-0.000 000 000 000 176 553 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 553 6(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
-0.000 000 000 000 176 553 6(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:

|-0.000 000 000 000 176 553 6| = 0.000 000 000 000 176 553 6


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: 0.000 000 000 000 176 553 6.

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) 0.000 000 000 000 176 553 6 × 2 = 0 + 0.000 000 000 000 353 107 2;
  • 2) 0.000 000 000 000 353 107 2 × 2 = 0 + 0.000 000 000 000 706 214 4;
  • 3) 0.000 000 000 000 706 214 4 × 2 = 0 + 0.000 000 000 001 412 428 8;
  • 4) 0.000 000 000 001 412 428 8 × 2 = 0 + 0.000 000 000 002 824 857 6;
  • 5) 0.000 000 000 002 824 857 6 × 2 = 0 + 0.000 000 000 005 649 715 2;
  • 6) 0.000 000 000 005 649 715 2 × 2 = 0 + 0.000 000 000 011 299 430 4;
  • 7) 0.000 000 000 011 299 430 4 × 2 = 0 + 0.000 000 000 022 598 860 8;
  • 8) 0.000 000 000 022 598 860 8 × 2 = 0 + 0.000 000 000 045 197 721 6;
  • 9) 0.000 000 000 045 197 721 6 × 2 = 0 + 0.000 000 000 090 395 443 2;
  • 10) 0.000 000 000 090 395 443 2 × 2 = 0 + 0.000 000 000 180 790 886 4;
  • 11) 0.000 000 000 180 790 886 4 × 2 = 0 + 0.000 000 000 361 581 772 8;
  • 12) 0.000 000 000 361 581 772 8 × 2 = 0 + 0.000 000 000 723 163 545 6;
  • 13) 0.000 000 000 723 163 545 6 × 2 = 0 + 0.000 000 001 446 327 091 2;
  • 14) 0.000 000 001 446 327 091 2 × 2 = 0 + 0.000 000 002 892 654 182 4;
  • 15) 0.000 000 002 892 654 182 4 × 2 = 0 + 0.000 000 005 785 308 364 8;
  • 16) 0.000 000 005 785 308 364 8 × 2 = 0 + 0.000 000 011 570 616 729 6;
  • 17) 0.000 000 011 570 616 729 6 × 2 = 0 + 0.000 000 023 141 233 459 2;
  • 18) 0.000 000 023 141 233 459 2 × 2 = 0 + 0.000 000 046 282 466 918 4;
  • 19) 0.000 000 046 282 466 918 4 × 2 = 0 + 0.000 000 092 564 933 836 8;
  • 20) 0.000 000 092 564 933 836 8 × 2 = 0 + 0.000 000 185 129 867 673 6;
  • 21) 0.000 000 185 129 867 673 6 × 2 = 0 + 0.000 000 370 259 735 347 2;
  • 22) 0.000 000 370 259 735 347 2 × 2 = 0 + 0.000 000 740 519 470 694 4;
  • 23) 0.000 000 740 519 470 694 4 × 2 = 0 + 0.000 001 481 038 941 388 8;
  • 24) 0.000 001 481 038 941 388 8 × 2 = 0 + 0.000 002 962 077 882 777 6;
  • 25) 0.000 002 962 077 882 777 6 × 2 = 0 + 0.000 005 924 155 765 555 2;
  • 26) 0.000 005 924 155 765 555 2 × 2 = 0 + 0.000 011 848 311 531 110 4;
  • 27) 0.000 011 848 311 531 110 4 × 2 = 0 + 0.000 023 696 623 062 220 8;
  • 28) 0.000 023 696 623 062 220 8 × 2 = 0 + 0.000 047 393 246 124 441 6;
  • 29) 0.000 047 393 246 124 441 6 × 2 = 0 + 0.000 094 786 492 248 883 2;
  • 30) 0.000 094 786 492 248 883 2 × 2 = 0 + 0.000 189 572 984 497 766 4;
  • 31) 0.000 189 572 984 497 766 4 × 2 = 0 + 0.000 379 145 968 995 532 8;
  • 32) 0.000 379 145 968 995 532 8 × 2 = 0 + 0.000 758 291 937 991 065 6;
  • 33) 0.000 758 291 937 991 065 6 × 2 = 0 + 0.001 516 583 875 982 131 2;
  • 34) 0.001 516 583 875 982 131 2 × 2 = 0 + 0.003 033 167 751 964 262 4;
  • 35) 0.003 033 167 751 964 262 4 × 2 = 0 + 0.006 066 335 503 928 524 8;
  • 36) 0.006 066 335 503 928 524 8 × 2 = 0 + 0.012 132 671 007 857 049 6;
  • 37) 0.012 132 671 007 857 049 6 × 2 = 0 + 0.024 265 342 015 714 099 2;
  • 38) 0.024 265 342 015 714 099 2 × 2 = 0 + 0.048 530 684 031 428 198 4;
  • 39) 0.048 530 684 031 428 198 4 × 2 = 0 + 0.097 061 368 062 856 396 8;
  • 40) 0.097 061 368 062 856 396 8 × 2 = 0 + 0.194 122 736 125 712 793 6;
  • 41) 0.194 122 736 125 712 793 6 × 2 = 0 + 0.388 245 472 251 425 587 2;
  • 42) 0.388 245 472 251 425 587 2 × 2 = 0 + 0.776 490 944 502 851 174 4;
  • 43) 0.776 490 944 502 851 174 4 × 2 = 1 + 0.552 981 889 005 702 348 8;
  • 44) 0.552 981 889 005 702 348 8 × 2 = 1 + 0.105 963 778 011 404 697 6;
  • 45) 0.105 963 778 011 404 697 6 × 2 = 0 + 0.211 927 556 022 809 395 2;
  • 46) 0.211 927 556 022 809 395 2 × 2 = 0 + 0.423 855 112 045 618 790 4;
  • 47) 0.423 855 112 045 618 790 4 × 2 = 0 + 0.847 710 224 091 237 580 8;
  • 48) 0.847 710 224 091 237 580 8 × 2 = 1 + 0.695 420 448 182 475 161 6;
  • 49) 0.695 420 448 182 475 161 6 × 2 = 1 + 0.390 840 896 364 950 323 2;
  • 50) 0.390 840 896 364 950 323 2 × 2 = 0 + 0.781 681 792 729 900 646 4;
  • 51) 0.781 681 792 729 900 646 4 × 2 = 1 + 0.563 363 585 459 801 292 8;
  • 52) 0.563 363 585 459 801 292 8 × 2 = 1 + 0.126 727 170 919 602 585 6;
  • 53) 0.126 727 170 919 602 585 6 × 2 = 0 + 0.253 454 341 839 205 171 2;
  • 54) 0.253 454 341 839 205 171 2 × 2 = 0 + 0.506 908 683 678 410 342 4;
  • 55) 0.506 908 683 678 410 342 4 × 2 = 1 + 0.013 817 367 356 820 684 8;
  • 56) 0.013 817 367 356 820 684 8 × 2 = 0 + 0.027 634 734 713 641 369 6;
  • 57) 0.027 634 734 713 641 369 6 × 2 = 0 + 0.055 269 469 427 282 739 2;
  • 58) 0.055 269 469 427 282 739 2 × 2 = 0 + 0.110 538 938 854 565 478 4;
  • 59) 0.110 538 938 854 565 478 4 × 2 = 0 + 0.221 077 877 709 130 956 8;
  • 60) 0.221 077 877 709 130 956 8 × 2 = 0 + 0.442 155 755 418 261 913 6;
  • 61) 0.442 155 755 418 261 913 6 × 2 = 0 + 0.884 311 510 836 523 827 2;
  • 62) 0.884 311 510 836 523 827 2 × 2 = 1 + 0.768 623 021 673 047 654 4;
  • 63) 0.768 623 021 673 047 654 4 × 2 = 1 + 0.537 246 043 346 095 308 8;
  • 64) 0.537 246 043 346 095 308 8 × 2 = 1 + 0.074 492 086 692 190 617 6;
  • 65) 0.074 492 086 692 190 617 6 × 2 = 0 + 0.148 984 173 384 381 235 2;
  • 66) 0.148 984 173 384 381 235 2 × 2 = 0 + 0.297 968 346 768 762 470 4;
  • 67) 0.297 968 346 768 762 470 4 × 2 = 0 + 0.595 936 693 537 524 940 8;
  • 68) 0.595 936 693 537 524 940 8 × 2 = 1 + 0.191 873 387 075 049 881 6;
  • 69) 0.191 873 387 075 049 881 6 × 2 = 0 + 0.383 746 774 150 099 763 2;
  • 70) 0.383 746 774 150 099 763 2 × 2 = 0 + 0.767 493 548 300 199 526 4;
  • 71) 0.767 493 548 300 199 526 4 × 2 = 1 + 0.534 987 096 600 399 052 8;
  • 72) 0.534 987 096 600 399 052 8 × 2 = 1 + 0.069 974 193 200 798 105 6;
  • 73) 0.069 974 193 200 798 105 6 × 2 = 0 + 0.139 948 386 401 596 211 2;
  • 74) 0.139 948 386 401 596 211 2 × 2 = 0 + 0.279 896 772 803 192 422 4;
  • 75) 0.279 896 772 803 192 422 4 × 2 = 0 + 0.559 793 545 606 384 844 8;
  • 76) 0.559 793 545 606 384 844 8 × 2 = 1 + 0.119 587 091 212 769 689 6;
  • 77) 0.119 587 091 212 769 689 6 × 2 = 0 + 0.239 174 182 425 539 379 2;
  • 78) 0.239 174 182 425 539 379 2 × 2 = 0 + 0.478 348 364 851 078 758 4;
  • 79) 0.478 348 364 851 078 758 4 × 2 = 0 + 0.956 696 729 702 157 516 8;
  • 80) 0.956 696 729 702 157 516 8 × 2 = 1 + 0.913 393 459 404 315 033 6;
  • 81) 0.913 393 459 404 315 033 6 × 2 = 1 + 0.826 786 918 808 630 067 2;
  • 82) 0.826 786 918 808 630 067 2 × 2 = 1 + 0.653 573 837 617 260 134 4;
  • 83) 0.653 573 837 617 260 134 4 × 2 = 1 + 0.307 147 675 234 520 268 8;
  • 84) 0.307 147 675 234 520 268 8 × 2 = 0 + 0.614 295 350 469 040 537 6;
  • 85) 0.614 295 350 469 040 537 6 × 2 = 1 + 0.228 590 700 938 081 075 2;
  • 86) 0.228 590 700 938 081 075 2 × 2 = 0 + 0.457 181 401 876 162 150 4;
  • 87) 0.457 181 401 876 162 150 4 × 2 = 0 + 0.914 362 803 752 324 300 8;
  • 88) 0.914 362 803 752 324 300 8 × 2 = 1 + 0.828 725 607 504 648 601 6;
  • 89) 0.828 725 607 504 648 601 6 × 2 = 1 + 0.657 451 215 009 297 203 2;
  • 90) 0.657 451 215 009 297 203 2 × 2 = 1 + 0.314 902 430 018 594 406 4;
  • 91) 0.314 902 430 018 594 406 4 × 2 = 0 + 0.629 804 860 037 188 812 8;
  • 92) 0.629 804 860 037 188 812 8 × 2 = 1 + 0.259 609 720 074 377 625 6;
  • 93) 0.259 609 720 074 377 625 6 × 2 = 0 + 0.519 219 440 148 755 251 2;
  • 94) 0.519 219 440 148 755 251 2 × 2 = 1 + 0.038 438 880 297 510 502 4;
  • 95) 0.038 438 880 297 510 502 4 × 2 = 0 + 0.076 877 760 595 021 004 8;

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:


0.000 000 000 000 176 553 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0000 0111 0001 0011 0001 0001 1110 1001 1101 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 553 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0000 0111 0001 0011 0001 0001 1110 1001 1101 010(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:


0.000 000 000 000 176 553 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0000 0111 0001 0011 0001 0001 1110 1001 1101 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0000 0111 0001 0011 0001 0001 1110 1001 1101 010(2) × 20 =

1.1000 1101 1001 0000 0011 1000 1001 1000 1000 1111 0100 1110 1010(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 0000 0011 1000 1001 1000 1000 1111 0100 1110 1010


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 0000 0011 1000 1001 1000 1000 1111 0100 1110 1010 =

1000 1101 1001 0000 0011 1000 1001 1000 1000 1111 0100 1110 1010


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 0000 0011 1000 1001 1000 1000 1111 0100 1110 1010


Decimal number -0.000 000 000 000 176 553 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0000 0011 1000 1001 1000 1000 1111 0100 1110 1010

Share

» Convert decimal -0.000 000 000 000 176 562 4 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: Pull vs. Push Leadership: The Invisible Power. 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