-0.000 000 000 000 176 555 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 555 9(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 9(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 9| = 0.000 000 000 000 176 555 9


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 9.

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 9 × 2 = 0 + 0.000 000 000 000 353 111 8;
  • 2) 0.000 000 000 000 353 111 8 × 2 = 0 + 0.000 000 000 000 706 223 6;
  • 3) 0.000 000 000 000 706 223 6 × 2 = 0 + 0.000 000 000 001 412 447 2;
  • 4) 0.000 000 000 001 412 447 2 × 2 = 0 + 0.000 000 000 002 824 894 4;
  • 5) 0.000 000 000 002 824 894 4 × 2 = 0 + 0.000 000 000 005 649 788 8;
  • 6) 0.000 000 000 005 649 788 8 × 2 = 0 + 0.000 000 000 011 299 577 6;
  • 7) 0.000 000 000 011 299 577 6 × 2 = 0 + 0.000 000 000 022 599 155 2;
  • 8) 0.000 000 000 022 599 155 2 × 2 = 0 + 0.000 000 000 045 198 310 4;
  • 9) 0.000 000 000 045 198 310 4 × 2 = 0 + 0.000 000 000 090 396 620 8;
  • 10) 0.000 000 000 090 396 620 8 × 2 = 0 + 0.000 000 000 180 793 241 6;
  • 11) 0.000 000 000 180 793 241 6 × 2 = 0 + 0.000 000 000 361 586 483 2;
  • 12) 0.000 000 000 361 586 483 2 × 2 = 0 + 0.000 000 000 723 172 966 4;
  • 13) 0.000 000 000 723 172 966 4 × 2 = 0 + 0.000 000 001 446 345 932 8;
  • 14) 0.000 000 001 446 345 932 8 × 2 = 0 + 0.000 000 002 892 691 865 6;
  • 15) 0.000 000 002 892 691 865 6 × 2 = 0 + 0.000 000 005 785 383 731 2;
  • 16) 0.000 000 005 785 383 731 2 × 2 = 0 + 0.000 000 011 570 767 462 4;
  • 17) 0.000 000 011 570 767 462 4 × 2 = 0 + 0.000 000 023 141 534 924 8;
  • 18) 0.000 000 023 141 534 924 8 × 2 = 0 + 0.000 000 046 283 069 849 6;
  • 19) 0.000 000 046 283 069 849 6 × 2 = 0 + 0.000 000 092 566 139 699 2;
  • 20) 0.000 000 092 566 139 699 2 × 2 = 0 + 0.000 000 185 132 279 398 4;
  • 21) 0.000 000 185 132 279 398 4 × 2 = 0 + 0.000 000 370 264 558 796 8;
  • 22) 0.000 000 370 264 558 796 8 × 2 = 0 + 0.000 000 740 529 117 593 6;
  • 23) 0.000 000 740 529 117 593 6 × 2 = 0 + 0.000 001 481 058 235 187 2;
  • 24) 0.000 001 481 058 235 187 2 × 2 = 0 + 0.000 002 962 116 470 374 4;
  • 25) 0.000 002 962 116 470 374 4 × 2 = 0 + 0.000 005 924 232 940 748 8;
  • 26) 0.000 005 924 232 940 748 8 × 2 = 0 + 0.000 011 848 465 881 497 6;
  • 27) 0.000 011 848 465 881 497 6 × 2 = 0 + 0.000 023 696 931 762 995 2;
  • 28) 0.000 023 696 931 762 995 2 × 2 = 0 + 0.000 047 393 863 525 990 4;
  • 29) 0.000 047 393 863 525 990 4 × 2 = 0 + 0.000 094 787 727 051 980 8;
  • 30) 0.000 094 787 727 051 980 8 × 2 = 0 + 0.000 189 575 454 103 961 6;
  • 31) 0.000 189 575 454 103 961 6 × 2 = 0 + 0.000 379 150 908 207 923 2;
  • 32) 0.000 379 150 908 207 923 2 × 2 = 0 + 0.000 758 301 816 415 846 4;
  • 33) 0.000 758 301 816 415 846 4 × 2 = 0 + 0.001 516 603 632 831 692 8;
  • 34) 0.001 516 603 632 831 692 8 × 2 = 0 + 0.003 033 207 265 663 385 6;
  • 35) 0.003 033 207 265 663 385 6 × 2 = 0 + 0.006 066 414 531 326 771 2;
  • 36) 0.006 066 414 531 326 771 2 × 2 = 0 + 0.012 132 829 062 653 542 4;
  • 37) 0.012 132 829 062 653 542 4 × 2 = 0 + 0.024 265 658 125 307 084 8;
  • 38) 0.024 265 658 125 307 084 8 × 2 = 0 + 0.048 531 316 250 614 169 6;
  • 39) 0.048 531 316 250 614 169 6 × 2 = 0 + 0.097 062 632 501 228 339 2;
  • 40) 0.097 062 632 501 228 339 2 × 2 = 0 + 0.194 125 265 002 456 678 4;
  • 41) 0.194 125 265 002 456 678 4 × 2 = 0 + 0.388 250 530 004 913 356 8;
  • 42) 0.388 250 530 004 913 356 8 × 2 = 0 + 0.776 501 060 009 826 713 6;
  • 43) 0.776 501 060 009 826 713 6 × 2 = 1 + 0.553 002 120 019 653 427 2;
  • 44) 0.553 002 120 019 653 427 2 × 2 = 1 + 0.106 004 240 039 306 854 4;
  • 45) 0.106 004 240 039 306 854 4 × 2 = 0 + 0.212 008 480 078 613 708 8;
  • 46) 0.212 008 480 078 613 708 8 × 2 = 0 + 0.424 016 960 157 227 417 6;
  • 47) 0.424 016 960 157 227 417 6 × 2 = 0 + 0.848 033 920 314 454 835 2;
  • 48) 0.848 033 920 314 454 835 2 × 2 = 1 + 0.696 067 840 628 909 670 4;
  • 49) 0.696 067 840 628 909 670 4 × 2 = 1 + 0.392 135 681 257 819 340 8;
  • 50) 0.392 135 681 257 819 340 8 × 2 = 0 + 0.784 271 362 515 638 681 6;
  • 51) 0.784 271 362 515 638 681 6 × 2 = 1 + 0.568 542 725 031 277 363 2;
  • 52) 0.568 542 725 031 277 363 2 × 2 = 1 + 0.137 085 450 062 554 726 4;
  • 53) 0.137 085 450 062 554 726 4 × 2 = 0 + 0.274 170 900 125 109 452 8;
  • 54) 0.274 170 900 125 109 452 8 × 2 = 0 + 0.548 341 800 250 218 905 6;
  • 55) 0.548 341 800 250 218 905 6 × 2 = 1 + 0.096 683 600 500 437 811 2;
  • 56) 0.096 683 600 500 437 811 2 × 2 = 0 + 0.193 367 201 000 875 622 4;
  • 57) 0.193 367 201 000 875 622 4 × 2 = 0 + 0.386 734 402 001 751 244 8;
  • 58) 0.386 734 402 001 751 244 8 × 2 = 0 + 0.773 468 804 003 502 489 6;
  • 59) 0.773 468 804 003 502 489 6 × 2 = 1 + 0.546 937 608 007 004 979 2;
  • 60) 0.546 937 608 007 004 979 2 × 2 = 1 + 0.093 875 216 014 009 958 4;
  • 61) 0.093 875 216 014 009 958 4 × 2 = 0 + 0.187 750 432 028 019 916 8;
  • 62) 0.187 750 432 028 019 916 8 × 2 = 0 + 0.375 500 864 056 039 833 6;
  • 63) 0.375 500 864 056 039 833 6 × 2 = 0 + 0.751 001 728 112 079 667 2;
  • 64) 0.751 001 728 112 079 667 2 × 2 = 1 + 0.502 003 456 224 159 334 4;
  • 65) 0.502 003 456 224 159 334 4 × 2 = 1 + 0.004 006 912 448 318 668 8;
  • 66) 0.004 006 912 448 318 668 8 × 2 = 0 + 0.008 013 824 896 637 337 6;
  • 67) 0.008 013 824 896 637 337 6 × 2 = 0 + 0.016 027 649 793 274 675 2;
  • 68) 0.016 027 649 793 274 675 2 × 2 = 0 + 0.032 055 299 586 549 350 4;
  • 69) 0.032 055 299 586 549 350 4 × 2 = 0 + 0.064 110 599 173 098 700 8;
  • 70) 0.064 110 599 173 098 700 8 × 2 = 0 + 0.128 221 198 346 197 401 6;
  • 71) 0.128 221 198 346 197 401 6 × 2 = 0 + 0.256 442 396 692 394 803 2;
  • 72) 0.256 442 396 692 394 803 2 × 2 = 0 + 0.512 884 793 384 789 606 4;
  • 73) 0.512 884 793 384 789 606 4 × 2 = 1 + 0.025 769 586 769 579 212 8;
  • 74) 0.025 769 586 769 579 212 8 × 2 = 0 + 0.051 539 173 539 158 425 6;
  • 75) 0.051 539 173 539 158 425 6 × 2 = 0 + 0.103 078 347 078 316 851 2;
  • 76) 0.103 078 347 078 316 851 2 × 2 = 0 + 0.206 156 694 156 633 702 4;
  • 77) 0.206 156 694 156 633 702 4 × 2 = 0 + 0.412 313 388 313 267 404 8;
  • 78) 0.412 313 388 313 267 404 8 × 2 = 0 + 0.824 626 776 626 534 809 6;
  • 79) 0.824 626 776 626 534 809 6 × 2 = 1 + 0.649 253 553 253 069 619 2;
  • 80) 0.649 253 553 253 069 619 2 × 2 = 1 + 0.298 507 106 506 139 238 4;
  • 81) 0.298 507 106 506 139 238 4 × 2 = 0 + 0.597 014 213 012 278 476 8;
  • 82) 0.597 014 213 012 278 476 8 × 2 = 1 + 0.194 028 426 024 556 953 6;
  • 83) 0.194 028 426 024 556 953 6 × 2 = 0 + 0.388 056 852 049 113 907 2;
  • 84) 0.388 056 852 049 113 907 2 × 2 = 0 + 0.776 113 704 098 227 814 4;
  • 85) 0.776 113 704 098 227 814 4 × 2 = 1 + 0.552 227 408 196 455 628 8;
  • 86) 0.552 227 408 196 455 628 8 × 2 = 1 + 0.104 454 816 392 911 257 6;
  • 87) 0.104 454 816 392 911 257 6 × 2 = 0 + 0.208 909 632 785 822 515 2;
  • 88) 0.208 909 632 785 822 515 2 × 2 = 0 + 0.417 819 265 571 645 030 4;
  • 89) 0.417 819 265 571 645 030 4 × 2 = 0 + 0.835 638 531 143 290 060 8;
  • 90) 0.835 638 531 143 290 060 8 × 2 = 1 + 0.671 277 062 286 580 121 6;
  • 91) 0.671 277 062 286 580 121 6 × 2 = 1 + 0.342 554 124 573 160 243 2;
  • 92) 0.342 554 124 573 160 243 2 × 2 = 0 + 0.685 108 249 146 320 486 4;
  • 93) 0.685 108 249 146 320 486 4 × 2 = 1 + 0.370 216 498 292 640 972 8;
  • 94) 0.370 216 498 292 640 972 8 × 2 = 0 + 0.740 432 996 585 281 945 6;
  • 95) 0.740 432 996 585 281 945 6 × 2 = 1 + 0.480 865 993 170 563 891 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:


0.000 000 000 000 176 555 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0001 1000 0000 1000 0011 0100 1100 0110 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 555 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0001 1000 0000 1000 0011 0100 1100 0110 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 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0001 1000 0000 1000 0011 0100 1100 0110 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0001 1000 0000 1000 0011 0100 1100 0110 101(2) × 20 =

1.1000 1101 1001 0001 1000 1100 0000 0100 0001 1010 0110 0011 0101(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 1000 1100 0000 0100 0001 1010 0110 0011 0101


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 1000 1100 0000 0100 0001 1010 0110 0011 0101 =

1000 1101 1001 0001 1000 1100 0000 0100 0001 1010 0110 0011 0101


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 1000 1100 0000 0100 0001 1010 0110 0011 0101


Decimal number -0.000 000 000 000 176 555 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 1000 1100 0000 0100 0001 1010 0110 0011 0101

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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