-0.000 000 000 000 176 516 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 516 4(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
-0.000 000 000 000 176 516 4(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:

|-0.000 000 000 000 176 516 4| = 0.000 000 000 000 176 516 4


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: 0.000 000 000 000 176 516 4.

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) 0.000 000 000 000 176 516 4 × 2 = 0 + 0.000 000 000 000 353 032 8;
  • 2) 0.000 000 000 000 353 032 8 × 2 = 0 + 0.000 000 000 000 706 065 6;
  • 3) 0.000 000 000 000 706 065 6 × 2 = 0 + 0.000 000 000 001 412 131 2;
  • 4) 0.000 000 000 001 412 131 2 × 2 = 0 + 0.000 000 000 002 824 262 4;
  • 5) 0.000 000 000 002 824 262 4 × 2 = 0 + 0.000 000 000 005 648 524 8;
  • 6) 0.000 000 000 005 648 524 8 × 2 = 0 + 0.000 000 000 011 297 049 6;
  • 7) 0.000 000 000 011 297 049 6 × 2 = 0 + 0.000 000 000 022 594 099 2;
  • 8) 0.000 000 000 022 594 099 2 × 2 = 0 + 0.000 000 000 045 188 198 4;
  • 9) 0.000 000 000 045 188 198 4 × 2 = 0 + 0.000 000 000 090 376 396 8;
  • 10) 0.000 000 000 090 376 396 8 × 2 = 0 + 0.000 000 000 180 752 793 6;
  • 11) 0.000 000 000 180 752 793 6 × 2 = 0 + 0.000 000 000 361 505 587 2;
  • 12) 0.000 000 000 361 505 587 2 × 2 = 0 + 0.000 000 000 723 011 174 4;
  • 13) 0.000 000 000 723 011 174 4 × 2 = 0 + 0.000 000 001 446 022 348 8;
  • 14) 0.000 000 001 446 022 348 8 × 2 = 0 + 0.000 000 002 892 044 697 6;
  • 15) 0.000 000 002 892 044 697 6 × 2 = 0 + 0.000 000 005 784 089 395 2;
  • 16) 0.000 000 005 784 089 395 2 × 2 = 0 + 0.000 000 011 568 178 790 4;
  • 17) 0.000 000 011 568 178 790 4 × 2 = 0 + 0.000 000 023 136 357 580 8;
  • 18) 0.000 000 023 136 357 580 8 × 2 = 0 + 0.000 000 046 272 715 161 6;
  • 19) 0.000 000 046 272 715 161 6 × 2 = 0 + 0.000 000 092 545 430 323 2;
  • 20) 0.000 000 092 545 430 323 2 × 2 = 0 + 0.000 000 185 090 860 646 4;
  • 21) 0.000 000 185 090 860 646 4 × 2 = 0 + 0.000 000 370 181 721 292 8;
  • 22) 0.000 000 370 181 721 292 8 × 2 = 0 + 0.000 000 740 363 442 585 6;
  • 23) 0.000 000 740 363 442 585 6 × 2 = 0 + 0.000 001 480 726 885 171 2;
  • 24) 0.000 001 480 726 885 171 2 × 2 = 0 + 0.000 002 961 453 770 342 4;
  • 25) 0.000 002 961 453 770 342 4 × 2 = 0 + 0.000 005 922 907 540 684 8;
  • 26) 0.000 005 922 907 540 684 8 × 2 = 0 + 0.000 011 845 815 081 369 6;
  • 27) 0.000 011 845 815 081 369 6 × 2 = 0 + 0.000 023 691 630 162 739 2;
  • 28) 0.000 023 691 630 162 739 2 × 2 = 0 + 0.000 047 383 260 325 478 4;
  • 29) 0.000 047 383 260 325 478 4 × 2 = 0 + 0.000 094 766 520 650 956 8;
  • 30) 0.000 094 766 520 650 956 8 × 2 = 0 + 0.000 189 533 041 301 913 6;
  • 31) 0.000 189 533 041 301 913 6 × 2 = 0 + 0.000 379 066 082 603 827 2;
  • 32) 0.000 379 066 082 603 827 2 × 2 = 0 + 0.000 758 132 165 207 654 4;
  • 33) 0.000 758 132 165 207 654 4 × 2 = 0 + 0.001 516 264 330 415 308 8;
  • 34) 0.001 516 264 330 415 308 8 × 2 = 0 + 0.003 032 528 660 830 617 6;
  • 35) 0.003 032 528 660 830 617 6 × 2 = 0 + 0.006 065 057 321 661 235 2;
  • 36) 0.006 065 057 321 661 235 2 × 2 = 0 + 0.012 130 114 643 322 470 4;
  • 37) 0.012 130 114 643 322 470 4 × 2 = 0 + 0.024 260 229 286 644 940 8;
  • 38) 0.024 260 229 286 644 940 8 × 2 = 0 + 0.048 520 458 573 289 881 6;
  • 39) 0.048 520 458 573 289 881 6 × 2 = 0 + 0.097 040 917 146 579 763 2;
  • 40) 0.097 040 917 146 579 763 2 × 2 = 0 + 0.194 081 834 293 159 526 4;
  • 41) 0.194 081 834 293 159 526 4 × 2 = 0 + 0.388 163 668 586 319 052 8;
  • 42) 0.388 163 668 586 319 052 8 × 2 = 0 + 0.776 327 337 172 638 105 6;
  • 43) 0.776 327 337 172 638 105 6 × 2 = 1 + 0.552 654 674 345 276 211 2;
  • 44) 0.552 654 674 345 276 211 2 × 2 = 1 + 0.105 309 348 690 552 422 4;
  • 45) 0.105 309 348 690 552 422 4 × 2 = 0 + 0.210 618 697 381 104 844 8;
  • 46) 0.210 618 697 381 104 844 8 × 2 = 0 + 0.421 237 394 762 209 689 6;
  • 47) 0.421 237 394 762 209 689 6 × 2 = 0 + 0.842 474 789 524 419 379 2;
  • 48) 0.842 474 789 524 419 379 2 × 2 = 1 + 0.684 949 579 048 838 758 4;
  • 49) 0.684 949 579 048 838 758 4 × 2 = 1 + 0.369 899 158 097 677 516 8;
  • 50) 0.369 899 158 097 677 516 8 × 2 = 0 + 0.739 798 316 195 355 033 6;
  • 51) 0.739 798 316 195 355 033 6 × 2 = 1 + 0.479 596 632 390 710 067 2;
  • 52) 0.479 596 632 390 710 067 2 × 2 = 0 + 0.959 193 264 781 420 134 4;
  • 53) 0.959 193 264 781 420 134 4 × 2 = 1 + 0.918 386 529 562 840 268 8;
  • 54) 0.918 386 529 562 840 268 8 × 2 = 1 + 0.836 773 059 125 680 537 6;
  • 55) 0.836 773 059 125 680 537 6 × 2 = 1 + 0.673 546 118 251 361 075 2;
  • 56) 0.673 546 118 251 361 075 2 × 2 = 1 + 0.347 092 236 502 722 150 4;
  • 57) 0.347 092 236 502 722 150 4 × 2 = 0 + 0.694 184 473 005 444 300 8;
  • 58) 0.694 184 473 005 444 300 8 × 2 = 1 + 0.388 368 946 010 888 601 6;
  • 59) 0.388 368 946 010 888 601 6 × 2 = 0 + 0.776 737 892 021 777 203 2;
  • 60) 0.776 737 892 021 777 203 2 × 2 = 1 + 0.553 475 784 043 554 406 4;
  • 61) 0.553 475 784 043 554 406 4 × 2 = 1 + 0.106 951 568 087 108 812 8;
  • 62) 0.106 951 568 087 108 812 8 × 2 = 0 + 0.213 903 136 174 217 625 6;
  • 63) 0.213 903 136 174 217 625 6 × 2 = 0 + 0.427 806 272 348 435 251 2;
  • 64) 0.427 806 272 348 435 251 2 × 2 = 0 + 0.855 612 544 696 870 502 4;
  • 65) 0.855 612 544 696 870 502 4 × 2 = 1 + 0.711 225 089 393 741 004 8;
  • 66) 0.711 225 089 393 741 004 8 × 2 = 1 + 0.422 450 178 787 482 009 6;
  • 67) 0.422 450 178 787 482 009 6 × 2 = 0 + 0.844 900 357 574 964 019 2;
  • 68) 0.844 900 357 574 964 019 2 × 2 = 1 + 0.689 800 715 149 928 038 4;
  • 69) 0.689 800 715 149 928 038 4 × 2 = 1 + 0.379 601 430 299 856 076 8;
  • 70) 0.379 601 430 299 856 076 8 × 2 = 0 + 0.759 202 860 599 712 153 6;
  • 71) 0.759 202 860 599 712 153 6 × 2 = 1 + 0.518 405 721 199 424 307 2;
  • 72) 0.518 405 721 199 424 307 2 × 2 = 1 + 0.036 811 442 398 848 614 4;
  • 73) 0.036 811 442 398 848 614 4 × 2 = 0 + 0.073 622 884 797 697 228 8;
  • 74) 0.073 622 884 797 697 228 8 × 2 = 0 + 0.147 245 769 595 394 457 6;
  • 75) 0.147 245 769 595 394 457 6 × 2 = 0 + 0.294 491 539 190 788 915 2;
  • 76) 0.294 491 539 190 788 915 2 × 2 = 0 + 0.588 983 078 381 577 830 4;
  • 77) 0.588 983 078 381 577 830 4 × 2 = 1 + 0.177 966 156 763 155 660 8;
  • 78) 0.177 966 156 763 155 660 8 × 2 = 0 + 0.355 932 313 526 311 321 6;
  • 79) 0.355 932 313 526 311 321 6 × 2 = 0 + 0.711 864 627 052 622 643 2;
  • 80) 0.711 864 627 052 622 643 2 × 2 = 1 + 0.423 729 254 105 245 286 4;
  • 81) 0.423 729 254 105 245 286 4 × 2 = 0 + 0.847 458 508 210 490 572 8;
  • 82) 0.847 458 508 210 490 572 8 × 2 = 1 + 0.694 917 016 420 981 145 6;
  • 83) 0.694 917 016 420 981 145 6 × 2 = 1 + 0.389 834 032 841 962 291 2;
  • 84) 0.389 834 032 841 962 291 2 × 2 = 0 + 0.779 668 065 683 924 582 4;
  • 85) 0.779 668 065 683 924 582 4 × 2 = 1 + 0.559 336 131 367 849 164 8;
  • 86) 0.559 336 131 367 849 164 8 × 2 = 1 + 0.118 672 262 735 698 329 6;
  • 87) 0.118 672 262 735 698 329 6 × 2 = 0 + 0.237 344 525 471 396 659 2;
  • 88) 0.237 344 525 471 396 659 2 × 2 = 0 + 0.474 689 050 942 793 318 4;
  • 89) 0.474 689 050 942 793 318 4 × 2 = 0 + 0.949 378 101 885 586 636 8;
  • 90) 0.949 378 101 885 586 636 8 × 2 = 1 + 0.898 756 203 771 173 273 6;
  • 91) 0.898 756 203 771 173 273 6 × 2 = 1 + 0.797 512 407 542 346 547 2;
  • 92) 0.797 512 407 542 346 547 2 × 2 = 1 + 0.595 024 815 084 693 094 4;
  • 93) 0.595 024 815 084 693 094 4 × 2 = 1 + 0.190 049 630 169 386 188 8;
  • 94) 0.190 049 630 169 386 188 8 × 2 = 0 + 0.380 099 260 338 772 377 6;
  • 95) 0.380 099 260 338 772 377 6 × 2 = 0 + 0.760 198 520 677 544 755 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:


0.000 000 000 000 176 516 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1111 0101 1000 1101 1011 0000 1001 0110 1100 0111 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 516 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1111 0101 1000 1101 1011 0000 1001 0110 1100 0111 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:


0.000 000 000 000 176 516 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1111 0101 1000 1101 1011 0000 1001 0110 1100 0111 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1111 0101 1000 1101 1011 0000 1001 0110 1100 0111 100(2) × 20 =

1.1000 1101 0111 1010 1100 0110 1101 1000 0100 1011 0110 0011 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 0111 1010 1100 0110 1101 1000 0100 1011 0110 0011 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 0111 1010 1100 0110 1101 1000 0100 1011 0110 0011 1100 =

1000 1101 0111 1010 1100 0110 1101 1000 0100 1011 0110 0011 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 0111 1010 1100 0110 1101 1000 0100 1011 0110 0011 1100


Decimal number -0.000 000 000 000 176 516 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 0111 1010 1100 0110 1101 1000 0100 1011 0110 0011 1100

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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