-0.000 000 000 000 013 46 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 013 46(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
-0.000 000 000 000 013 46(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:

|-0.000 000 000 000 013 46| = 0.000 000 000 000 013 46


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: 0.000 000 000 000 013 46.

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) 0.000 000 000 000 013 46 × 2 = 0 + 0.000 000 000 000 026 92;
  • 2) 0.000 000 000 000 026 92 × 2 = 0 + 0.000 000 000 000 053 84;
  • 3) 0.000 000 000 000 053 84 × 2 = 0 + 0.000 000 000 000 107 68;
  • 4) 0.000 000 000 000 107 68 × 2 = 0 + 0.000 000 000 000 215 36;
  • 5) 0.000 000 000 000 215 36 × 2 = 0 + 0.000 000 000 000 430 72;
  • 6) 0.000 000 000 000 430 72 × 2 = 0 + 0.000 000 000 000 861 44;
  • 7) 0.000 000 000 000 861 44 × 2 = 0 + 0.000 000 000 001 722 88;
  • 8) 0.000 000 000 001 722 88 × 2 = 0 + 0.000 000 000 003 445 76;
  • 9) 0.000 000 000 003 445 76 × 2 = 0 + 0.000 000 000 006 891 52;
  • 10) 0.000 000 000 006 891 52 × 2 = 0 + 0.000 000 000 013 783 04;
  • 11) 0.000 000 000 013 783 04 × 2 = 0 + 0.000 000 000 027 566 08;
  • 12) 0.000 000 000 027 566 08 × 2 = 0 + 0.000 000 000 055 132 16;
  • 13) 0.000 000 000 055 132 16 × 2 = 0 + 0.000 000 000 110 264 32;
  • 14) 0.000 000 000 110 264 32 × 2 = 0 + 0.000 000 000 220 528 64;
  • 15) 0.000 000 000 220 528 64 × 2 = 0 + 0.000 000 000 441 057 28;
  • 16) 0.000 000 000 441 057 28 × 2 = 0 + 0.000 000 000 882 114 56;
  • 17) 0.000 000 000 882 114 56 × 2 = 0 + 0.000 000 001 764 229 12;
  • 18) 0.000 000 001 764 229 12 × 2 = 0 + 0.000 000 003 528 458 24;
  • 19) 0.000 000 003 528 458 24 × 2 = 0 + 0.000 000 007 056 916 48;
  • 20) 0.000 000 007 056 916 48 × 2 = 0 + 0.000 000 014 113 832 96;
  • 21) 0.000 000 014 113 832 96 × 2 = 0 + 0.000 000 028 227 665 92;
  • 22) 0.000 000 028 227 665 92 × 2 = 0 + 0.000 000 056 455 331 84;
  • 23) 0.000 000 056 455 331 84 × 2 = 0 + 0.000 000 112 910 663 68;
  • 24) 0.000 000 112 910 663 68 × 2 = 0 + 0.000 000 225 821 327 36;
  • 25) 0.000 000 225 821 327 36 × 2 = 0 + 0.000 000 451 642 654 72;
  • 26) 0.000 000 451 642 654 72 × 2 = 0 + 0.000 000 903 285 309 44;
  • 27) 0.000 000 903 285 309 44 × 2 = 0 + 0.000 001 806 570 618 88;
  • 28) 0.000 001 806 570 618 88 × 2 = 0 + 0.000 003 613 141 237 76;
  • 29) 0.000 003 613 141 237 76 × 2 = 0 + 0.000 007 226 282 475 52;
  • 30) 0.000 007 226 282 475 52 × 2 = 0 + 0.000 014 452 564 951 04;
  • 31) 0.000 014 452 564 951 04 × 2 = 0 + 0.000 028 905 129 902 08;
  • 32) 0.000 028 905 129 902 08 × 2 = 0 + 0.000 057 810 259 804 16;
  • 33) 0.000 057 810 259 804 16 × 2 = 0 + 0.000 115 620 519 608 32;
  • 34) 0.000 115 620 519 608 32 × 2 = 0 + 0.000 231 241 039 216 64;
  • 35) 0.000 231 241 039 216 64 × 2 = 0 + 0.000 462 482 078 433 28;
  • 36) 0.000 462 482 078 433 28 × 2 = 0 + 0.000 924 964 156 866 56;
  • 37) 0.000 924 964 156 866 56 × 2 = 0 + 0.001 849 928 313 733 12;
  • 38) 0.001 849 928 313 733 12 × 2 = 0 + 0.003 699 856 627 466 24;
  • 39) 0.003 699 856 627 466 24 × 2 = 0 + 0.007 399 713 254 932 48;
  • 40) 0.007 399 713 254 932 48 × 2 = 0 + 0.014 799 426 509 864 96;
  • 41) 0.014 799 426 509 864 96 × 2 = 0 + 0.029 598 853 019 729 92;
  • 42) 0.029 598 853 019 729 92 × 2 = 0 + 0.059 197 706 039 459 84;
  • 43) 0.059 197 706 039 459 84 × 2 = 0 + 0.118 395 412 078 919 68;
  • 44) 0.118 395 412 078 919 68 × 2 = 0 + 0.236 790 824 157 839 36;
  • 45) 0.236 790 824 157 839 36 × 2 = 0 + 0.473 581 648 315 678 72;
  • 46) 0.473 581 648 315 678 72 × 2 = 0 + 0.947 163 296 631 357 44;
  • 47) 0.947 163 296 631 357 44 × 2 = 1 + 0.894 326 593 262 714 88;
  • 48) 0.894 326 593 262 714 88 × 2 = 1 + 0.788 653 186 525 429 76;
  • 49) 0.788 653 186 525 429 76 × 2 = 1 + 0.577 306 373 050 859 52;
  • 50) 0.577 306 373 050 859 52 × 2 = 1 + 0.154 612 746 101 719 04;
  • 51) 0.154 612 746 101 719 04 × 2 = 0 + 0.309 225 492 203 438 08;
  • 52) 0.309 225 492 203 438 08 × 2 = 0 + 0.618 450 984 406 876 16;
  • 53) 0.618 450 984 406 876 16 × 2 = 1 + 0.236 901 968 813 752 32;
  • 54) 0.236 901 968 813 752 32 × 2 = 0 + 0.473 803 937 627 504 64;
  • 55) 0.473 803 937 627 504 64 × 2 = 0 + 0.947 607 875 255 009 28;
  • 56) 0.947 607 875 255 009 28 × 2 = 1 + 0.895 215 750 510 018 56;
  • 57) 0.895 215 750 510 018 56 × 2 = 1 + 0.790 431 501 020 037 12;
  • 58) 0.790 431 501 020 037 12 × 2 = 1 + 0.580 863 002 040 074 24;
  • 59) 0.580 863 002 040 074 24 × 2 = 1 + 0.161 726 004 080 148 48;
  • 60) 0.161 726 004 080 148 48 × 2 = 0 + 0.323 452 008 160 296 96;
  • 61) 0.323 452 008 160 296 96 × 2 = 0 + 0.646 904 016 320 593 92;
  • 62) 0.646 904 016 320 593 92 × 2 = 1 + 0.293 808 032 641 187 84;
  • 63) 0.293 808 032 641 187 84 × 2 = 0 + 0.587 616 065 282 375 68;
  • 64) 0.587 616 065 282 375 68 × 2 = 1 + 0.175 232 130 564 751 36;
  • 65) 0.175 232 130 564 751 36 × 2 = 0 + 0.350 464 261 129 502 72;
  • 66) 0.350 464 261 129 502 72 × 2 = 0 + 0.700 928 522 259 005 44;
  • 67) 0.700 928 522 259 005 44 × 2 = 1 + 0.401 857 044 518 010 88;
  • 68) 0.401 857 044 518 010 88 × 2 = 0 + 0.803 714 089 036 021 76;
  • 69) 0.803 714 089 036 021 76 × 2 = 1 + 0.607 428 178 072 043 52;
  • 70) 0.607 428 178 072 043 52 × 2 = 1 + 0.214 856 356 144 087 04;
  • 71) 0.214 856 356 144 087 04 × 2 = 0 + 0.429 712 712 288 174 08;
  • 72) 0.429 712 712 288 174 08 × 2 = 0 + 0.859 425 424 576 348 16;
  • 73) 0.859 425 424 576 348 16 × 2 = 1 + 0.718 850 849 152 696 32;
  • 74) 0.718 850 849 152 696 32 × 2 = 1 + 0.437 701 698 305 392 64;
  • 75) 0.437 701 698 305 392 64 × 2 = 0 + 0.875 403 396 610 785 28;
  • 76) 0.875 403 396 610 785 28 × 2 = 1 + 0.750 806 793 221 570 56;
  • 77) 0.750 806 793 221 570 56 × 2 = 1 + 0.501 613 586 443 141 12;
  • 78) 0.501 613 586 443 141 12 × 2 = 1 + 0.003 227 172 886 282 24;
  • 79) 0.003 227 172 886 282 24 × 2 = 0 + 0.006 454 345 772 564 48;
  • 80) 0.006 454 345 772 564 48 × 2 = 0 + 0.012 908 691 545 128 96;
  • 81) 0.012 908 691 545 128 96 × 2 = 0 + 0.025 817 383 090 257 92;
  • 82) 0.025 817 383 090 257 92 × 2 = 0 + 0.051 634 766 180 515 84;
  • 83) 0.051 634 766 180 515 84 × 2 = 0 + 0.103 269 532 361 031 68;
  • 84) 0.103 269 532 361 031 68 × 2 = 0 + 0.206 539 064 722 063 36;
  • 85) 0.206 539 064 722 063 36 × 2 = 0 + 0.413 078 129 444 126 72;
  • 86) 0.413 078 129 444 126 72 × 2 = 0 + 0.826 156 258 888 253 44;
  • 87) 0.826 156 258 888 253 44 × 2 = 1 + 0.652 312 517 776 506 88;
  • 88) 0.652 312 517 776 506 88 × 2 = 1 + 0.304 625 035 553 013 76;
  • 89) 0.304 625 035 553 013 76 × 2 = 0 + 0.609 250 071 106 027 52;
  • 90) 0.609 250 071 106 027 52 × 2 = 1 + 0.218 500 142 212 055 04;
  • 91) 0.218 500 142 212 055 04 × 2 = 0 + 0.437 000 284 424 110 08;
  • 92) 0.437 000 284 424 110 08 × 2 = 0 + 0.874 000 568 848 220 16;
  • 93) 0.874 000 568 848 220 16 × 2 = 1 + 0.748 001 137 696 440 32;
  • 94) 0.748 001 137 696 440 32 × 2 = 1 + 0.496 002 275 392 880 64;
  • 95) 0.496 002 275 392 880 64 × 2 = 0 + 0.992 004 550 785 761 28;
  • 96) 0.992 004 550 785 761 28 × 2 = 1 + 0.984 009 101 571 522 56;
  • 97) 0.984 009 101 571 522 56 × 2 = 1 + 0.968 018 203 143 045 12;
  • 98) 0.968 018 203 143 045 12 × 2 = 1 + 0.936 036 406 286 090 24;
  • 99) 0.936 036 406 286 090 24 × 2 = 1 + 0.872 072 812 572 180 48;

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:


0.000 000 000 000 013 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1100 1001 1110 0101 0010 1100 1101 1100 0000 0011 0100 1101 111(2)

6. Positive number before normalization:

0.000 000 000 000 013 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1100 1001 1110 0101 0010 1100 1101 1100 0000 0011 0100 1101 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 47 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 013 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1100 1001 1110 0101 0010 1100 1101 1100 0000 0011 0100 1101 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1100 1001 1110 0101 0010 1100 1101 1100 0000 0011 0100 1101 111(2) × 20 =

1.1110 0100 1111 0010 1001 0110 0110 1110 0000 0001 1010 0110 1111(2) × 2-47


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): -47

Mantissa (not normalized):
1.1110 0100 1111 0010 1001 0110 0110 1110 0000 0001 1010 0110 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-47 + 2(11-1) - 1 =

(-47 + 1 023)(10) =

976(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;
  • 976 ÷ 2 = 488 + 0;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 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) =

976(10) =

011 1101 0000(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. 1110 0100 1111 0010 1001 0110 0110 1110 0000 0001 1010 0110 1111 =

1110 0100 1111 0010 1001 0110 0110 1110 0000 0001 1010 0110 1111


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 0000

Mantissa (52 bits) =
1110 0100 1111 0010 1001 0110 0110 1110 0000 0001 1010 0110 1111


Decimal number -0.000 000 000 000 013 46 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0000 - 1110 0100 1111 0010 1001 0110 0110 1110 0000 0001 1010 0110 1111

» Convert decimal -0.000 000 000 000 014 31 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 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - 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