-0.000 000 000 000 000 000 67 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 000 67(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
-0.000 000 000 000 000 000 67(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:

|-0.000 000 000 000 000 000 67| = 0.000 000 000 000 000 000 67


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: 0.000 000 000 000 000 000 67.

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) 0.000 000 000 000 000 000 67 × 2 = 0 + 0.000 000 000 000 000 001 34;
  • 2) 0.000 000 000 000 000 001 34 × 2 = 0 + 0.000 000 000 000 000 002 68;
  • 3) 0.000 000 000 000 000 002 68 × 2 = 0 + 0.000 000 000 000 000 005 36;
  • 4) 0.000 000 000 000 000 005 36 × 2 = 0 + 0.000 000 000 000 000 010 72;
  • 5) 0.000 000 000 000 000 010 72 × 2 = 0 + 0.000 000 000 000 000 021 44;
  • 6) 0.000 000 000 000 000 021 44 × 2 = 0 + 0.000 000 000 000 000 042 88;
  • 7) 0.000 000 000 000 000 042 88 × 2 = 0 + 0.000 000 000 000 000 085 76;
  • 8) 0.000 000 000 000 000 085 76 × 2 = 0 + 0.000 000 000 000 000 171 52;
  • 9) 0.000 000 000 000 000 171 52 × 2 = 0 + 0.000 000 000 000 000 343 04;
  • 10) 0.000 000 000 000 000 343 04 × 2 = 0 + 0.000 000 000 000 000 686 08;
  • 11) 0.000 000 000 000 000 686 08 × 2 = 0 + 0.000 000 000 000 001 372 16;
  • 12) 0.000 000 000 000 001 372 16 × 2 = 0 + 0.000 000 000 000 002 744 32;
  • 13) 0.000 000 000 000 002 744 32 × 2 = 0 + 0.000 000 000 000 005 488 64;
  • 14) 0.000 000 000 000 005 488 64 × 2 = 0 + 0.000 000 000 000 010 977 28;
  • 15) 0.000 000 000 000 010 977 28 × 2 = 0 + 0.000 000 000 000 021 954 56;
  • 16) 0.000 000 000 000 021 954 56 × 2 = 0 + 0.000 000 000 000 043 909 12;
  • 17) 0.000 000 000 000 043 909 12 × 2 = 0 + 0.000 000 000 000 087 818 24;
  • 18) 0.000 000 000 000 087 818 24 × 2 = 0 + 0.000 000 000 000 175 636 48;
  • 19) 0.000 000 000 000 175 636 48 × 2 = 0 + 0.000 000 000 000 351 272 96;
  • 20) 0.000 000 000 000 351 272 96 × 2 = 0 + 0.000 000 000 000 702 545 92;
  • 21) 0.000 000 000 000 702 545 92 × 2 = 0 + 0.000 000 000 001 405 091 84;
  • 22) 0.000 000 000 001 405 091 84 × 2 = 0 + 0.000 000 000 002 810 183 68;
  • 23) 0.000 000 000 002 810 183 68 × 2 = 0 + 0.000 000 000 005 620 367 36;
  • 24) 0.000 000 000 005 620 367 36 × 2 = 0 + 0.000 000 000 011 240 734 72;
  • 25) 0.000 000 000 011 240 734 72 × 2 = 0 + 0.000 000 000 022 481 469 44;
  • 26) 0.000 000 000 022 481 469 44 × 2 = 0 + 0.000 000 000 044 962 938 88;
  • 27) 0.000 000 000 044 962 938 88 × 2 = 0 + 0.000 000 000 089 925 877 76;
  • 28) 0.000 000 000 089 925 877 76 × 2 = 0 + 0.000 000 000 179 851 755 52;
  • 29) 0.000 000 000 179 851 755 52 × 2 = 0 + 0.000 000 000 359 703 511 04;
  • 30) 0.000 000 000 359 703 511 04 × 2 = 0 + 0.000 000 000 719 407 022 08;
  • 31) 0.000 000 000 719 407 022 08 × 2 = 0 + 0.000 000 001 438 814 044 16;
  • 32) 0.000 000 001 438 814 044 16 × 2 = 0 + 0.000 000 002 877 628 088 32;
  • 33) 0.000 000 002 877 628 088 32 × 2 = 0 + 0.000 000 005 755 256 176 64;
  • 34) 0.000 000 005 755 256 176 64 × 2 = 0 + 0.000 000 011 510 512 353 28;
  • 35) 0.000 000 011 510 512 353 28 × 2 = 0 + 0.000 000 023 021 024 706 56;
  • 36) 0.000 000 023 021 024 706 56 × 2 = 0 + 0.000 000 046 042 049 413 12;
  • 37) 0.000 000 046 042 049 413 12 × 2 = 0 + 0.000 000 092 084 098 826 24;
  • 38) 0.000 000 092 084 098 826 24 × 2 = 0 + 0.000 000 184 168 197 652 48;
  • 39) 0.000 000 184 168 197 652 48 × 2 = 0 + 0.000 000 368 336 395 304 96;
  • 40) 0.000 000 368 336 395 304 96 × 2 = 0 + 0.000 000 736 672 790 609 92;
  • 41) 0.000 000 736 672 790 609 92 × 2 = 0 + 0.000 001 473 345 581 219 84;
  • 42) 0.000 001 473 345 581 219 84 × 2 = 0 + 0.000 002 946 691 162 439 68;
  • 43) 0.000 002 946 691 162 439 68 × 2 = 0 + 0.000 005 893 382 324 879 36;
  • 44) 0.000 005 893 382 324 879 36 × 2 = 0 + 0.000 011 786 764 649 758 72;
  • 45) 0.000 011 786 764 649 758 72 × 2 = 0 + 0.000 023 573 529 299 517 44;
  • 46) 0.000 023 573 529 299 517 44 × 2 = 0 + 0.000 047 147 058 599 034 88;
  • 47) 0.000 047 147 058 599 034 88 × 2 = 0 + 0.000 094 294 117 198 069 76;
  • 48) 0.000 094 294 117 198 069 76 × 2 = 0 + 0.000 188 588 234 396 139 52;
  • 49) 0.000 188 588 234 396 139 52 × 2 = 0 + 0.000 377 176 468 792 279 04;
  • 50) 0.000 377 176 468 792 279 04 × 2 = 0 + 0.000 754 352 937 584 558 08;
  • 51) 0.000 754 352 937 584 558 08 × 2 = 0 + 0.001 508 705 875 169 116 16;
  • 52) 0.001 508 705 875 169 116 16 × 2 = 0 + 0.003 017 411 750 338 232 32;
  • 53) 0.003 017 411 750 338 232 32 × 2 = 0 + 0.006 034 823 500 676 464 64;
  • 54) 0.006 034 823 500 676 464 64 × 2 = 0 + 0.012 069 647 001 352 929 28;
  • 55) 0.012 069 647 001 352 929 28 × 2 = 0 + 0.024 139 294 002 705 858 56;
  • 56) 0.024 139 294 002 705 858 56 × 2 = 0 + 0.048 278 588 005 411 717 12;
  • 57) 0.048 278 588 005 411 717 12 × 2 = 0 + 0.096 557 176 010 823 434 24;
  • 58) 0.096 557 176 010 823 434 24 × 2 = 0 + 0.193 114 352 021 646 868 48;
  • 59) 0.193 114 352 021 646 868 48 × 2 = 0 + 0.386 228 704 043 293 736 96;
  • 60) 0.386 228 704 043 293 736 96 × 2 = 0 + 0.772 457 408 086 587 473 92;
  • 61) 0.772 457 408 086 587 473 92 × 2 = 1 + 0.544 914 816 173 174 947 84;
  • 62) 0.544 914 816 173 174 947 84 × 2 = 1 + 0.089 829 632 346 349 895 68;
  • 63) 0.089 829 632 346 349 895 68 × 2 = 0 + 0.179 659 264 692 699 791 36;
  • 64) 0.179 659 264 692 699 791 36 × 2 = 0 + 0.359 318 529 385 399 582 72;
  • 65) 0.359 318 529 385 399 582 72 × 2 = 0 + 0.718 637 058 770 799 165 44;
  • 66) 0.718 637 058 770 799 165 44 × 2 = 1 + 0.437 274 117 541 598 330 88;
  • 67) 0.437 274 117 541 598 330 88 × 2 = 0 + 0.874 548 235 083 196 661 76;
  • 68) 0.874 548 235 083 196 661 76 × 2 = 1 + 0.749 096 470 166 393 323 52;
  • 69) 0.749 096 470 166 393 323 52 × 2 = 1 + 0.498 192 940 332 786 647 04;
  • 70) 0.498 192 940 332 786 647 04 × 2 = 0 + 0.996 385 880 665 573 294 08;
  • 71) 0.996 385 880 665 573 294 08 × 2 = 1 + 0.992 771 761 331 146 588 16;
  • 72) 0.992 771 761 331 146 588 16 × 2 = 1 + 0.985 543 522 662 293 176 32;
  • 73) 0.985 543 522 662 293 176 32 × 2 = 1 + 0.971 087 045 324 586 352 64;
  • 74) 0.971 087 045 324 586 352 64 × 2 = 1 + 0.942 174 090 649 172 705 28;
  • 75) 0.942 174 090 649 172 705 28 × 2 = 1 + 0.884 348 181 298 345 410 56;
  • 76) 0.884 348 181 298 345 410 56 × 2 = 1 + 0.768 696 362 596 690 821 12;
  • 77) 0.768 696 362 596 690 821 12 × 2 = 1 + 0.537 392 725 193 381 642 24;
  • 78) 0.537 392 725 193 381 642 24 × 2 = 1 + 0.074 785 450 386 763 284 48;
  • 79) 0.074 785 450 386 763 284 48 × 2 = 0 + 0.149 570 900 773 526 568 96;
  • 80) 0.149 570 900 773 526 568 96 × 2 = 0 + 0.299 141 801 547 053 137 92;
  • 81) 0.299 141 801 547 053 137 92 × 2 = 0 + 0.598 283 603 094 106 275 84;
  • 82) 0.598 283 603 094 106 275 84 × 2 = 1 + 0.196 567 206 188 212 551 68;
  • 83) 0.196 567 206 188 212 551 68 × 2 = 0 + 0.393 134 412 376 425 103 36;
  • 84) 0.393 134 412 376 425 103 36 × 2 = 0 + 0.786 268 824 752 850 206 72;
  • 85) 0.786 268 824 752 850 206 72 × 2 = 1 + 0.572 537 649 505 700 413 44;
  • 86) 0.572 537 649 505 700 413 44 × 2 = 1 + 0.145 075 299 011 400 826 88;
  • 87) 0.145 075 299 011 400 826 88 × 2 = 0 + 0.290 150 598 022 801 653 76;
  • 88) 0.290 150 598 022 801 653 76 × 2 = 0 + 0.580 301 196 045 603 307 52;
  • 89) 0.580 301 196 045 603 307 52 × 2 = 1 + 0.160 602 392 091 206 615 04;
  • 90) 0.160 602 392 091 206 615 04 × 2 = 0 + 0.321 204 784 182 413 230 08;
  • 91) 0.321 204 784 182 413 230 08 × 2 = 0 + 0.642 409 568 364 826 460 16;
  • 92) 0.642 409 568 364 826 460 16 × 2 = 1 + 0.284 819 136 729 652 920 32;
  • 93) 0.284 819 136 729 652 920 32 × 2 = 0 + 0.569 638 273 459 305 840 64;
  • 94) 0.569 638 273 459 305 840 64 × 2 = 1 + 0.139 276 546 918 611 681 28;
  • 95) 0.139 276 546 918 611 681 28 × 2 = 0 + 0.278 553 093 837 223 362 56;
  • 96) 0.278 553 093 837 223 362 56 × 2 = 0 + 0.557 106 187 674 446 725 12;
  • 97) 0.557 106 187 674 446 725 12 × 2 = 1 + 0.114 212 375 348 893 450 24;
  • 98) 0.114 212 375 348 893 450 24 × 2 = 0 + 0.228 424 750 697 786 900 48;
  • 99) 0.228 424 750 697 786 900 48 × 2 = 0 + 0.456 849 501 395 573 800 96;
  • 100) 0.456 849 501 395 573 800 96 × 2 = 0 + 0.913 699 002 791 147 601 92;
  • 101) 0.913 699 002 791 147 601 92 × 2 = 1 + 0.827 398 005 582 295 203 84;
  • 102) 0.827 398 005 582 295 203 84 × 2 = 1 + 0.654 796 011 164 590 407 68;
  • 103) 0.654 796 011 164 590 407 68 × 2 = 1 + 0.309 592 022 329 180 815 36;
  • 104) 0.309 592 022 329 180 815 36 × 2 = 0 + 0.619 184 044 658 361 630 72;
  • 105) 0.619 184 044 658 361 630 72 × 2 = 1 + 0.238 368 089 316 723 261 44;
  • 106) 0.238 368 089 316 723 261 44 × 2 = 0 + 0.476 736 178 633 446 522 88;
  • 107) 0.476 736 178 633 446 522 88 × 2 = 0 + 0.953 472 357 266 893 045 76;
  • 108) 0.953 472 357 266 893 045 76 × 2 = 1 + 0.906 944 714 533 786 091 52;
  • 109) 0.906 944 714 533 786 091 52 × 2 = 1 + 0.813 889 429 067 572 183 04;
  • 110) 0.813 889 429 067 572 183 04 × 2 = 1 + 0.627 778 858 135 144 366 08;
  • 111) 0.627 778 858 135 144 366 08 × 2 = 1 + 0.255 557 716 270 288 732 16;
  • 112) 0.255 557 716 270 288 732 16 × 2 = 0 + 0.511 115 432 540 577 464 32;
  • 113) 0.511 115 432 540 577 464 32 × 2 = 1 + 0.022 230 865 081 154 928 64;

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:


0.000 000 000 000 000 000 67(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0101 1011 1111 1100 0100 1100 1001 0100 1000 1110 1001 1110 1(2)

6. Positive number before normalization:

0.000 000 000 000 000 000 67(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0101 1011 1111 1100 0100 1100 1001 0100 1000 1110 1001 1110 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 61 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 67(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0101 1011 1111 1100 0100 1100 1001 0100 1000 1110 1001 1110 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0101 1011 1111 1100 0100 1100 1001 0100 1000 1110 1001 1110 1(2) × 20 =

1.1000 1011 0111 1111 1000 1001 1001 0010 1001 0001 1101 0011 1101(2) × 2-61


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): -61

Mantissa (not normalized):
1.1000 1011 0111 1111 1000 1001 1001 0010 1001 0001 1101 0011 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-61 + 2(11-1) - 1 =

(-61 + 1 023)(10) =

962(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;
  • 962 ÷ 2 = 481 + 0;
  • 481 ÷ 2 = 240 + 1;
  • 240 ÷ 2 = 120 + 0;
  • 120 ÷ 2 = 60 + 0;
  • 60 ÷ 2 = 30 + 0;
  • 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) =

962(10) =

011 1100 0010(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 1011 0111 1111 1000 1001 1001 0010 1001 0001 1101 0011 1101 =

1000 1011 0111 1111 1000 1001 1001 0010 1001 0001 1101 0011 1101


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 1100 0010

Mantissa (52 bits) =
1000 1011 0111 1111 1000 1001 1001 0010 1001 0001 1101 0011 1101


Decimal number -0.000 000 000 000 000 000 67 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1100 0010 - 1000 1011 0111 1111 1000 1001 1001 0010 1001 0001 1101 0011 1101

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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