0.052 631 578 947 368 418 130 992 040 460 114 367 306 232 452 391 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.052 631 578 947 368 418 130 992 040 460 114 367 306 232 452 391(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
0.052 631 578 947 368 418 130 992 040 460 114 367 306 232 452 391(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: 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;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.052 631 578 947 368 418 130 992 040 460 114 367 306 232 452 391.

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) 0.052 631 578 947 368 418 130 992 040 460 114 367 306 232 452 391 × 2 = 0 + 0.105 263 157 894 736 836 261 984 080 920 228 734 612 464 904 782;
  • 2) 0.105 263 157 894 736 836 261 984 080 920 228 734 612 464 904 782 × 2 = 0 + 0.210 526 315 789 473 672 523 968 161 840 457 469 224 929 809 564;
  • 3) 0.210 526 315 789 473 672 523 968 161 840 457 469 224 929 809 564 × 2 = 0 + 0.421 052 631 578 947 345 047 936 323 680 914 938 449 859 619 128;
  • 4) 0.421 052 631 578 947 345 047 936 323 680 914 938 449 859 619 128 × 2 = 0 + 0.842 105 263 157 894 690 095 872 647 361 829 876 899 719 238 256;
  • 5) 0.842 105 263 157 894 690 095 872 647 361 829 876 899 719 238 256 × 2 = 1 + 0.684 210 526 315 789 380 191 745 294 723 659 753 799 438 476 512;
  • 6) 0.684 210 526 315 789 380 191 745 294 723 659 753 799 438 476 512 × 2 = 1 + 0.368 421 052 631 578 760 383 490 589 447 319 507 598 876 953 024;
  • 7) 0.368 421 052 631 578 760 383 490 589 447 319 507 598 876 953 024 × 2 = 0 + 0.736 842 105 263 157 520 766 981 178 894 639 015 197 753 906 048;
  • 8) 0.736 842 105 263 157 520 766 981 178 894 639 015 197 753 906 048 × 2 = 1 + 0.473 684 210 526 315 041 533 962 357 789 278 030 395 507 812 096;
  • 9) 0.473 684 210 526 315 041 533 962 357 789 278 030 395 507 812 096 × 2 = 0 + 0.947 368 421 052 630 083 067 924 715 578 556 060 791 015 624 192;
  • 10) 0.947 368 421 052 630 083 067 924 715 578 556 060 791 015 624 192 × 2 = 1 + 0.894 736 842 105 260 166 135 849 431 157 112 121 582 031 248 384;
  • 11) 0.894 736 842 105 260 166 135 849 431 157 112 121 582 031 248 384 × 2 = 1 + 0.789 473 684 210 520 332 271 698 862 314 224 243 164 062 496 768;
  • 12) 0.789 473 684 210 520 332 271 698 862 314 224 243 164 062 496 768 × 2 = 1 + 0.578 947 368 421 040 664 543 397 724 628 448 486 328 124 993 536;
  • 13) 0.578 947 368 421 040 664 543 397 724 628 448 486 328 124 993 536 × 2 = 1 + 0.157 894 736 842 081 329 086 795 449 256 896 972 656 249 987 072;
  • 14) 0.157 894 736 842 081 329 086 795 449 256 896 972 656 249 987 072 × 2 = 0 + 0.315 789 473 684 162 658 173 590 898 513 793 945 312 499 974 144;
  • 15) 0.315 789 473 684 162 658 173 590 898 513 793 945 312 499 974 144 × 2 = 0 + 0.631 578 947 368 325 316 347 181 797 027 587 890 624 999 948 288;
  • 16) 0.631 578 947 368 325 316 347 181 797 027 587 890 624 999 948 288 × 2 = 1 + 0.263 157 894 736 650 632 694 363 594 055 175 781 249 999 896 576;
  • 17) 0.263 157 894 736 650 632 694 363 594 055 175 781 249 999 896 576 × 2 = 0 + 0.526 315 789 473 301 265 388 727 188 110 351 562 499 999 793 152;
  • 18) 0.526 315 789 473 301 265 388 727 188 110 351 562 499 999 793 152 × 2 = 1 + 0.052 631 578 946 602 530 777 454 376 220 703 124 999 999 586 304;
  • 19) 0.052 631 578 946 602 530 777 454 376 220 703 124 999 999 586 304 × 2 = 0 + 0.105 263 157 893 205 061 554 908 752 441 406 249 999 999 172 608;
  • 20) 0.105 263 157 893 205 061 554 908 752 441 406 249 999 999 172 608 × 2 = 0 + 0.210 526 315 786 410 123 109 817 504 882 812 499 999 998 345 216;
  • 21) 0.210 526 315 786 410 123 109 817 504 882 812 499 999 998 345 216 × 2 = 0 + 0.421 052 631 572 820 246 219 635 009 765 624 999 999 996 690 432;
  • 22) 0.421 052 631 572 820 246 219 635 009 765 624 999 999 996 690 432 × 2 = 0 + 0.842 105 263 145 640 492 439 270 019 531 249 999 999 993 380 864;
  • 23) 0.842 105 263 145 640 492 439 270 019 531 249 999 999 993 380 864 × 2 = 1 + 0.684 210 526 291 280 984 878 540 039 062 499 999 999 986 761 728;
  • 24) 0.684 210 526 291 280 984 878 540 039 062 499 999 999 986 761 728 × 2 = 1 + 0.368 421 052 582 561 969 757 080 078 124 999 999 999 973 523 456;
  • 25) 0.368 421 052 582 561 969 757 080 078 124 999 999 999 973 523 456 × 2 = 0 + 0.736 842 105 165 123 939 514 160 156 249 999 999 999 947 046 912;
  • 26) 0.736 842 105 165 123 939 514 160 156 249 999 999 999 947 046 912 × 2 = 1 + 0.473 684 210 330 247 879 028 320 312 499 999 999 999 894 093 824;
  • 27) 0.473 684 210 330 247 879 028 320 312 499 999 999 999 894 093 824 × 2 = 0 + 0.947 368 420 660 495 758 056 640 624 999 999 999 999 788 187 648;
  • 28) 0.947 368 420 660 495 758 056 640 624 999 999 999 999 788 187 648 × 2 = 1 + 0.894 736 841 320 991 516 113 281 249 999 999 999 999 576 375 296;
  • 29) 0.894 736 841 320 991 516 113 281 249 999 999 999 999 576 375 296 × 2 = 1 + 0.789 473 682 641 983 032 226 562 499 999 999 999 999 152 750 592;
  • 30) 0.789 473 682 641 983 032 226 562 499 999 999 999 999 152 750 592 × 2 = 1 + 0.578 947 365 283 966 064 453 124 999 999 999 999 998 305 501 184;
  • 31) 0.578 947 365 283 966 064 453 124 999 999 999 999 998 305 501 184 × 2 = 1 + 0.157 894 730 567 932 128 906 249 999 999 999 999 996 611 002 368;
  • 32) 0.157 894 730 567 932 128 906 249 999 999 999 999 996 611 002 368 × 2 = 0 + 0.315 789 461 135 864 257 812 499 999 999 999 999 993 222 004 736;
  • 33) 0.315 789 461 135 864 257 812 499 999 999 999 999 993 222 004 736 × 2 = 0 + 0.631 578 922 271 728 515 624 999 999 999 999 999 986 444 009 472;
  • 34) 0.631 578 922 271 728 515 624 999 999 999 999 999 986 444 009 472 × 2 = 1 + 0.263 157 844 543 457 031 249 999 999 999 999 999 972 888 018 944;
  • 35) 0.263 157 844 543 457 031 249 999 999 999 999 999 972 888 018 944 × 2 = 0 + 0.526 315 689 086 914 062 499 999 999 999 999 999 945 776 037 888;
  • 36) 0.526 315 689 086 914 062 499 999 999 999 999 999 945 776 037 888 × 2 = 1 + 0.052 631 378 173 828 124 999 999 999 999 999 999 891 552 075 776;
  • 37) 0.052 631 378 173 828 124 999 999 999 999 999 999 891 552 075 776 × 2 = 0 + 0.105 262 756 347 656 249 999 999 999 999 999 999 783 104 151 552;
  • 38) 0.105 262 756 347 656 249 999 999 999 999 999 999 783 104 151 552 × 2 = 0 + 0.210 525 512 695 312 499 999 999 999 999 999 999 566 208 303 104;
  • 39) 0.210 525 512 695 312 499 999 999 999 999 999 999 566 208 303 104 × 2 = 0 + 0.421 051 025 390 624 999 999 999 999 999 999 999 132 416 606 208;
  • 40) 0.421 051 025 390 624 999 999 999 999 999 999 999 132 416 606 208 × 2 = 0 + 0.842 102 050 781 249 999 999 999 999 999 999 998 264 833 212 416;
  • 41) 0.842 102 050 781 249 999 999 999 999 999 999 998 264 833 212 416 × 2 = 1 + 0.684 204 101 562 499 999 999 999 999 999 999 996 529 666 424 832;
  • 42) 0.684 204 101 562 499 999 999 999 999 999 999 996 529 666 424 832 × 2 = 1 + 0.368 408 203 124 999 999 999 999 999 999 999 993 059 332 849 664;
  • 43) 0.368 408 203 124 999 999 999 999 999 999 999 993 059 332 849 664 × 2 = 0 + 0.736 816 406 249 999 999 999 999 999 999 999 986 118 665 699 328;
  • 44) 0.736 816 406 249 999 999 999 999 999 999 999 986 118 665 699 328 × 2 = 1 + 0.473 632 812 499 999 999 999 999 999 999 999 972 237 331 398 656;
  • 45) 0.473 632 812 499 999 999 999 999 999 999 999 972 237 331 398 656 × 2 = 0 + 0.947 265 624 999 999 999 999 999 999 999 999 944 474 662 797 312;
  • 46) 0.947 265 624 999 999 999 999 999 999 999 999 944 474 662 797 312 × 2 = 1 + 0.894 531 249 999 999 999 999 999 999 999 999 888 949 325 594 624;
  • 47) 0.894 531 249 999 999 999 999 999 999 999 999 888 949 325 594 624 × 2 = 1 + 0.789 062 499 999 999 999 999 999 999 999 999 777 898 651 189 248;
  • 48) 0.789 062 499 999 999 999 999 999 999 999 999 777 898 651 189 248 × 2 = 1 + 0.578 124 999 999 999 999 999 999 999 999 999 555 797 302 378 496;
  • 49) 0.578 124 999 999 999 999 999 999 999 999 999 555 797 302 378 496 × 2 = 1 + 0.156 249 999 999 999 999 999 999 999 999 999 111 594 604 756 992;
  • 50) 0.156 249 999 999 999 999 999 999 999 999 999 111 594 604 756 992 × 2 = 0 + 0.312 499 999 999 999 999 999 999 999 999 998 223 189 209 513 984;
  • 51) 0.312 499 999 999 999 999 999 999 999 999 998 223 189 209 513 984 × 2 = 0 + 0.624 999 999 999 999 999 999 999 999 999 996 446 378 419 027 968;
  • 52) 0.624 999 999 999 999 999 999 999 999 999 996 446 378 419 027 968 × 2 = 1 + 0.249 999 999 999 999 999 999 999 999 999 992 892 756 838 055 936;
  • 53) 0.249 999 999 999 999 999 999 999 999 999 992 892 756 838 055 936 × 2 = 0 + 0.499 999 999 999 999 999 999 999 999 999 985 785 513 676 111 872;
  • 54) 0.499 999 999 999 999 999 999 999 999 999 985 785 513 676 111 872 × 2 = 0 + 0.999 999 999 999 999 999 999 999 999 999 971 571 027 352 223 744;
  • 55) 0.999 999 999 999 999 999 999 999 999 999 971 571 027 352 223 744 × 2 = 1 + 0.999 999 999 999 999 999 999 999 999 999 943 142 054 704 447 488;
  • 56) 0.999 999 999 999 999 999 999 999 999 999 943 142 054 704 447 488 × 2 = 1 + 0.999 999 999 999 999 999 999 999 999 999 886 284 109 408 894 976;
  • 57) 0.999 999 999 999 999 999 999 999 999 999 886 284 109 408 894 976 × 2 = 1 + 0.999 999 999 999 999 999 999 999 999 999 772 568 218 817 789 952;

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:


0.052 631 578 947 368 418 130 992 040 460 114 367 306 232 452 391(10) =

0.0000 1101 0111 1001 0100 0011 0101 1110 0101 0000 1101 0111 1001 0011 1(2)

5. Positive number before normalization:

0.052 631 578 947 368 418 130 992 040 460 114 367 306 232 452 391(10) =

0.0000 1101 0111 1001 0100 0011 0101 1110 0101 0000 1101 0111 1001 0011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.052 631 578 947 368 418 130 992 040 460 114 367 306 232 452 391(10) =

0.0000 1101 0111 1001 0100 0011 0101 1110 0101 0000 1101 0111 1001 0011 1(2) =

0.0000 1101 0111 1001 0100 0011 0101 1110 0101 0000 1101 0111 1001 0011 1(2) × 20 =

1.1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 0111(2) × 2-5


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): -5

Mantissa (not normalized):
1.1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 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;

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) =

1018(10) =

011 1111 1010(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 0111 =

1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 0111


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 1010

Mantissa (52 bits) =
1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 0111


Decimal number 0.052 631 578 947 368 418 130 992 040 460 114 367 306 232 452 391 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1010 - 1010 1111 0010 1000 0110 1011 1100 1010 0001 1010 1111 0010 0111

Share

» Convert decimal 0.052 631 578 947 368 418 130 992 040 460 114 367 306 232 452 314 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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