-0.000 000 000 000 293 890 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 293 890 8(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
-0.000 000 000 000 293 890 8(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:

|-0.000 000 000 000 293 890 8| = 0.000 000 000 000 293 890 8


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: 0.000 000 000 000 293 890 8.

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) 0.000 000 000 000 293 890 8 × 2 = 0 + 0.000 000 000 000 587 781 6;
  • 2) 0.000 000 000 000 587 781 6 × 2 = 0 + 0.000 000 000 001 175 563 2;
  • 3) 0.000 000 000 001 175 563 2 × 2 = 0 + 0.000 000 000 002 351 126 4;
  • 4) 0.000 000 000 002 351 126 4 × 2 = 0 + 0.000 000 000 004 702 252 8;
  • 5) 0.000 000 000 004 702 252 8 × 2 = 0 + 0.000 000 000 009 404 505 6;
  • 6) 0.000 000 000 009 404 505 6 × 2 = 0 + 0.000 000 000 018 809 011 2;
  • 7) 0.000 000 000 018 809 011 2 × 2 = 0 + 0.000 000 000 037 618 022 4;
  • 8) 0.000 000 000 037 618 022 4 × 2 = 0 + 0.000 000 000 075 236 044 8;
  • 9) 0.000 000 000 075 236 044 8 × 2 = 0 + 0.000 000 000 150 472 089 6;
  • 10) 0.000 000 000 150 472 089 6 × 2 = 0 + 0.000 000 000 300 944 179 2;
  • 11) 0.000 000 000 300 944 179 2 × 2 = 0 + 0.000 000 000 601 888 358 4;
  • 12) 0.000 000 000 601 888 358 4 × 2 = 0 + 0.000 000 001 203 776 716 8;
  • 13) 0.000 000 001 203 776 716 8 × 2 = 0 + 0.000 000 002 407 553 433 6;
  • 14) 0.000 000 002 407 553 433 6 × 2 = 0 + 0.000 000 004 815 106 867 2;
  • 15) 0.000 000 004 815 106 867 2 × 2 = 0 + 0.000 000 009 630 213 734 4;
  • 16) 0.000 000 009 630 213 734 4 × 2 = 0 + 0.000 000 019 260 427 468 8;
  • 17) 0.000 000 019 260 427 468 8 × 2 = 0 + 0.000 000 038 520 854 937 6;
  • 18) 0.000 000 038 520 854 937 6 × 2 = 0 + 0.000 000 077 041 709 875 2;
  • 19) 0.000 000 077 041 709 875 2 × 2 = 0 + 0.000 000 154 083 419 750 4;
  • 20) 0.000 000 154 083 419 750 4 × 2 = 0 + 0.000 000 308 166 839 500 8;
  • 21) 0.000 000 308 166 839 500 8 × 2 = 0 + 0.000 000 616 333 679 001 6;
  • 22) 0.000 000 616 333 679 001 6 × 2 = 0 + 0.000 001 232 667 358 003 2;
  • 23) 0.000 001 232 667 358 003 2 × 2 = 0 + 0.000 002 465 334 716 006 4;
  • 24) 0.000 002 465 334 716 006 4 × 2 = 0 + 0.000 004 930 669 432 012 8;
  • 25) 0.000 004 930 669 432 012 8 × 2 = 0 + 0.000 009 861 338 864 025 6;
  • 26) 0.000 009 861 338 864 025 6 × 2 = 0 + 0.000 019 722 677 728 051 2;
  • 27) 0.000 019 722 677 728 051 2 × 2 = 0 + 0.000 039 445 355 456 102 4;
  • 28) 0.000 039 445 355 456 102 4 × 2 = 0 + 0.000 078 890 710 912 204 8;
  • 29) 0.000 078 890 710 912 204 8 × 2 = 0 + 0.000 157 781 421 824 409 6;
  • 30) 0.000 157 781 421 824 409 6 × 2 = 0 + 0.000 315 562 843 648 819 2;
  • 31) 0.000 315 562 843 648 819 2 × 2 = 0 + 0.000 631 125 687 297 638 4;
  • 32) 0.000 631 125 687 297 638 4 × 2 = 0 + 0.001 262 251 374 595 276 8;
  • 33) 0.001 262 251 374 595 276 8 × 2 = 0 + 0.002 524 502 749 190 553 6;
  • 34) 0.002 524 502 749 190 553 6 × 2 = 0 + 0.005 049 005 498 381 107 2;
  • 35) 0.005 049 005 498 381 107 2 × 2 = 0 + 0.010 098 010 996 762 214 4;
  • 36) 0.010 098 010 996 762 214 4 × 2 = 0 + 0.020 196 021 993 524 428 8;
  • 37) 0.020 196 021 993 524 428 8 × 2 = 0 + 0.040 392 043 987 048 857 6;
  • 38) 0.040 392 043 987 048 857 6 × 2 = 0 + 0.080 784 087 974 097 715 2;
  • 39) 0.080 784 087 974 097 715 2 × 2 = 0 + 0.161 568 175 948 195 430 4;
  • 40) 0.161 568 175 948 195 430 4 × 2 = 0 + 0.323 136 351 896 390 860 8;
  • 41) 0.323 136 351 896 390 860 8 × 2 = 0 + 0.646 272 703 792 781 721 6;
  • 42) 0.646 272 703 792 781 721 6 × 2 = 1 + 0.292 545 407 585 563 443 2;
  • 43) 0.292 545 407 585 563 443 2 × 2 = 0 + 0.585 090 815 171 126 886 4;
  • 44) 0.585 090 815 171 126 886 4 × 2 = 1 + 0.170 181 630 342 253 772 8;
  • 45) 0.170 181 630 342 253 772 8 × 2 = 0 + 0.340 363 260 684 507 545 6;
  • 46) 0.340 363 260 684 507 545 6 × 2 = 0 + 0.680 726 521 369 015 091 2;
  • 47) 0.680 726 521 369 015 091 2 × 2 = 1 + 0.361 453 042 738 030 182 4;
  • 48) 0.361 453 042 738 030 182 4 × 2 = 0 + 0.722 906 085 476 060 364 8;
  • 49) 0.722 906 085 476 060 364 8 × 2 = 1 + 0.445 812 170 952 120 729 6;
  • 50) 0.445 812 170 952 120 729 6 × 2 = 0 + 0.891 624 341 904 241 459 2;
  • 51) 0.891 624 341 904 241 459 2 × 2 = 1 + 0.783 248 683 808 482 918 4;
  • 52) 0.783 248 683 808 482 918 4 × 2 = 1 + 0.566 497 367 616 965 836 8;
  • 53) 0.566 497 367 616 965 836 8 × 2 = 1 + 0.132 994 735 233 931 673 6;
  • 54) 0.132 994 735 233 931 673 6 × 2 = 0 + 0.265 989 470 467 863 347 2;
  • 55) 0.265 989 470 467 863 347 2 × 2 = 0 + 0.531 978 940 935 726 694 4;
  • 56) 0.531 978 940 935 726 694 4 × 2 = 1 + 0.063 957 881 871 453 388 8;
  • 57) 0.063 957 881 871 453 388 8 × 2 = 0 + 0.127 915 763 742 906 777 6;
  • 58) 0.127 915 763 742 906 777 6 × 2 = 0 + 0.255 831 527 485 813 555 2;
  • 59) 0.255 831 527 485 813 555 2 × 2 = 0 + 0.511 663 054 971 627 110 4;
  • 60) 0.511 663 054 971 627 110 4 × 2 = 1 + 0.023 326 109 943 254 220 8;
  • 61) 0.023 326 109 943 254 220 8 × 2 = 0 + 0.046 652 219 886 508 441 6;
  • 62) 0.046 652 219 886 508 441 6 × 2 = 0 + 0.093 304 439 773 016 883 2;
  • 63) 0.093 304 439 773 016 883 2 × 2 = 0 + 0.186 608 879 546 033 766 4;
  • 64) 0.186 608 879 546 033 766 4 × 2 = 0 + 0.373 217 759 092 067 532 8;
  • 65) 0.373 217 759 092 067 532 8 × 2 = 0 + 0.746 435 518 184 135 065 6;
  • 66) 0.746 435 518 184 135 065 6 × 2 = 1 + 0.492 871 036 368 270 131 2;
  • 67) 0.492 871 036 368 270 131 2 × 2 = 0 + 0.985 742 072 736 540 262 4;
  • 68) 0.985 742 072 736 540 262 4 × 2 = 1 + 0.971 484 145 473 080 524 8;
  • 69) 0.971 484 145 473 080 524 8 × 2 = 1 + 0.942 968 290 946 161 049 6;
  • 70) 0.942 968 290 946 161 049 6 × 2 = 1 + 0.885 936 581 892 322 099 2;
  • 71) 0.885 936 581 892 322 099 2 × 2 = 1 + 0.771 873 163 784 644 198 4;
  • 72) 0.771 873 163 784 644 198 4 × 2 = 1 + 0.543 746 327 569 288 396 8;
  • 73) 0.543 746 327 569 288 396 8 × 2 = 1 + 0.087 492 655 138 576 793 6;
  • 74) 0.087 492 655 138 576 793 6 × 2 = 0 + 0.174 985 310 277 153 587 2;
  • 75) 0.174 985 310 277 153 587 2 × 2 = 0 + 0.349 970 620 554 307 174 4;
  • 76) 0.349 970 620 554 307 174 4 × 2 = 0 + 0.699 941 241 108 614 348 8;
  • 77) 0.699 941 241 108 614 348 8 × 2 = 1 + 0.399 882 482 217 228 697 6;
  • 78) 0.399 882 482 217 228 697 6 × 2 = 0 + 0.799 764 964 434 457 395 2;
  • 79) 0.799 764 964 434 457 395 2 × 2 = 1 + 0.599 529 928 868 914 790 4;
  • 80) 0.599 529 928 868 914 790 4 × 2 = 1 + 0.199 059 857 737 829 580 8;
  • 81) 0.199 059 857 737 829 580 8 × 2 = 0 + 0.398 119 715 475 659 161 6;
  • 82) 0.398 119 715 475 659 161 6 × 2 = 0 + 0.796 239 430 951 318 323 2;
  • 83) 0.796 239 430 951 318 323 2 × 2 = 1 + 0.592 478 861 902 636 646 4;
  • 84) 0.592 478 861 902 636 646 4 × 2 = 1 + 0.184 957 723 805 273 292 8;
  • 85) 0.184 957 723 805 273 292 8 × 2 = 0 + 0.369 915 447 610 546 585 6;
  • 86) 0.369 915 447 610 546 585 6 × 2 = 0 + 0.739 830 895 221 093 171 2;
  • 87) 0.739 830 895 221 093 171 2 × 2 = 1 + 0.479 661 790 442 186 342 4;
  • 88) 0.479 661 790 442 186 342 4 × 2 = 0 + 0.959 323 580 884 372 684 8;
  • 89) 0.959 323 580 884 372 684 8 × 2 = 1 + 0.918 647 161 768 745 369 6;
  • 90) 0.918 647 161 768 745 369 6 × 2 = 1 + 0.837 294 323 537 490 739 2;
  • 91) 0.837 294 323 537 490 739 2 × 2 = 1 + 0.674 588 647 074 981 478 4;
  • 92) 0.674 588 647 074 981 478 4 × 2 = 1 + 0.349 177 294 149 962 956 8;
  • 93) 0.349 177 294 149 962 956 8 × 2 = 0 + 0.698 354 588 299 925 913 6;
  • 94) 0.698 354 588 299 925 913 6 × 2 = 1 + 0.396 709 176 599 851 827 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).


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:


0.000 000 000 000 293 890 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1001 0001 0000 0101 1111 1000 1011 0011 0010 1111 01(2)

6. Positive number before normalization:

0.000 000 000 000 293 890 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1001 0001 0000 0101 1111 1000 1011 0011 0010 1111 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 42 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 293 890 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1001 0001 0000 0101 1111 1000 1011 0011 0010 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1001 0001 0000 0101 1111 1000 1011 0011 0010 1111 01(2) × 20 =

1.0100 1010 1110 0100 0100 0001 0111 1110 0010 1100 1100 1011 1101(2) × 2-42


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): -42

Mantissa (not normalized):
1.0100 1010 1110 0100 0100 0001 0111 1110 0010 1100 1100 1011 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-42 + 2(11-1) - 1 =

(-42 + 1 023)(10) =

981(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;
  • 981 ÷ 2 = 490 + 1;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 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) =

981(10) =

011 1101 0101(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. 0100 1010 1110 0100 0100 0001 0111 1110 0010 1100 1100 1011 1101 =

0100 1010 1110 0100 0100 0001 0111 1110 0010 1100 1100 1011 1101


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 0101

Mantissa (52 bits) =
0100 1010 1110 0100 0100 0001 0111 1110 0010 1100 1100 1011 1101


Decimal number -0.000 000 000 000 293 890 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0101 - 0100 1010 1110 0100 0100 0001 0111 1110 0010 1100 1100 1011 1101

» Convert decimal -0.000 000 000 000 293 892 2 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 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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