-0.000 000 000 000 176 555 69 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 555 69(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
-0.000 000 000 000 176 555 69(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:

|-0.000 000 000 000 176 555 69| = 0.000 000 000 000 176 555 69


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: 0.000 000 000 000 176 555 69.

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) 0.000 000 000 000 176 555 69 × 2 = 0 + 0.000 000 000 000 353 111 38;
  • 2) 0.000 000 000 000 353 111 38 × 2 = 0 + 0.000 000 000 000 706 222 76;
  • 3) 0.000 000 000 000 706 222 76 × 2 = 0 + 0.000 000 000 001 412 445 52;
  • 4) 0.000 000 000 001 412 445 52 × 2 = 0 + 0.000 000 000 002 824 891 04;
  • 5) 0.000 000 000 002 824 891 04 × 2 = 0 + 0.000 000 000 005 649 782 08;
  • 6) 0.000 000 000 005 649 782 08 × 2 = 0 + 0.000 000 000 011 299 564 16;
  • 7) 0.000 000 000 011 299 564 16 × 2 = 0 + 0.000 000 000 022 599 128 32;
  • 8) 0.000 000 000 022 599 128 32 × 2 = 0 + 0.000 000 000 045 198 256 64;
  • 9) 0.000 000 000 045 198 256 64 × 2 = 0 + 0.000 000 000 090 396 513 28;
  • 10) 0.000 000 000 090 396 513 28 × 2 = 0 + 0.000 000 000 180 793 026 56;
  • 11) 0.000 000 000 180 793 026 56 × 2 = 0 + 0.000 000 000 361 586 053 12;
  • 12) 0.000 000 000 361 586 053 12 × 2 = 0 + 0.000 000 000 723 172 106 24;
  • 13) 0.000 000 000 723 172 106 24 × 2 = 0 + 0.000 000 001 446 344 212 48;
  • 14) 0.000 000 001 446 344 212 48 × 2 = 0 + 0.000 000 002 892 688 424 96;
  • 15) 0.000 000 002 892 688 424 96 × 2 = 0 + 0.000 000 005 785 376 849 92;
  • 16) 0.000 000 005 785 376 849 92 × 2 = 0 + 0.000 000 011 570 753 699 84;
  • 17) 0.000 000 011 570 753 699 84 × 2 = 0 + 0.000 000 023 141 507 399 68;
  • 18) 0.000 000 023 141 507 399 68 × 2 = 0 + 0.000 000 046 283 014 799 36;
  • 19) 0.000 000 046 283 014 799 36 × 2 = 0 + 0.000 000 092 566 029 598 72;
  • 20) 0.000 000 092 566 029 598 72 × 2 = 0 + 0.000 000 185 132 059 197 44;
  • 21) 0.000 000 185 132 059 197 44 × 2 = 0 + 0.000 000 370 264 118 394 88;
  • 22) 0.000 000 370 264 118 394 88 × 2 = 0 + 0.000 000 740 528 236 789 76;
  • 23) 0.000 000 740 528 236 789 76 × 2 = 0 + 0.000 001 481 056 473 579 52;
  • 24) 0.000 001 481 056 473 579 52 × 2 = 0 + 0.000 002 962 112 947 159 04;
  • 25) 0.000 002 962 112 947 159 04 × 2 = 0 + 0.000 005 924 225 894 318 08;
  • 26) 0.000 005 924 225 894 318 08 × 2 = 0 + 0.000 011 848 451 788 636 16;
  • 27) 0.000 011 848 451 788 636 16 × 2 = 0 + 0.000 023 696 903 577 272 32;
  • 28) 0.000 023 696 903 577 272 32 × 2 = 0 + 0.000 047 393 807 154 544 64;
  • 29) 0.000 047 393 807 154 544 64 × 2 = 0 + 0.000 094 787 614 309 089 28;
  • 30) 0.000 094 787 614 309 089 28 × 2 = 0 + 0.000 189 575 228 618 178 56;
  • 31) 0.000 189 575 228 618 178 56 × 2 = 0 + 0.000 379 150 457 236 357 12;
  • 32) 0.000 379 150 457 236 357 12 × 2 = 0 + 0.000 758 300 914 472 714 24;
  • 33) 0.000 758 300 914 472 714 24 × 2 = 0 + 0.001 516 601 828 945 428 48;
  • 34) 0.001 516 601 828 945 428 48 × 2 = 0 + 0.003 033 203 657 890 856 96;
  • 35) 0.003 033 203 657 890 856 96 × 2 = 0 + 0.006 066 407 315 781 713 92;
  • 36) 0.006 066 407 315 781 713 92 × 2 = 0 + 0.012 132 814 631 563 427 84;
  • 37) 0.012 132 814 631 563 427 84 × 2 = 0 + 0.024 265 629 263 126 855 68;
  • 38) 0.024 265 629 263 126 855 68 × 2 = 0 + 0.048 531 258 526 253 711 36;
  • 39) 0.048 531 258 526 253 711 36 × 2 = 0 + 0.097 062 517 052 507 422 72;
  • 40) 0.097 062 517 052 507 422 72 × 2 = 0 + 0.194 125 034 105 014 845 44;
  • 41) 0.194 125 034 105 014 845 44 × 2 = 0 + 0.388 250 068 210 029 690 88;
  • 42) 0.388 250 068 210 029 690 88 × 2 = 0 + 0.776 500 136 420 059 381 76;
  • 43) 0.776 500 136 420 059 381 76 × 2 = 1 + 0.553 000 272 840 118 763 52;
  • 44) 0.553 000 272 840 118 763 52 × 2 = 1 + 0.106 000 545 680 237 527 04;
  • 45) 0.106 000 545 680 237 527 04 × 2 = 0 + 0.212 001 091 360 475 054 08;
  • 46) 0.212 001 091 360 475 054 08 × 2 = 0 + 0.424 002 182 720 950 108 16;
  • 47) 0.424 002 182 720 950 108 16 × 2 = 0 + 0.848 004 365 441 900 216 32;
  • 48) 0.848 004 365 441 900 216 32 × 2 = 1 + 0.696 008 730 883 800 432 64;
  • 49) 0.696 008 730 883 800 432 64 × 2 = 1 + 0.392 017 461 767 600 865 28;
  • 50) 0.392 017 461 767 600 865 28 × 2 = 0 + 0.784 034 923 535 201 730 56;
  • 51) 0.784 034 923 535 201 730 56 × 2 = 1 + 0.568 069 847 070 403 461 12;
  • 52) 0.568 069 847 070 403 461 12 × 2 = 1 + 0.136 139 694 140 806 922 24;
  • 53) 0.136 139 694 140 806 922 24 × 2 = 0 + 0.272 279 388 281 613 844 48;
  • 54) 0.272 279 388 281 613 844 48 × 2 = 0 + 0.544 558 776 563 227 688 96;
  • 55) 0.544 558 776 563 227 688 96 × 2 = 1 + 0.089 117 553 126 455 377 92;
  • 56) 0.089 117 553 126 455 377 92 × 2 = 0 + 0.178 235 106 252 910 755 84;
  • 57) 0.178 235 106 252 910 755 84 × 2 = 0 + 0.356 470 212 505 821 511 68;
  • 58) 0.356 470 212 505 821 511 68 × 2 = 0 + 0.712 940 425 011 643 023 36;
  • 59) 0.712 940 425 011 643 023 36 × 2 = 1 + 0.425 880 850 023 286 046 72;
  • 60) 0.425 880 850 023 286 046 72 × 2 = 0 + 0.851 761 700 046 572 093 44;
  • 61) 0.851 761 700 046 572 093 44 × 2 = 1 + 0.703 523 400 093 144 186 88;
  • 62) 0.703 523 400 093 144 186 88 × 2 = 1 + 0.407 046 800 186 288 373 76;
  • 63) 0.407 046 800 186 288 373 76 × 2 = 0 + 0.814 093 600 372 576 747 52;
  • 64) 0.814 093 600 372 576 747 52 × 2 = 1 + 0.628 187 200 745 153 495 04;
  • 65) 0.628 187 200 745 153 495 04 × 2 = 1 + 0.256 374 401 490 306 990 08;
  • 66) 0.256 374 401 490 306 990 08 × 2 = 0 + 0.512 748 802 980 613 980 16;
  • 67) 0.512 748 802 980 613 980 16 × 2 = 1 + 0.025 497 605 961 227 960 32;
  • 68) 0.025 497 605 961 227 960 32 × 2 = 0 + 0.050 995 211 922 455 920 64;
  • 69) 0.050 995 211 922 455 920 64 × 2 = 0 + 0.101 990 423 844 911 841 28;
  • 70) 0.101 990 423 844 911 841 28 × 2 = 0 + 0.203 980 847 689 823 682 56;
  • 71) 0.203 980 847 689 823 682 56 × 2 = 0 + 0.407 961 695 379 647 365 12;
  • 72) 0.407 961 695 379 647 365 12 × 2 = 0 + 0.815 923 390 759 294 730 24;
  • 73) 0.815 923 390 759 294 730 24 × 2 = 1 + 0.631 846 781 518 589 460 48;
  • 74) 0.631 846 781 518 589 460 48 × 2 = 1 + 0.263 693 563 037 178 920 96;
  • 75) 0.263 693 563 037 178 920 96 × 2 = 0 + 0.527 387 126 074 357 841 92;
  • 76) 0.527 387 126 074 357 841 92 × 2 = 1 + 0.054 774 252 148 715 683 84;
  • 77) 0.054 774 252 148 715 683 84 × 2 = 0 + 0.109 548 504 297 431 367 68;
  • 78) 0.109 548 504 297 431 367 68 × 2 = 0 + 0.219 097 008 594 862 735 36;
  • 79) 0.219 097 008 594 862 735 36 × 2 = 0 + 0.438 194 017 189 725 470 72;
  • 80) 0.438 194 017 189 725 470 72 × 2 = 0 + 0.876 388 034 379 450 941 44;
  • 81) 0.876 388 034 379 450 941 44 × 2 = 1 + 0.752 776 068 758 901 882 88;
  • 82) 0.752 776 068 758 901 882 88 × 2 = 1 + 0.505 552 137 517 803 765 76;
  • 83) 0.505 552 137 517 803 765 76 × 2 = 1 + 0.011 104 275 035 607 531 52;
  • 84) 0.011 104 275 035 607 531 52 × 2 = 0 + 0.022 208 550 071 215 063 04;
  • 85) 0.022 208 550 071 215 063 04 × 2 = 0 + 0.044 417 100 142 430 126 08;
  • 86) 0.044 417 100 142 430 126 08 × 2 = 0 + 0.088 834 200 284 860 252 16;
  • 87) 0.088 834 200 284 860 252 16 × 2 = 0 + 0.177 668 400 569 720 504 32;
  • 88) 0.177 668 400 569 720 504 32 × 2 = 0 + 0.355 336 801 139 441 008 64;
  • 89) 0.355 336 801 139 441 008 64 × 2 = 0 + 0.710 673 602 278 882 017 28;
  • 90) 0.710 673 602 278 882 017 28 × 2 = 1 + 0.421 347 204 557 764 034 56;
  • 91) 0.421 347 204 557 764 034 56 × 2 = 0 + 0.842 694 409 115 528 069 12;
  • 92) 0.842 694 409 115 528 069 12 × 2 = 1 + 0.685 388 818 231 056 138 24;
  • 93) 0.685 388 818 231 056 138 24 × 2 = 1 + 0.370 777 636 462 112 276 48;
  • 94) 0.370 777 636 462 112 276 48 × 2 = 0 + 0.741 555 272 924 224 552 96;
  • 95) 0.741 555 272 924 224 552 96 × 2 = 1 + 0.483 110 545 848 449 105 92;

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:


0.000 000 000 000 176 555 69(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1101 1010 0000 1101 0000 1110 0000 0101 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 555 69(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1101 1010 0000 1101 0000 1110 0000 0101 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 555 69(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1101 1010 0000 1101 0000 1110 0000 0101 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1101 1010 0000 1101 0000 1110 0000 0101 101(2) × 20 =

1.1000 1101 1001 0001 0110 1101 0000 0110 1000 0111 0000 0010 1101(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 1001 0001 0110 1101 0000 0110 1000 0111 0000 0010 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 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) =

980(10) =

011 1101 0100(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. 1000 1101 1001 0001 0110 1101 0000 0110 1000 0111 0000 0010 1101 =

1000 1101 1001 0001 0110 1101 0000 0110 1000 0111 0000 0010 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 0100

Mantissa (52 bits) =
1000 1101 1001 0001 0110 1101 0000 0110 1000 0111 0000 0010 1101


Decimal number -0.000 000 000 000 176 555 69 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 0110 1101 0000 0110 1000 0111 0000 0010 1101

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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