-0.000 000 000 000 176 557 385 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 385(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
-0.000 000 000 000 176 557 385(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:

|-0.000 000 000 000 176 557 385| = 0.000 000 000 000 176 557 385


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: 0.000 000 000 000 176 557 385.

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) 0.000 000 000 000 176 557 385 × 2 = 0 + 0.000 000 000 000 353 114 77;
  • 2) 0.000 000 000 000 353 114 77 × 2 = 0 + 0.000 000 000 000 706 229 54;
  • 3) 0.000 000 000 000 706 229 54 × 2 = 0 + 0.000 000 000 001 412 459 08;
  • 4) 0.000 000 000 001 412 459 08 × 2 = 0 + 0.000 000 000 002 824 918 16;
  • 5) 0.000 000 000 002 824 918 16 × 2 = 0 + 0.000 000 000 005 649 836 32;
  • 6) 0.000 000 000 005 649 836 32 × 2 = 0 + 0.000 000 000 011 299 672 64;
  • 7) 0.000 000 000 011 299 672 64 × 2 = 0 + 0.000 000 000 022 599 345 28;
  • 8) 0.000 000 000 022 599 345 28 × 2 = 0 + 0.000 000 000 045 198 690 56;
  • 9) 0.000 000 000 045 198 690 56 × 2 = 0 + 0.000 000 000 090 397 381 12;
  • 10) 0.000 000 000 090 397 381 12 × 2 = 0 + 0.000 000 000 180 794 762 24;
  • 11) 0.000 000 000 180 794 762 24 × 2 = 0 + 0.000 000 000 361 589 524 48;
  • 12) 0.000 000 000 361 589 524 48 × 2 = 0 + 0.000 000 000 723 179 048 96;
  • 13) 0.000 000 000 723 179 048 96 × 2 = 0 + 0.000 000 001 446 358 097 92;
  • 14) 0.000 000 001 446 358 097 92 × 2 = 0 + 0.000 000 002 892 716 195 84;
  • 15) 0.000 000 002 892 716 195 84 × 2 = 0 + 0.000 000 005 785 432 391 68;
  • 16) 0.000 000 005 785 432 391 68 × 2 = 0 + 0.000 000 011 570 864 783 36;
  • 17) 0.000 000 011 570 864 783 36 × 2 = 0 + 0.000 000 023 141 729 566 72;
  • 18) 0.000 000 023 141 729 566 72 × 2 = 0 + 0.000 000 046 283 459 133 44;
  • 19) 0.000 000 046 283 459 133 44 × 2 = 0 + 0.000 000 092 566 918 266 88;
  • 20) 0.000 000 092 566 918 266 88 × 2 = 0 + 0.000 000 185 133 836 533 76;
  • 21) 0.000 000 185 133 836 533 76 × 2 = 0 + 0.000 000 370 267 673 067 52;
  • 22) 0.000 000 370 267 673 067 52 × 2 = 0 + 0.000 000 740 535 346 135 04;
  • 23) 0.000 000 740 535 346 135 04 × 2 = 0 + 0.000 001 481 070 692 270 08;
  • 24) 0.000 001 481 070 692 270 08 × 2 = 0 + 0.000 002 962 141 384 540 16;
  • 25) 0.000 002 962 141 384 540 16 × 2 = 0 + 0.000 005 924 282 769 080 32;
  • 26) 0.000 005 924 282 769 080 32 × 2 = 0 + 0.000 011 848 565 538 160 64;
  • 27) 0.000 011 848 565 538 160 64 × 2 = 0 + 0.000 023 697 131 076 321 28;
  • 28) 0.000 023 697 131 076 321 28 × 2 = 0 + 0.000 047 394 262 152 642 56;
  • 29) 0.000 047 394 262 152 642 56 × 2 = 0 + 0.000 094 788 524 305 285 12;
  • 30) 0.000 094 788 524 305 285 12 × 2 = 0 + 0.000 189 577 048 610 570 24;
  • 31) 0.000 189 577 048 610 570 24 × 2 = 0 + 0.000 379 154 097 221 140 48;
  • 32) 0.000 379 154 097 221 140 48 × 2 = 0 + 0.000 758 308 194 442 280 96;
  • 33) 0.000 758 308 194 442 280 96 × 2 = 0 + 0.001 516 616 388 884 561 92;
  • 34) 0.001 516 616 388 884 561 92 × 2 = 0 + 0.003 033 232 777 769 123 84;
  • 35) 0.003 033 232 777 769 123 84 × 2 = 0 + 0.006 066 465 555 538 247 68;
  • 36) 0.006 066 465 555 538 247 68 × 2 = 0 + 0.012 132 931 111 076 495 36;
  • 37) 0.012 132 931 111 076 495 36 × 2 = 0 + 0.024 265 862 222 152 990 72;
  • 38) 0.024 265 862 222 152 990 72 × 2 = 0 + 0.048 531 724 444 305 981 44;
  • 39) 0.048 531 724 444 305 981 44 × 2 = 0 + 0.097 063 448 888 611 962 88;
  • 40) 0.097 063 448 888 611 962 88 × 2 = 0 + 0.194 126 897 777 223 925 76;
  • 41) 0.194 126 897 777 223 925 76 × 2 = 0 + 0.388 253 795 554 447 851 52;
  • 42) 0.388 253 795 554 447 851 52 × 2 = 0 + 0.776 507 591 108 895 703 04;
  • 43) 0.776 507 591 108 895 703 04 × 2 = 1 + 0.553 015 182 217 791 406 08;
  • 44) 0.553 015 182 217 791 406 08 × 2 = 1 + 0.106 030 364 435 582 812 16;
  • 45) 0.106 030 364 435 582 812 16 × 2 = 0 + 0.212 060 728 871 165 624 32;
  • 46) 0.212 060 728 871 165 624 32 × 2 = 0 + 0.424 121 457 742 331 248 64;
  • 47) 0.424 121 457 742 331 248 64 × 2 = 0 + 0.848 242 915 484 662 497 28;
  • 48) 0.848 242 915 484 662 497 28 × 2 = 1 + 0.696 485 830 969 324 994 56;
  • 49) 0.696 485 830 969 324 994 56 × 2 = 1 + 0.392 971 661 938 649 989 12;
  • 50) 0.392 971 661 938 649 989 12 × 2 = 0 + 0.785 943 323 877 299 978 24;
  • 51) 0.785 943 323 877 299 978 24 × 2 = 1 + 0.571 886 647 754 599 956 48;
  • 52) 0.571 886 647 754 599 956 48 × 2 = 1 + 0.143 773 295 509 199 912 96;
  • 53) 0.143 773 295 509 199 912 96 × 2 = 0 + 0.287 546 591 018 399 825 92;
  • 54) 0.287 546 591 018 399 825 92 × 2 = 0 + 0.575 093 182 036 799 651 84;
  • 55) 0.575 093 182 036 799 651 84 × 2 = 1 + 0.150 186 364 073 599 303 68;
  • 56) 0.150 186 364 073 599 303 68 × 2 = 0 + 0.300 372 728 147 198 607 36;
  • 57) 0.300 372 728 147 198 607 36 × 2 = 0 + 0.600 745 456 294 397 214 72;
  • 58) 0.600 745 456 294 397 214 72 × 2 = 1 + 0.201 490 912 588 794 429 44;
  • 59) 0.201 490 912 588 794 429 44 × 2 = 0 + 0.402 981 825 177 588 858 88;
  • 60) 0.402 981 825 177 588 858 88 × 2 = 0 + 0.805 963 650 355 177 717 76;
  • 61) 0.805 963 650 355 177 717 76 × 2 = 1 + 0.611 927 300 710 355 435 52;
  • 62) 0.611 927 300 710 355 435 52 × 2 = 1 + 0.223 854 601 420 710 871 04;
  • 63) 0.223 854 601 420 710 871 04 × 2 = 0 + 0.447 709 202 841 421 742 08;
  • 64) 0.447 709 202 841 421 742 08 × 2 = 0 + 0.895 418 405 682 843 484 16;
  • 65) 0.895 418 405 682 843 484 16 × 2 = 1 + 0.790 836 811 365 686 968 32;
  • 66) 0.790 836 811 365 686 968 32 × 2 = 1 + 0.581 673 622 731 373 936 64;
  • 67) 0.581 673 622 731 373 936 64 × 2 = 1 + 0.163 347 245 462 747 873 28;
  • 68) 0.163 347 245 462 747 873 28 × 2 = 0 + 0.326 694 490 925 495 746 56;
  • 69) 0.326 694 490 925 495 746 56 × 2 = 0 + 0.653 388 981 850 991 493 12;
  • 70) 0.653 388 981 850 991 493 12 × 2 = 1 + 0.306 777 963 701 982 986 24;
  • 71) 0.306 777 963 701 982 986 24 × 2 = 0 + 0.613 555 927 403 965 972 48;
  • 72) 0.613 555 927 403 965 972 48 × 2 = 1 + 0.227 111 854 807 931 944 96;
  • 73) 0.227 111 854 807 931 944 96 × 2 = 0 + 0.454 223 709 615 863 889 92;
  • 74) 0.454 223 709 615 863 889 92 × 2 = 0 + 0.908 447 419 231 727 779 84;
  • 75) 0.908 447 419 231 727 779 84 × 2 = 1 + 0.816 894 838 463 455 559 68;
  • 76) 0.816 894 838 463 455 559 68 × 2 = 1 + 0.633 789 676 926 911 119 36;
  • 77) 0.633 789 676 926 911 119 36 × 2 = 1 + 0.267 579 353 853 822 238 72;
  • 78) 0.267 579 353 853 822 238 72 × 2 = 0 + 0.535 158 707 707 644 477 44;
  • 79) 0.535 158 707 707 644 477 44 × 2 = 1 + 0.070 317 415 415 288 954 88;
  • 80) 0.070 317 415 415 288 954 88 × 2 = 0 + 0.140 634 830 830 577 909 76;
  • 81) 0.140 634 830 830 577 909 76 × 2 = 0 + 0.281 269 661 661 155 819 52;
  • 82) 0.281 269 661 661 155 819 52 × 2 = 0 + 0.562 539 323 322 311 639 04;
  • 83) 0.562 539 323 322 311 639 04 × 2 = 1 + 0.125 078 646 644 623 278 08;
  • 84) 0.125 078 646 644 623 278 08 × 2 = 0 + 0.250 157 293 289 246 556 16;
  • 85) 0.250 157 293 289 246 556 16 × 2 = 0 + 0.500 314 586 578 493 112 32;
  • 86) 0.500 314 586 578 493 112 32 × 2 = 1 + 0.000 629 173 156 986 224 64;
  • 87) 0.000 629 173 156 986 224 64 × 2 = 0 + 0.001 258 346 313 972 449 28;
  • 88) 0.001 258 346 313 972 449 28 × 2 = 0 + 0.002 516 692 627 944 898 56;
  • 89) 0.002 516 692 627 944 898 56 × 2 = 0 + 0.005 033 385 255 889 797 12;
  • 90) 0.005 033 385 255 889 797 12 × 2 = 0 + 0.010 066 770 511 779 594 24;
  • 91) 0.010 066 770 511 779 594 24 × 2 = 0 + 0.020 133 541 023 559 188 48;
  • 92) 0.020 133 541 023 559 188 48 × 2 = 0 + 0.040 267 082 047 118 376 96;
  • 93) 0.040 267 082 047 118 376 96 × 2 = 0 + 0.080 534 164 094 236 753 92;
  • 94) 0.080 534 164 094 236 753 92 × 2 = 0 + 0.161 068 328 188 473 507 84;
  • 95) 0.161 068 328 188 473 507 84 × 2 = 0 + 0.322 136 656 376 947 015 68;

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:


0.000 000 000 000 176 557 385(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1110 0101 0011 1010 0010 0100 0000 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 385(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1110 0101 0011 1010 0010 0100 0000 000(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:


0.000 000 000 000 176 557 385(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1110 0101 0011 1010 0010 0100 0000 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1110 0101 0011 1010 0010 0100 0000 000(2) × 20 =

1.1000 1101 1001 0010 0110 0111 0010 1001 1101 0001 0010 0000 0000(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 0010 0110 0111 0010 1001 1101 0001 0010 0000 0000


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 0010 0110 0111 0010 1001 1101 0001 0010 0000 0000 =

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


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 0010 0110 0111 0010 1001 1101 0001 0010 0000 0000


Decimal number -0.000 000 000 000 176 557 385 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

Share

» Convert decimal -0.000 000 000 000 176 557 439 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 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Magnetic Leadership: The Invisible Power of the Pull. NotebookLM YouTube Video of 11.31 minutes

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