-0.000 000 000 000 176 570 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 570 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
-0.000 000 000 000 176 570 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:

|-0.000 000 000 000 176 570 4| = 0.000 000 000 000 176 570 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: 0.000 000 000 000 176 570 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) 0.000 000 000 000 176 570 4 × 2 = 0 + 0.000 000 000 000 353 140 8;
  • 2) 0.000 000 000 000 353 140 8 × 2 = 0 + 0.000 000 000 000 706 281 6;
  • 3) 0.000 000 000 000 706 281 6 × 2 = 0 + 0.000 000 000 001 412 563 2;
  • 4) 0.000 000 000 001 412 563 2 × 2 = 0 + 0.000 000 000 002 825 126 4;
  • 5) 0.000 000 000 002 825 126 4 × 2 = 0 + 0.000 000 000 005 650 252 8;
  • 6) 0.000 000 000 005 650 252 8 × 2 = 0 + 0.000 000 000 011 300 505 6;
  • 7) 0.000 000 000 011 300 505 6 × 2 = 0 + 0.000 000 000 022 601 011 2;
  • 8) 0.000 000 000 022 601 011 2 × 2 = 0 + 0.000 000 000 045 202 022 4;
  • 9) 0.000 000 000 045 202 022 4 × 2 = 0 + 0.000 000 000 090 404 044 8;
  • 10) 0.000 000 000 090 404 044 8 × 2 = 0 + 0.000 000 000 180 808 089 6;
  • 11) 0.000 000 000 180 808 089 6 × 2 = 0 + 0.000 000 000 361 616 179 2;
  • 12) 0.000 000 000 361 616 179 2 × 2 = 0 + 0.000 000 000 723 232 358 4;
  • 13) 0.000 000 000 723 232 358 4 × 2 = 0 + 0.000 000 001 446 464 716 8;
  • 14) 0.000 000 001 446 464 716 8 × 2 = 0 + 0.000 000 002 892 929 433 6;
  • 15) 0.000 000 002 892 929 433 6 × 2 = 0 + 0.000 000 005 785 858 867 2;
  • 16) 0.000 000 005 785 858 867 2 × 2 = 0 + 0.000 000 011 571 717 734 4;
  • 17) 0.000 000 011 571 717 734 4 × 2 = 0 + 0.000 000 023 143 435 468 8;
  • 18) 0.000 000 023 143 435 468 8 × 2 = 0 + 0.000 000 046 286 870 937 6;
  • 19) 0.000 000 046 286 870 937 6 × 2 = 0 + 0.000 000 092 573 741 875 2;
  • 20) 0.000 000 092 573 741 875 2 × 2 = 0 + 0.000 000 185 147 483 750 4;
  • 21) 0.000 000 185 147 483 750 4 × 2 = 0 + 0.000 000 370 294 967 500 8;
  • 22) 0.000 000 370 294 967 500 8 × 2 = 0 + 0.000 000 740 589 935 001 6;
  • 23) 0.000 000 740 589 935 001 6 × 2 = 0 + 0.000 001 481 179 870 003 2;
  • 24) 0.000 001 481 179 870 003 2 × 2 = 0 + 0.000 002 962 359 740 006 4;
  • 25) 0.000 002 962 359 740 006 4 × 2 = 0 + 0.000 005 924 719 480 012 8;
  • 26) 0.000 005 924 719 480 012 8 × 2 = 0 + 0.000 011 849 438 960 025 6;
  • 27) 0.000 011 849 438 960 025 6 × 2 = 0 + 0.000 023 698 877 920 051 2;
  • 28) 0.000 023 698 877 920 051 2 × 2 = 0 + 0.000 047 397 755 840 102 4;
  • 29) 0.000 047 397 755 840 102 4 × 2 = 0 + 0.000 094 795 511 680 204 8;
  • 30) 0.000 094 795 511 680 204 8 × 2 = 0 + 0.000 189 591 023 360 409 6;
  • 31) 0.000 189 591 023 360 409 6 × 2 = 0 + 0.000 379 182 046 720 819 2;
  • 32) 0.000 379 182 046 720 819 2 × 2 = 0 + 0.000 758 364 093 441 638 4;
  • 33) 0.000 758 364 093 441 638 4 × 2 = 0 + 0.001 516 728 186 883 276 8;
  • 34) 0.001 516 728 186 883 276 8 × 2 = 0 + 0.003 033 456 373 766 553 6;
  • 35) 0.003 033 456 373 766 553 6 × 2 = 0 + 0.006 066 912 747 533 107 2;
  • 36) 0.006 066 912 747 533 107 2 × 2 = 0 + 0.012 133 825 495 066 214 4;
  • 37) 0.012 133 825 495 066 214 4 × 2 = 0 + 0.024 267 650 990 132 428 8;
  • 38) 0.024 267 650 990 132 428 8 × 2 = 0 + 0.048 535 301 980 264 857 6;
  • 39) 0.048 535 301 980 264 857 6 × 2 = 0 + 0.097 070 603 960 529 715 2;
  • 40) 0.097 070 603 960 529 715 2 × 2 = 0 + 0.194 141 207 921 059 430 4;
  • 41) 0.194 141 207 921 059 430 4 × 2 = 0 + 0.388 282 415 842 118 860 8;
  • 42) 0.388 282 415 842 118 860 8 × 2 = 0 + 0.776 564 831 684 237 721 6;
  • 43) 0.776 564 831 684 237 721 6 × 2 = 1 + 0.553 129 663 368 475 443 2;
  • 44) 0.553 129 663 368 475 443 2 × 2 = 1 + 0.106 259 326 736 950 886 4;
  • 45) 0.106 259 326 736 950 886 4 × 2 = 0 + 0.212 518 653 473 901 772 8;
  • 46) 0.212 518 653 473 901 772 8 × 2 = 0 + 0.425 037 306 947 803 545 6;
  • 47) 0.425 037 306 947 803 545 6 × 2 = 0 + 0.850 074 613 895 607 091 2;
  • 48) 0.850 074 613 895 607 091 2 × 2 = 1 + 0.700 149 227 791 214 182 4;
  • 49) 0.700 149 227 791 214 182 4 × 2 = 1 + 0.400 298 455 582 428 364 8;
  • 50) 0.400 298 455 582 428 364 8 × 2 = 0 + 0.800 596 911 164 856 729 6;
  • 51) 0.800 596 911 164 856 729 6 × 2 = 1 + 0.601 193 822 329 713 459 2;
  • 52) 0.601 193 822 329 713 459 2 × 2 = 1 + 0.202 387 644 659 426 918 4;
  • 53) 0.202 387 644 659 426 918 4 × 2 = 0 + 0.404 775 289 318 853 836 8;
  • 54) 0.404 775 289 318 853 836 8 × 2 = 0 + 0.809 550 578 637 707 673 6;
  • 55) 0.809 550 578 637 707 673 6 × 2 = 1 + 0.619 101 157 275 415 347 2;
  • 56) 0.619 101 157 275 415 347 2 × 2 = 1 + 0.238 202 314 550 830 694 4;
  • 57) 0.238 202 314 550 830 694 4 × 2 = 0 + 0.476 404 629 101 661 388 8;
  • 58) 0.476 404 629 101 661 388 8 × 2 = 0 + 0.952 809 258 203 322 777 6;
  • 59) 0.952 809 258 203 322 777 6 × 2 = 1 + 0.905 618 516 406 645 555 2;
  • 60) 0.905 618 516 406 645 555 2 × 2 = 1 + 0.811 237 032 813 291 110 4;
  • 61) 0.811 237 032 813 291 110 4 × 2 = 1 + 0.622 474 065 626 582 220 8;
  • 62) 0.622 474 065 626 582 220 8 × 2 = 1 + 0.244 948 131 253 164 441 6;
  • 63) 0.244 948 131 253 164 441 6 × 2 = 0 + 0.489 896 262 506 328 883 2;
  • 64) 0.489 896 262 506 328 883 2 × 2 = 0 + 0.979 792 525 012 657 766 4;
  • 65) 0.979 792 525 012 657 766 4 × 2 = 1 + 0.959 585 050 025 315 532 8;
  • 66) 0.959 585 050 025 315 532 8 × 2 = 1 + 0.919 170 100 050 631 065 6;
  • 67) 0.919 170 100 050 631 065 6 × 2 = 1 + 0.838 340 200 101 262 131 2;
  • 68) 0.838 340 200 101 262 131 2 × 2 = 1 + 0.676 680 400 202 524 262 4;
  • 69) 0.676 680 400 202 524 262 4 × 2 = 1 + 0.353 360 800 405 048 524 8;
  • 70) 0.353 360 800 405 048 524 8 × 2 = 0 + 0.706 721 600 810 097 049 6;
  • 71) 0.706 721 600 810 097 049 6 × 2 = 1 + 0.413 443 201 620 194 099 2;
  • 72) 0.413 443 201 620 194 099 2 × 2 = 0 + 0.826 886 403 240 388 198 4;
  • 73) 0.826 886 403 240 388 198 4 × 2 = 1 + 0.653 772 806 480 776 396 8;
  • 74) 0.653 772 806 480 776 396 8 × 2 = 1 + 0.307 545 612 961 552 793 6;
  • 75) 0.307 545 612 961 552 793 6 × 2 = 0 + 0.615 091 225 923 105 587 2;
  • 76) 0.615 091 225 923 105 587 2 × 2 = 1 + 0.230 182 451 846 211 174 4;
  • 77) 0.230 182 451 846 211 174 4 × 2 = 0 + 0.460 364 903 692 422 348 8;
  • 78) 0.460 364 903 692 422 348 8 × 2 = 0 + 0.920 729 807 384 844 697 6;
  • 79) 0.920 729 807 384 844 697 6 × 2 = 1 + 0.841 459 614 769 689 395 2;
  • 80) 0.841 459 614 769 689 395 2 × 2 = 1 + 0.682 919 229 539 378 790 4;
  • 81) 0.682 919 229 539 378 790 4 × 2 = 1 + 0.365 838 459 078 757 580 8;
  • 82) 0.365 838 459 078 757 580 8 × 2 = 0 + 0.731 676 918 157 515 161 6;
  • 83) 0.731 676 918 157 515 161 6 × 2 = 1 + 0.463 353 836 315 030 323 2;
  • 84) 0.463 353 836 315 030 323 2 × 2 = 0 + 0.926 707 672 630 060 646 4;
  • 85) 0.926 707 672 630 060 646 4 × 2 = 1 + 0.853 415 345 260 121 292 8;
  • 86) 0.853 415 345 260 121 292 8 × 2 = 1 + 0.706 830 690 520 242 585 6;
  • 87) 0.706 830 690 520 242 585 6 × 2 = 1 + 0.413 661 381 040 485 171 2;
  • 88) 0.413 661 381 040 485 171 2 × 2 = 0 + 0.827 322 762 080 970 342 4;
  • 89) 0.827 322 762 080 970 342 4 × 2 = 1 + 0.654 645 524 161 940 684 8;
  • 90) 0.654 645 524 161 940 684 8 × 2 = 1 + 0.309 291 048 323 881 369 6;
  • 91) 0.309 291 048 323 881 369 6 × 2 = 0 + 0.618 582 096 647 762 739 2;
  • 92) 0.618 582 096 647 762 739 2 × 2 = 1 + 0.237 164 193 295 525 478 4;
  • 93) 0.237 164 193 295 525 478 4 × 2 = 0 + 0.474 328 386 591 050 956 8;
  • 94) 0.474 328 386 591 050 956 8 × 2 = 0 + 0.948 656 773 182 101 913 6;
  • 95) 0.948 656 773 182 101 913 6 × 2 = 1 + 0.897 313 546 364 203 827 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:


0.000 000 000 000 176 570 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0011 1100 1111 1010 1101 0011 1010 1110 1101 001(2)

6. Positive number before normalization:

0.000 000 000 000 176 570 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0011 1100 1111 1010 1101 0011 1010 1110 1101 001(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:


0.000 000 000 000 176 570 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0011 1100 1111 1010 1101 0011 1010 1110 1101 001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0011 0011 1100 1111 1010 1101 0011 1010 1110 1101 001(2) × 20 =

1.1000 1101 1001 1001 1110 0111 1101 0110 1001 1101 0111 0110 1001(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 1001 1110 0111 1101 0110 1001 1101 0111 0110 1001


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 1001 1110 0111 1101 0110 1001 1101 0111 0110 1001 =

1000 1101 1001 1001 1110 0111 1101 0110 1001 1101 0111 0110 1001


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 1001 1110 0111 1101 0110 1001 1101 0111 0110 1001


Decimal number -0.000 000 000 000 176 570 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 1001 1110 0111 1101 0110 1001 1101 0111 0110 1001

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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