802 380 238 023 802 380 238 023 525 662 72 × 2 = 0 + 0.047 604 760 476 047 604 760 476 047 604 760 476 047 051 325 44;
  • 28) 0.047 604 760 476 047 604 760 476 047 604 760 476 047 051 325 44 × 2 = 0 + 0.095 209 520 952 095 209 520 952 095 209 520 952 094 102 650 88;
  • 29) 0.095 209 520 952 095 209 520 952 095 209 520 952 094 102 650 88 × 2 = 0 + 0.190 419 041 904 190 419 041 904 190 419 041 904 188 205 301 76;
  • 30) 0.190 419 041 904 190 419 041 904 190 419 041 904 188 205 301 76 × 2 = 0 + 0.380 838 083 808 380 838 083 808 380 838 083 808 376 410 603 52;
  • 31) 0.380 838 083 808 380 838 083 808 380 838 083 808 376 410 603 52 × 2 = 0 + 0.761 676 167 616 761 676 167 616 761 676 167 616 752 821 207 04;
  • 32) 0.761 676 167 616 761 676 167 616 761 676 167 616 752 821 207 04 × 2 = 1 + 0.523 352 335 233 523 352 335 233 523 352 335 233 505 642 414 08;
  • 33) 0.523 352 335 233 523 352 335 233 523 352 335 233 505 642 414 08 × 2 = 1 + 0.046 704 670 467 046 704 670 467 046 704 670 467 011 284 828 16;
  • 34) 0.046 704 670 467 046 704 670 467 046 704 670 467 011 284 828 16 × 2 = 0 + 0.093 409 340 934 093 409 340 934 093 409 340 934 022 569 656 32;
  • 35) 0.093 409 340 934 093 409 340 934 093 409 340 934 022 569 656 32 × 2 = 0 + 0.186 818 681 868 186 818 681 868 186 818 681 868 045 139 312 64;
  • 36) 0.186 818 681 868 186 818 681 868 186 818 681 868 045 139 312 64 × 2 = 0 + 0.373 637 363 736 373 637 363 736 373 637 363 736 090 278 625 28;
  • 37) 0.373 637 363 736 373 637 363 736 373 637 363 736 090 278 625 28 × 2 = 0 + 0.747 274 727 472 747 274 727 472 747 274 727 472 180 557 250 56;
  • 38) 0.747 274 727 472 747 274 727 472 747 274 727 472 180 557 250 56 × 2 = 1 + 0.494 549 454 945 494 549 454 945 494 549 454 944 361 114 501 12;
  • 39) 0.494 549 454 945 494 549 454 945 494 549 454 944 361 114 501 12 × 2 = 0 + 0.989 098 909 890 989 098 909 890 989 098 909 888 722 229 002 24;
  • 40) 0.989 098 909 890 989 098 909 890 989 098 909 888 722 229 002 24 × 2 = 1 + 0.978 197 819 781 978 197 819 781 978 197 819 777 444 458 004 48;
  • 41) 0.978 197 819 781 978 197 819 781 978 197 819 777 444 458 004 48 × 2 = 1 + 0.956 395 639 563 956 395 639 563 956 395 639 554 888 916 008 96;
  • 42) 0.956 395 639 563 956 395 639 563 956 395 639 554 888 916 008 96 × 2 = 1 + 0.912 791 279 127 912 791 279 127 912 791 279 109 777 832 017 92;
  • 43) 0.912 791 279 127 912 791 279 127 912 791 279 109 777 832 017 92 × 2 = 1 + 0.825 582 558 255 825 582 558 255 825 582 558 219 555 664 035 84;
  • 44) 0.825 582 558 255 825 582 558 255 825 582 558 219 555 664 035 84 × 2 = 1 + 0.651 165 116 511 651 165 116 511 651 165 116 439 111 328 071 68;
  • 45) 0.651 165 116 511 651 165 116 511 651 165 116 439 111 328 071 68 × 2 = 1 + 0.302 330 233 023 302 330 233 023 302 330 232 878 222 656 143 36;
  • 46) 0.302 330 233 023 302 330 233 023 302 330 232 878 222 656 143 36 × 2 = 0 + 0.604 660 466 046 604 660 466 046 604 660 465 756 445 312 286 72;
  • 47) 0.604 660 466 046 604 660 466 046 604 660 465 756 445 312 286 72 × 2 = 1 + 0.209 320 932 093 209 320 932 093 209 320 931 512 890 624 573 44;
  • 48) 0.209 320 932 093 209 320 932 093 209 320 931 512 890 624 573 44 × 2 = 0 + 0.418 641 864 186 418 641 864 186 418 641 863 025 781 249 146 88;
  • 49) 0.418 641 864 186 418 641 864 186 418 641 863 025 781 249 146 88 × 2 = 0 + 0.837 283 728 372 837 283 728 372 837 283 726 051 562 498 293 76;
  • 50) 0.837 283 728 372 837 283 728 372 837 283 726 051 562 498 293 76 × 2 = 1 + 0.674 567 456 745 674 567 456 745 674 567 452 103 124 996 587 52;
  • 51) 0.674 567 456 745 674 567 456 745 674 567 452 103 124 996 587 52 × 2 = 1 + 0.349 134 913 491 349 134 913 491 349 134 904 206 249 993 175 04;
  • 52) 0.349 134 913 491 349 134 913 491 349 134 904 206 249 993 175 04 × 2 = 0 + 0.698 269 826 982 698 269 826 982 698 269 808 412 499 986 350 08;
  • 53) 0.698 269 826 982 698 269 826 982 698 269 808 412 499 986 350 08 × 2 = 1 + 0.396 539 653 965 396 539 653 965 396 539 616 824 999 972 700 16;

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...)


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:


0.123 412 341 234 123 412 341 234 123 412 341 234 123 412 341 23(10) =

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


5. Positive number before normalization:

1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 341 23(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:


1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 341 23(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


Number 1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 341 23 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1111 - 0001 1111 1001 0111 1111 0011 1000 0001 1000 0101 1111 1010 0110

(64 bits IEEE 754)

More operations of this kind:

1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 341 22 = ? ... 1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 341 24 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

1.123 412 341 234 123 412 341 234 123 412 341 234 123 412 341 23 to 64 bit double precision IEEE 754 binary floating point = ? Mar 24 09:04 UTC (GMT)
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10 001 110 011 001 111 000 101 110 010 011 001 110 011 010 100 110 010 011 011 010 011 010 010 001 001 000 100 111 101 001 101 101 110 101 011 100 110 101 001 100 100 110 110 111 111 110 111 110 100 110 101 100 111 011 110 010 011 001 000 100 100 011 000 101 111 101 101 111 001 100 100 111 110 101 010 011 100 110 010 111 111 011 010 095 to 64 bit double precision IEEE 754 binary floating point = ? Mar 24 08:58 UTC (GMT)
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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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