-0.000 000 000 000 176 552 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 552 8(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
-0.000 000 000 000 176 552 8(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:

|-0.000 000 000 000 176 552 8| = 0.000 000 000 000 176 552 8


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: 0.000 000 000 000 176 552 8.

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) 0.000 000 000 000 176 552 8 × 2 = 0 + 0.000 000 000 000 353 105 6;
  • 2) 0.000 000 000 000 353 105 6 × 2 = 0 + 0.000 000 000 000 706 211 2;
  • 3) 0.000 000 000 000 706 211 2 × 2 = 0 + 0.000 000 000 001 412 422 4;
  • 4) 0.000 000 000 001 412 422 4 × 2 = 0 + 0.000 000 000 002 824 844 8;
  • 5) 0.000 000 000 002 824 844 8 × 2 = 0 + 0.000 000 000 005 649 689 6;
  • 6) 0.000 000 000 005 649 689 6 × 2 = 0 + 0.000 000 000 011 299 379 2;
  • 7) 0.000 000 000 011 299 379 2 × 2 = 0 + 0.000 000 000 022 598 758 4;
  • 8) 0.000 000 000 022 598 758 4 × 2 = 0 + 0.000 000 000 045 197 516 8;
  • 9) 0.000 000 000 045 197 516 8 × 2 = 0 + 0.000 000 000 090 395 033 6;
  • 10) 0.000 000 000 090 395 033 6 × 2 = 0 + 0.000 000 000 180 790 067 2;
  • 11) 0.000 000 000 180 790 067 2 × 2 = 0 + 0.000 000 000 361 580 134 4;
  • 12) 0.000 000 000 361 580 134 4 × 2 = 0 + 0.000 000 000 723 160 268 8;
  • 13) 0.000 000 000 723 160 268 8 × 2 = 0 + 0.000 000 001 446 320 537 6;
  • 14) 0.000 000 001 446 320 537 6 × 2 = 0 + 0.000 000 002 892 641 075 2;
  • 15) 0.000 000 002 892 641 075 2 × 2 = 0 + 0.000 000 005 785 282 150 4;
  • 16) 0.000 000 005 785 282 150 4 × 2 = 0 + 0.000 000 011 570 564 300 8;
  • 17) 0.000 000 011 570 564 300 8 × 2 = 0 + 0.000 000 023 141 128 601 6;
  • 18) 0.000 000 023 141 128 601 6 × 2 = 0 + 0.000 000 046 282 257 203 2;
  • 19) 0.000 000 046 282 257 203 2 × 2 = 0 + 0.000 000 092 564 514 406 4;
  • 20) 0.000 000 092 564 514 406 4 × 2 = 0 + 0.000 000 185 129 028 812 8;
  • 21) 0.000 000 185 129 028 812 8 × 2 = 0 + 0.000 000 370 258 057 625 6;
  • 22) 0.000 000 370 258 057 625 6 × 2 = 0 + 0.000 000 740 516 115 251 2;
  • 23) 0.000 000 740 516 115 251 2 × 2 = 0 + 0.000 001 481 032 230 502 4;
  • 24) 0.000 001 481 032 230 502 4 × 2 = 0 + 0.000 002 962 064 461 004 8;
  • 25) 0.000 002 962 064 461 004 8 × 2 = 0 + 0.000 005 924 128 922 009 6;
  • 26) 0.000 005 924 128 922 009 6 × 2 = 0 + 0.000 011 848 257 844 019 2;
  • 27) 0.000 011 848 257 844 019 2 × 2 = 0 + 0.000 023 696 515 688 038 4;
  • 28) 0.000 023 696 515 688 038 4 × 2 = 0 + 0.000 047 393 031 376 076 8;
  • 29) 0.000 047 393 031 376 076 8 × 2 = 0 + 0.000 094 786 062 752 153 6;
  • 30) 0.000 094 786 062 752 153 6 × 2 = 0 + 0.000 189 572 125 504 307 2;
  • 31) 0.000 189 572 125 504 307 2 × 2 = 0 + 0.000 379 144 251 008 614 4;
  • 32) 0.000 379 144 251 008 614 4 × 2 = 0 + 0.000 758 288 502 017 228 8;
  • 33) 0.000 758 288 502 017 228 8 × 2 = 0 + 0.001 516 577 004 034 457 6;
  • 34) 0.001 516 577 004 034 457 6 × 2 = 0 + 0.003 033 154 008 068 915 2;
  • 35) 0.003 033 154 008 068 915 2 × 2 = 0 + 0.006 066 308 016 137 830 4;
  • 36) 0.006 066 308 016 137 830 4 × 2 = 0 + 0.012 132 616 032 275 660 8;
  • 37) 0.012 132 616 032 275 660 8 × 2 = 0 + 0.024 265 232 064 551 321 6;
  • 38) 0.024 265 232 064 551 321 6 × 2 = 0 + 0.048 530 464 129 102 643 2;
  • 39) 0.048 530 464 129 102 643 2 × 2 = 0 + 0.097 060 928 258 205 286 4;
  • 40) 0.097 060 928 258 205 286 4 × 2 = 0 + 0.194 121 856 516 410 572 8;
  • 41) 0.194 121 856 516 410 572 8 × 2 = 0 + 0.388 243 713 032 821 145 6;
  • 42) 0.388 243 713 032 821 145 6 × 2 = 0 + 0.776 487 426 065 642 291 2;
  • 43) 0.776 487 426 065 642 291 2 × 2 = 1 + 0.552 974 852 131 284 582 4;
  • 44) 0.552 974 852 131 284 582 4 × 2 = 1 + 0.105 949 704 262 569 164 8;
  • 45) 0.105 949 704 262 569 164 8 × 2 = 0 + 0.211 899 408 525 138 329 6;
  • 46) 0.211 899 408 525 138 329 6 × 2 = 0 + 0.423 798 817 050 276 659 2;
  • 47) 0.423 798 817 050 276 659 2 × 2 = 0 + 0.847 597 634 100 553 318 4;
  • 48) 0.847 597 634 100 553 318 4 × 2 = 1 + 0.695 195 268 201 106 636 8;
  • 49) 0.695 195 268 201 106 636 8 × 2 = 1 + 0.390 390 536 402 213 273 6;
  • 50) 0.390 390 536 402 213 273 6 × 2 = 0 + 0.780 781 072 804 426 547 2;
  • 51) 0.780 781 072 804 426 547 2 × 2 = 1 + 0.561 562 145 608 853 094 4;
  • 52) 0.561 562 145 608 853 094 4 × 2 = 1 + 0.123 124 291 217 706 188 8;
  • 53) 0.123 124 291 217 706 188 8 × 2 = 0 + 0.246 248 582 435 412 377 6;
  • 54) 0.246 248 582 435 412 377 6 × 2 = 0 + 0.492 497 164 870 824 755 2;
  • 55) 0.492 497 164 870 824 755 2 × 2 = 0 + 0.984 994 329 741 649 510 4;
  • 56) 0.984 994 329 741 649 510 4 × 2 = 1 + 0.969 988 659 483 299 020 8;
  • 57) 0.969 988 659 483 299 020 8 × 2 = 1 + 0.939 977 318 966 598 041 6;
  • 58) 0.939 977 318 966 598 041 6 × 2 = 1 + 0.879 954 637 933 196 083 2;
  • 59) 0.879 954 637 933 196 083 2 × 2 = 1 + 0.759 909 275 866 392 166 4;
  • 60) 0.759 909 275 866 392 166 4 × 2 = 1 + 0.519 818 551 732 784 332 8;
  • 61) 0.519 818 551 732 784 332 8 × 2 = 1 + 0.039 637 103 465 568 665 6;
  • 62) 0.039 637 103 465 568 665 6 × 2 = 0 + 0.079 274 206 931 137 331 2;
  • 63) 0.079 274 206 931 137 331 2 × 2 = 0 + 0.158 548 413 862 274 662 4;
  • 64) 0.158 548 413 862 274 662 4 × 2 = 0 + 0.317 096 827 724 549 324 8;
  • 65) 0.317 096 827 724 549 324 8 × 2 = 0 + 0.634 193 655 449 098 649 6;
  • 66) 0.634 193 655 449 098 649 6 × 2 = 1 + 0.268 387 310 898 197 299 2;
  • 67) 0.268 387 310 898 197 299 2 × 2 = 0 + 0.536 774 621 796 394 598 4;
  • 68) 0.536 774 621 796 394 598 4 × 2 = 1 + 0.073 549 243 592 789 196 8;
  • 69) 0.073 549 243 592 789 196 8 × 2 = 0 + 0.147 098 487 185 578 393 6;
  • 70) 0.147 098 487 185 578 393 6 × 2 = 0 + 0.294 196 974 371 156 787 2;
  • 71) 0.294 196 974 371 156 787 2 × 2 = 0 + 0.588 393 948 742 313 574 4;
  • 72) 0.588 393 948 742 313 574 4 × 2 = 1 + 0.176 787 897 484 627 148 8;
  • 73) 0.176 787 897 484 627 148 8 × 2 = 0 + 0.353 575 794 969 254 297 6;
  • 74) 0.353 575 794 969 254 297 6 × 2 = 0 + 0.707 151 589 938 508 595 2;
  • 75) 0.707 151 589 938 508 595 2 × 2 = 1 + 0.414 303 179 877 017 190 4;
  • 76) 0.414 303 179 877 017 190 4 × 2 = 0 + 0.828 606 359 754 034 380 8;
  • 77) 0.828 606 359 754 034 380 8 × 2 = 1 + 0.657 212 719 508 068 761 6;
  • 78) 0.657 212 719 508 068 761 6 × 2 = 1 + 0.314 425 439 016 137 523 2;
  • 79) 0.314 425 439 016 137 523 2 × 2 = 0 + 0.628 850 878 032 275 046 4;
  • 80) 0.628 850 878 032 275 046 4 × 2 = 1 + 0.257 701 756 064 550 092 8;
  • 81) 0.257 701 756 064 550 092 8 × 2 = 0 + 0.515 403 512 129 100 185 6;
  • 82) 0.515 403 512 129 100 185 6 × 2 = 1 + 0.030 807 024 258 200 371 2;
  • 83) 0.030 807 024 258 200 371 2 × 2 = 0 + 0.061 614 048 516 400 742 4;
  • 84) 0.061 614 048 516 400 742 4 × 2 = 0 + 0.123 228 097 032 801 484 8;
  • 85) 0.123 228 097 032 801 484 8 × 2 = 0 + 0.246 456 194 065 602 969 6;
  • 86) 0.246 456 194 065 602 969 6 × 2 = 0 + 0.492 912 388 131 205 939 2;
  • 87) 0.492 912 388 131 205 939 2 × 2 = 0 + 0.985 824 776 262 411 878 4;
  • 88) 0.985 824 776 262 411 878 4 × 2 = 1 + 0.971 649 552 524 823 756 8;
  • 89) 0.971 649 552 524 823 756 8 × 2 = 1 + 0.943 299 105 049 647 513 6;
  • 90) 0.943 299 105 049 647 513 6 × 2 = 1 + 0.886 598 210 099 295 027 2;
  • 91) 0.886 598 210 099 295 027 2 × 2 = 1 + 0.773 196 420 198 590 054 4;
  • 92) 0.773 196 420 198 590 054 4 × 2 = 1 + 0.546 392 840 397 180 108 8;
  • 93) 0.546 392 840 397 180 108 8 × 2 = 1 + 0.092 785 680 794 360 217 6;
  • 94) 0.092 785 680 794 360 217 6 × 2 = 0 + 0.185 571 361 588 720 435 2;
  • 95) 0.185 571 361 588 720 435 2 × 2 = 0 + 0.371 142 723 177 440 870 4;

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:


0.000 000 000 000 176 552 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1111 1000 0101 0001 0010 1101 0100 0001 1111 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 552 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1111 1000 0101 0001 0010 1101 0100 0001 1111 100(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:


0.000 000 000 000 176 552 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1111 1000 0101 0001 0010 1101 0100 0001 1111 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1111 1000 0101 0001 0010 1101 0100 0001 1111 100(2) × 20 =

1.1000 1101 1000 1111 1100 0010 1000 1001 0110 1010 0000 1111 1100(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 1000 1111 1100 0010 1000 1001 0110 1010 0000 1111 1100


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 1000 1111 1100 0010 1000 1001 0110 1010 0000 1111 1100 =

1000 1101 1000 1111 1100 0010 1000 1001 0110 1010 0000 1111 1100


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 1000 1111 1100 0010 1000 1001 0110 1010 0000 1111 1100


Decimal number -0.000 000 000 000 176 552 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1111 1100 0010 1000 1001 0110 1010 0000 1111 1100

Share

» Convert decimal -0.000 000 000 000 176 556 1 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