-0.085 969 999 999 999 990 869 525 845 482 712 611 556 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.085 969 999 999 999 990 869 525 845 482 712 611 556 3(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
-0.085 969 999 999 999 990 869 525 845 482 712 611 556 3(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:

|-0.085 969 999 999 999 990 869 525 845 482 712 611 556 3| = 0.085 969 999 999 999 990 869 525 845 482 712 611 556 3


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: 0.085 969 999 999 999 990 869 525 845 482 712 611 556 3.

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) 0.085 969 999 999 999 990 869 525 845 482 712 611 556 3 × 2 = 0 + 0.171 939 999 999 999 981 739 051 690 965 425 223 112 6;
  • 2) 0.171 939 999 999 999 981 739 051 690 965 425 223 112 6 × 2 = 0 + 0.343 879 999 999 999 963 478 103 381 930 850 446 225 2;
  • 3) 0.343 879 999 999 999 963 478 103 381 930 850 446 225 2 × 2 = 0 + 0.687 759 999 999 999 926 956 206 763 861 700 892 450 4;
  • 4) 0.687 759 999 999 999 926 956 206 763 861 700 892 450 4 × 2 = 1 + 0.375 519 999 999 999 853 912 413 527 723 401 784 900 8;
  • 5) 0.375 519 999 999 999 853 912 413 527 723 401 784 900 8 × 2 = 0 + 0.751 039 999 999 999 707 824 827 055 446 803 569 801 6;
  • 6) 0.751 039 999 999 999 707 824 827 055 446 803 569 801 6 × 2 = 1 + 0.502 079 999 999 999 415 649 654 110 893 607 139 603 2;
  • 7) 0.502 079 999 999 999 415 649 654 110 893 607 139 603 2 × 2 = 1 + 0.004 159 999 999 998 831 299 308 221 787 214 279 206 4;
  • 8) 0.004 159 999 999 998 831 299 308 221 787 214 279 206 4 × 2 = 0 + 0.008 319 999 999 997 662 598 616 443 574 428 558 412 8;
  • 9) 0.008 319 999 999 997 662 598 616 443 574 428 558 412 8 × 2 = 0 + 0.016 639 999 999 995 325 197 232 887 148 857 116 825 6;
  • 10) 0.016 639 999 999 995 325 197 232 887 148 857 116 825 6 × 2 = 0 + 0.033 279 999 999 990 650 394 465 774 297 714 233 651 2;
  • 11) 0.033 279 999 999 990 650 394 465 774 297 714 233 651 2 × 2 = 0 + 0.066 559 999 999 981 300 788 931 548 595 428 467 302 4;
  • 12) 0.066 559 999 999 981 300 788 931 548 595 428 467 302 4 × 2 = 0 + 0.133 119 999 999 962 601 577 863 097 190 856 934 604 8;
  • 13) 0.133 119 999 999 962 601 577 863 097 190 856 934 604 8 × 2 = 0 + 0.266 239 999 999 925 203 155 726 194 381 713 869 209 6;
  • 14) 0.266 239 999 999 925 203 155 726 194 381 713 869 209 6 × 2 = 0 + 0.532 479 999 999 850 406 311 452 388 763 427 738 419 2;
  • 15) 0.532 479 999 999 850 406 311 452 388 763 427 738 419 2 × 2 = 1 + 0.064 959 999 999 700 812 622 904 777 526 855 476 838 4;
  • 16) 0.064 959 999 999 700 812 622 904 777 526 855 476 838 4 × 2 = 0 + 0.129 919 999 999 401 625 245 809 555 053 710 953 676 8;
  • 17) 0.129 919 999 999 401 625 245 809 555 053 710 953 676 8 × 2 = 0 + 0.259 839 999 998 803 250 491 619 110 107 421 907 353 6;
  • 18) 0.259 839 999 998 803 250 491 619 110 107 421 907 353 6 × 2 = 0 + 0.519 679 999 997 606 500 983 238 220 214 843 814 707 2;
  • 19) 0.519 679 999 997 606 500 983 238 220 214 843 814 707 2 × 2 = 1 + 0.039 359 999 995 213 001 966 476 440 429 687 629 414 4;
  • 20) 0.039 359 999 995 213 001 966 476 440 429 687 629 414 4 × 2 = 0 + 0.078 719 999 990 426 003 932 952 880 859 375 258 828 8;
  • 21) 0.078 719 999 990 426 003 932 952 880 859 375 258 828 8 × 2 = 0 + 0.157 439 999 980 852 007 865 905 761 718 750 517 657 6;
  • 22) 0.157 439 999 980 852 007 865 905 761 718 750 517 657 6 × 2 = 0 + 0.314 879 999 961 704 015 731 811 523 437 501 035 315 2;
  • 23) 0.314 879 999 961 704 015 731 811 523 437 501 035 315 2 × 2 = 0 + 0.629 759 999 923 408 031 463 623 046 875 002 070 630 4;
  • 24) 0.629 759 999 923 408 031 463 623 046 875 002 070 630 4 × 2 = 1 + 0.259 519 999 846 816 062 927 246 093 750 004 141 260 8;
  • 25) 0.259 519 999 846 816 062 927 246 093 750 004 141 260 8 × 2 = 0 + 0.519 039 999 693 632 125 854 492 187 500 008 282 521 6;
  • 26) 0.519 039 999 693 632 125 854 492 187 500 008 282 521 6 × 2 = 1 + 0.038 079 999 387 264 251 708 984 375 000 016 565 043 2;
  • 27) 0.038 079 999 387 264 251 708 984 375 000 016 565 043 2 × 2 = 0 + 0.076 159 998 774 528 503 417 968 750 000 033 130 086 4;
  • 28) 0.076 159 998 774 528 503 417 968 750 000 033 130 086 4 × 2 = 0 + 0.152 319 997 549 057 006 835 937 500 000 066 260 172 8;
  • 29) 0.152 319 997 549 057 006 835 937 500 000 066 260 172 8 × 2 = 0 + 0.304 639 995 098 114 013 671 875 000 000 132 520 345 6;
  • 30) 0.304 639 995 098 114 013 671 875 000 000 132 520 345 6 × 2 = 0 + 0.609 279 990 196 228 027 343 750 000 000 265 040 691 2;
  • 31) 0.609 279 990 196 228 027 343 750 000 000 265 040 691 2 × 2 = 1 + 0.218 559 980 392 456 054 687 500 000 000 530 081 382 4;
  • 32) 0.218 559 980 392 456 054 687 500 000 000 530 081 382 4 × 2 = 0 + 0.437 119 960 784 912 109 375 000 000 001 060 162 764 8;
  • 33) 0.437 119 960 784 912 109 375 000 000 001 060 162 764 8 × 2 = 0 + 0.874 239 921 569 824 218 750 000 000 002 120 325 529 6;
  • 34) 0.874 239 921 569 824 218 750 000 000 002 120 325 529 6 × 2 = 1 + 0.748 479 843 139 648 437 500 000 000 004 240 651 059 2;
  • 35) 0.748 479 843 139 648 437 500 000 000 004 240 651 059 2 × 2 = 1 + 0.496 959 686 279 296 875 000 000 000 008 481 302 118 4;
  • 36) 0.496 959 686 279 296 875 000 000 000 008 481 302 118 4 × 2 = 0 + 0.993 919 372 558 593 750 000 000 000 016 962 604 236 8;
  • 37) 0.993 919 372 558 593 750 000 000 000 016 962 604 236 8 × 2 = 1 + 0.987 838 745 117 187 500 000 000 000 033 925 208 473 6;
  • 38) 0.987 838 745 117 187 500 000 000 000 033 925 208 473 6 × 2 = 1 + 0.975 677 490 234 375 000 000 000 000 067 850 416 947 2;
  • 39) 0.975 677 490 234 375 000 000 000 000 067 850 416 947 2 × 2 = 1 + 0.951 354 980 468 750 000 000 000 000 135 700 833 894 4;
  • 40) 0.951 354 980 468 750 000 000 000 000 135 700 833 894 4 × 2 = 1 + 0.902 709 960 937 500 000 000 000 000 271 401 667 788 8;
  • 41) 0.902 709 960 937 500 000 000 000 000 271 401 667 788 8 × 2 = 1 + 0.805 419 921 875 000 000 000 000 000 542 803 335 577 6;
  • 42) 0.805 419 921 875 000 000 000 000 000 542 803 335 577 6 × 2 = 1 + 0.610 839 843 750 000 000 000 000 001 085 606 671 155 2;
  • 43) 0.610 839 843 750 000 000 000 000 001 085 606 671 155 2 × 2 = 1 + 0.221 679 687 500 000 000 000 000 002 171 213 342 310 4;
  • 44) 0.221 679 687 500 000 000 000 000 002 171 213 342 310 4 × 2 = 0 + 0.443 359 375 000 000 000 000 000 004 342 426 684 620 8;
  • 45) 0.443 359 375 000 000 000 000 000 004 342 426 684 620 8 × 2 = 0 + 0.886 718 750 000 000 000 000 000 008 684 853 369 241 6;
  • 46) 0.886 718 750 000 000 000 000 000 008 684 853 369 241 6 × 2 = 1 + 0.773 437 500 000 000 000 000 000 017 369 706 738 483 2;
  • 47) 0.773 437 500 000 000 000 000 000 017 369 706 738 483 2 × 2 = 1 + 0.546 875 000 000 000 000 000 000 034 739 413 476 966 4;
  • 48) 0.546 875 000 000 000 000 000 000 034 739 413 476 966 4 × 2 = 1 + 0.093 750 000 000 000 000 000 000 069 478 826 953 932 8;
  • 49) 0.093 750 000 000 000 000 000 000 069 478 826 953 932 8 × 2 = 0 + 0.187 500 000 000 000 000 000 000 138 957 653 907 865 6;
  • 50) 0.187 500 000 000 000 000 000 000 138 957 653 907 865 6 × 2 = 0 + 0.375 000 000 000 000 000 000 000 277 915 307 815 731 2;
  • 51) 0.375 000 000 000 000 000 000 000 277 915 307 815 731 2 × 2 = 0 + 0.750 000 000 000 000 000 000 000 555 830 615 631 462 4;
  • 52) 0.750 000 000 000 000 000 000 000 555 830 615 631 462 4 × 2 = 1 + 0.500 000 000 000 000 000 000 001 111 661 231 262 924 8;
  • 53) 0.500 000 000 000 000 000 000 001 111 661 231 262 924 8 × 2 = 1 + 0.000 000 000 000 000 000 000 002 223 322 462 525 849 6;
  • 54) 0.000 000 000 000 000 000 000 002 223 322 462 525 849 6 × 2 = 0 + 0.000 000 000 000 000 000 000 004 446 644 925 051 699 2;
  • 55) 0.000 000 000 000 000 000 000 004 446 644 925 051 699 2 × 2 = 0 + 0.000 000 000 000 000 000 000 008 893 289 850 103 398 4;
  • 56) 0.000 000 000 000 000 000 000 008 893 289 850 103 398 4 × 2 = 0 + 0.000 000 000 000 000 000 000 017 786 579 700 206 796 8;

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:


0.085 969 999 999 999 990 869 525 845 482 712 611 556 3(10) =

0.0001 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0111 0001 1000(2)

6. Positive number before normalization:

0.085 969 999 999 999 990 869 525 845 482 712 611 556 3(10) =

0.0001 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0111 0001 1000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.085 969 999 999 999 990 869 525 845 482 712 611 556 3(10) =

0.0001 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0111 0001 1000(2) =

0.0001 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0111 0001 1000(2) × 20 =

1.0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0111 0001 1000(2) × 2-4


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):
1.0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0111 0001 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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;

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) =

1019(10) =

011 1111 1011(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. 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0111 0001 1000 =

0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0111 0001 1000


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 1111 1011

Mantissa (52 bits) =
0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0111 0001 1000


Decimal number -0.085 969 999 999 999 990 869 525 845 482 712 611 556 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0110 0000 0010 0010 0001 0100 0010 0110 1111 1110 0111 0001 1000

» Convert decimal -0.085 969 999 999 999 990 869 525 845 482 712 611 563 7 to 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