-0.000 000 000 000 006 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 006 1(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
-0.000 000 000 000 006 1(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:

|-0.000 000 000 000 006 1| = 0.000 000 000 000 006 1


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: 0.000 000 000 000 006 1.

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) 0.000 000 000 000 006 1 × 2 = 0 + 0.000 000 000 000 012 2;
  • 2) 0.000 000 000 000 012 2 × 2 = 0 + 0.000 000 000 000 024 4;
  • 3) 0.000 000 000 000 024 4 × 2 = 0 + 0.000 000 000 000 048 8;
  • 4) 0.000 000 000 000 048 8 × 2 = 0 + 0.000 000 000 000 097 6;
  • 5) 0.000 000 000 000 097 6 × 2 = 0 + 0.000 000 000 000 195 2;
  • 6) 0.000 000 000 000 195 2 × 2 = 0 + 0.000 000 000 000 390 4;
  • 7) 0.000 000 000 000 390 4 × 2 = 0 + 0.000 000 000 000 780 8;
  • 8) 0.000 000 000 000 780 8 × 2 = 0 + 0.000 000 000 001 561 6;
  • 9) 0.000 000 000 001 561 6 × 2 = 0 + 0.000 000 000 003 123 2;
  • 10) 0.000 000 000 003 123 2 × 2 = 0 + 0.000 000 000 006 246 4;
  • 11) 0.000 000 000 006 246 4 × 2 = 0 + 0.000 000 000 012 492 8;
  • 12) 0.000 000 000 012 492 8 × 2 = 0 + 0.000 000 000 024 985 6;
  • 13) 0.000 000 000 024 985 6 × 2 = 0 + 0.000 000 000 049 971 2;
  • 14) 0.000 000 000 049 971 2 × 2 = 0 + 0.000 000 000 099 942 4;
  • 15) 0.000 000 000 099 942 4 × 2 = 0 + 0.000 000 000 199 884 8;
  • 16) 0.000 000 000 199 884 8 × 2 = 0 + 0.000 000 000 399 769 6;
  • 17) 0.000 000 000 399 769 6 × 2 = 0 + 0.000 000 000 799 539 2;
  • 18) 0.000 000 000 799 539 2 × 2 = 0 + 0.000 000 001 599 078 4;
  • 19) 0.000 000 001 599 078 4 × 2 = 0 + 0.000 000 003 198 156 8;
  • 20) 0.000 000 003 198 156 8 × 2 = 0 + 0.000 000 006 396 313 6;
  • 21) 0.000 000 006 396 313 6 × 2 = 0 + 0.000 000 012 792 627 2;
  • 22) 0.000 000 012 792 627 2 × 2 = 0 + 0.000 000 025 585 254 4;
  • 23) 0.000 000 025 585 254 4 × 2 = 0 + 0.000 000 051 170 508 8;
  • 24) 0.000 000 051 170 508 8 × 2 = 0 + 0.000 000 102 341 017 6;
  • 25) 0.000 000 102 341 017 6 × 2 = 0 + 0.000 000 204 682 035 2;
  • 26) 0.000 000 204 682 035 2 × 2 = 0 + 0.000 000 409 364 070 4;
  • 27) 0.000 000 409 364 070 4 × 2 = 0 + 0.000 000 818 728 140 8;
  • 28) 0.000 000 818 728 140 8 × 2 = 0 + 0.000 001 637 456 281 6;
  • 29) 0.000 001 637 456 281 6 × 2 = 0 + 0.000 003 274 912 563 2;
  • 30) 0.000 003 274 912 563 2 × 2 = 0 + 0.000 006 549 825 126 4;
  • 31) 0.000 006 549 825 126 4 × 2 = 0 + 0.000 013 099 650 252 8;
  • 32) 0.000 013 099 650 252 8 × 2 = 0 + 0.000 026 199 300 505 6;
  • 33) 0.000 026 199 300 505 6 × 2 = 0 + 0.000 052 398 601 011 2;
  • 34) 0.000 052 398 601 011 2 × 2 = 0 + 0.000 104 797 202 022 4;
  • 35) 0.000 104 797 202 022 4 × 2 = 0 + 0.000 209 594 404 044 8;
  • 36) 0.000 209 594 404 044 8 × 2 = 0 + 0.000 419 188 808 089 6;
  • 37) 0.000 419 188 808 089 6 × 2 = 0 + 0.000 838 377 616 179 2;
  • 38) 0.000 838 377 616 179 2 × 2 = 0 + 0.001 676 755 232 358 4;
  • 39) 0.001 676 755 232 358 4 × 2 = 0 + 0.003 353 510 464 716 8;
  • 40) 0.003 353 510 464 716 8 × 2 = 0 + 0.006 707 020 929 433 6;
  • 41) 0.006 707 020 929 433 6 × 2 = 0 + 0.013 414 041 858 867 2;
  • 42) 0.013 414 041 858 867 2 × 2 = 0 + 0.026 828 083 717 734 4;
  • 43) 0.026 828 083 717 734 4 × 2 = 0 + 0.053 656 167 435 468 8;
  • 44) 0.053 656 167 435 468 8 × 2 = 0 + 0.107 312 334 870 937 6;
  • 45) 0.107 312 334 870 937 6 × 2 = 0 + 0.214 624 669 741 875 2;
  • 46) 0.214 624 669 741 875 2 × 2 = 0 + 0.429 249 339 483 750 4;
  • 47) 0.429 249 339 483 750 4 × 2 = 0 + 0.858 498 678 967 500 8;
  • 48) 0.858 498 678 967 500 8 × 2 = 1 + 0.716 997 357 935 001 6;
  • 49) 0.716 997 357 935 001 6 × 2 = 1 + 0.433 994 715 870 003 2;
  • 50) 0.433 994 715 870 003 2 × 2 = 0 + 0.867 989 431 740 006 4;
  • 51) 0.867 989 431 740 006 4 × 2 = 1 + 0.735 978 863 480 012 8;
  • 52) 0.735 978 863 480 012 8 × 2 = 1 + 0.471 957 726 960 025 6;
  • 53) 0.471 957 726 960 025 6 × 2 = 0 + 0.943 915 453 920 051 2;
  • 54) 0.943 915 453 920 051 2 × 2 = 1 + 0.887 830 907 840 102 4;
  • 55) 0.887 830 907 840 102 4 × 2 = 1 + 0.775 661 815 680 204 8;
  • 56) 0.775 661 815 680 204 8 × 2 = 1 + 0.551 323 631 360 409 6;
  • 57) 0.551 323 631 360 409 6 × 2 = 1 + 0.102 647 262 720 819 2;
  • 58) 0.102 647 262 720 819 2 × 2 = 0 + 0.205 294 525 441 638 4;
  • 59) 0.205 294 525 441 638 4 × 2 = 0 + 0.410 589 050 883 276 8;
  • 60) 0.410 589 050 883 276 8 × 2 = 0 + 0.821 178 101 766 553 6;
  • 61) 0.821 178 101 766 553 6 × 2 = 1 + 0.642 356 203 533 107 2;
  • 62) 0.642 356 203 533 107 2 × 2 = 1 + 0.284 712 407 066 214 4;
  • 63) 0.284 712 407 066 214 4 × 2 = 0 + 0.569 424 814 132 428 8;
  • 64) 0.569 424 814 132 428 8 × 2 = 1 + 0.138 849 628 264 857 6;
  • 65) 0.138 849 628 264 857 6 × 2 = 0 + 0.277 699 256 529 715 2;
  • 66) 0.277 699 256 529 715 2 × 2 = 0 + 0.555 398 513 059 430 4;
  • 67) 0.555 398 513 059 430 4 × 2 = 1 + 0.110 797 026 118 860 8;
  • 68) 0.110 797 026 118 860 8 × 2 = 0 + 0.221 594 052 237 721 6;
  • 69) 0.221 594 052 237 721 6 × 2 = 0 + 0.443 188 104 475 443 2;
  • 70) 0.443 188 104 475 443 2 × 2 = 0 + 0.886 376 208 950 886 4;
  • 71) 0.886 376 208 950 886 4 × 2 = 1 + 0.772 752 417 901 772 8;
  • 72) 0.772 752 417 901 772 8 × 2 = 1 + 0.545 504 835 803 545 6;
  • 73) 0.545 504 835 803 545 6 × 2 = 1 + 0.091 009 671 607 091 2;
  • 74) 0.091 009 671 607 091 2 × 2 = 0 + 0.182 019 343 214 182 4;
  • 75) 0.182 019 343 214 182 4 × 2 = 0 + 0.364 038 686 428 364 8;
  • 76) 0.364 038 686 428 364 8 × 2 = 0 + 0.728 077 372 856 729 6;
  • 77) 0.728 077 372 856 729 6 × 2 = 1 + 0.456 154 745 713 459 2;
  • 78) 0.456 154 745 713 459 2 × 2 = 0 + 0.912 309 491 426 918 4;
  • 79) 0.912 309 491 426 918 4 × 2 = 1 + 0.824 618 982 853 836 8;
  • 80) 0.824 618 982 853 836 8 × 2 = 1 + 0.649 237 965 707 673 6;
  • 81) 0.649 237 965 707 673 6 × 2 = 1 + 0.298 475 931 415 347 2;
  • 82) 0.298 475 931 415 347 2 × 2 = 0 + 0.596 951 862 830 694 4;
  • 83) 0.596 951 862 830 694 4 × 2 = 1 + 0.193 903 725 661 388 8;
  • 84) 0.193 903 725 661 388 8 × 2 = 0 + 0.387 807 451 322 777 6;
  • 85) 0.387 807 451 322 777 6 × 2 = 0 + 0.775 614 902 645 555 2;
  • 86) 0.775 614 902 645 555 2 × 2 = 1 + 0.551 229 805 291 110 4;
  • 87) 0.551 229 805 291 110 4 × 2 = 1 + 0.102 459 610 582 220 8;
  • 88) 0.102 459 610 582 220 8 × 2 = 0 + 0.204 919 221 164 441 6;
  • 89) 0.204 919 221 164 441 6 × 2 = 0 + 0.409 838 442 328 883 2;
  • 90) 0.409 838 442 328 883 2 × 2 = 0 + 0.819 676 884 657 766 4;
  • 91) 0.819 676 884 657 766 4 × 2 = 1 + 0.639 353 769 315 532 8;
  • 92) 0.639 353 769 315 532 8 × 2 = 1 + 0.278 707 538 631 065 6;
  • 93) 0.278 707 538 631 065 6 × 2 = 0 + 0.557 415 077 262 131 2;
  • 94) 0.557 415 077 262 131 2 × 2 = 1 + 0.114 830 154 524 262 4;
  • 95) 0.114 830 154 524 262 4 × 2 = 0 + 0.229 660 309 048 524 8;
  • 96) 0.229 660 309 048 524 8 × 2 = 0 + 0.459 320 618 097 049 6;
  • 97) 0.459 320 618 097 049 6 × 2 = 0 + 0.918 641 236 194 099 2;
  • 98) 0.918 641 236 194 099 2 × 2 = 1 + 0.837 282 472 388 198 4;
  • 99) 0.837 282 472 388 198 4 × 2 = 1 + 0.674 564 944 776 396 8;
  • 100) 0.674 564 944 776 396 8 × 2 = 1 + 0.349 129 889 552 793 6;

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:


0.000 000 000 000 006 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1011 0111 1000 1101 0010 0011 1000 1011 1010 0110 0011 0100 0111(2)

6. Positive number before normalization:

0.000 000 000 000 006 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1011 0111 1000 1101 0010 0011 1000 1011 1010 0110 0011 0100 0111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 48 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 006 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1011 0111 1000 1101 0010 0011 1000 1011 1010 0110 0011 0100 0111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1011 0111 1000 1101 0010 0011 1000 1011 1010 0110 0011 0100 0111(2) × 20 =

1.1011 0111 1000 1101 0010 0011 1000 1011 1010 0110 0011 0100 0111(2) × 2-48


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): -48

Mantissa (not normalized):
1.1011 0111 1000 1101 0010 0011 1000 1011 1010 0110 0011 0100 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-48 + 2(11-1) - 1 =

(-48 + 1 023)(10) =

975(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;
  • 975 ÷ 2 = 487 + 1;
  • 487 ÷ 2 = 243 + 1;
  • 243 ÷ 2 = 121 + 1;
  • 121 ÷ 2 = 60 + 1;
  • 60 ÷ 2 = 30 + 0;
  • 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) =

975(10) =

011 1100 1111(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. 1011 0111 1000 1101 0010 0011 1000 1011 1010 0110 0011 0100 0111 =

1011 0111 1000 1101 0010 0011 1000 1011 1010 0110 0011 0100 0111


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 1100 1111

Mantissa (52 bits) =
1011 0111 1000 1101 0010 0011 1000 1011 1010 0110 0011 0100 0111


Decimal number -0.000 000 000 000 006 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1100 1111 - 1011 0111 1000 1101 0010 0011 1000 1011 1010 0110 0011 0100 0111

» Convert decimal 0.000 000 000 000 003 8 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