-0.000 000 000 000 000 001 12 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 001 12(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
-0.000 000 000 000 000 001 12(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:

|-0.000 000 000 000 000 001 12| = 0.000 000 000 000 000 001 12


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: 0.000 000 000 000 000 001 12.

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) 0.000 000 000 000 000 001 12 × 2 = 0 + 0.000 000 000 000 000 002 24;
  • 2) 0.000 000 000 000 000 002 24 × 2 = 0 + 0.000 000 000 000 000 004 48;
  • 3) 0.000 000 000 000 000 004 48 × 2 = 0 + 0.000 000 000 000 000 008 96;
  • 4) 0.000 000 000 000 000 008 96 × 2 = 0 + 0.000 000 000 000 000 017 92;
  • 5) 0.000 000 000 000 000 017 92 × 2 = 0 + 0.000 000 000 000 000 035 84;
  • 6) 0.000 000 000 000 000 035 84 × 2 = 0 + 0.000 000 000 000 000 071 68;
  • 7) 0.000 000 000 000 000 071 68 × 2 = 0 + 0.000 000 000 000 000 143 36;
  • 8) 0.000 000 000 000 000 143 36 × 2 = 0 + 0.000 000 000 000 000 286 72;
  • 9) 0.000 000 000 000 000 286 72 × 2 = 0 + 0.000 000 000 000 000 573 44;
  • 10) 0.000 000 000 000 000 573 44 × 2 = 0 + 0.000 000 000 000 001 146 88;
  • 11) 0.000 000 000 000 001 146 88 × 2 = 0 + 0.000 000 000 000 002 293 76;
  • 12) 0.000 000 000 000 002 293 76 × 2 = 0 + 0.000 000 000 000 004 587 52;
  • 13) 0.000 000 000 000 004 587 52 × 2 = 0 + 0.000 000 000 000 009 175 04;
  • 14) 0.000 000 000 000 009 175 04 × 2 = 0 + 0.000 000 000 000 018 350 08;
  • 15) 0.000 000 000 000 018 350 08 × 2 = 0 + 0.000 000 000 000 036 700 16;
  • 16) 0.000 000 000 000 036 700 16 × 2 = 0 + 0.000 000 000 000 073 400 32;
  • 17) 0.000 000 000 000 073 400 32 × 2 = 0 + 0.000 000 000 000 146 800 64;
  • 18) 0.000 000 000 000 146 800 64 × 2 = 0 + 0.000 000 000 000 293 601 28;
  • 19) 0.000 000 000 000 293 601 28 × 2 = 0 + 0.000 000 000 000 587 202 56;
  • 20) 0.000 000 000 000 587 202 56 × 2 = 0 + 0.000 000 000 001 174 405 12;
  • 21) 0.000 000 000 001 174 405 12 × 2 = 0 + 0.000 000 000 002 348 810 24;
  • 22) 0.000 000 000 002 348 810 24 × 2 = 0 + 0.000 000 000 004 697 620 48;
  • 23) 0.000 000 000 004 697 620 48 × 2 = 0 + 0.000 000 000 009 395 240 96;
  • 24) 0.000 000 000 009 395 240 96 × 2 = 0 + 0.000 000 000 018 790 481 92;
  • 25) 0.000 000 000 018 790 481 92 × 2 = 0 + 0.000 000 000 037 580 963 84;
  • 26) 0.000 000 000 037 580 963 84 × 2 = 0 + 0.000 000 000 075 161 927 68;
  • 27) 0.000 000 000 075 161 927 68 × 2 = 0 + 0.000 000 000 150 323 855 36;
  • 28) 0.000 000 000 150 323 855 36 × 2 = 0 + 0.000 000 000 300 647 710 72;
  • 29) 0.000 000 000 300 647 710 72 × 2 = 0 + 0.000 000 000 601 295 421 44;
  • 30) 0.000 000 000 601 295 421 44 × 2 = 0 + 0.000 000 001 202 590 842 88;
  • 31) 0.000 000 001 202 590 842 88 × 2 = 0 + 0.000 000 002 405 181 685 76;
  • 32) 0.000 000 002 405 181 685 76 × 2 = 0 + 0.000 000 004 810 363 371 52;
  • 33) 0.000 000 004 810 363 371 52 × 2 = 0 + 0.000 000 009 620 726 743 04;
  • 34) 0.000 000 009 620 726 743 04 × 2 = 0 + 0.000 000 019 241 453 486 08;
  • 35) 0.000 000 019 241 453 486 08 × 2 = 0 + 0.000 000 038 482 906 972 16;
  • 36) 0.000 000 038 482 906 972 16 × 2 = 0 + 0.000 000 076 965 813 944 32;
  • 37) 0.000 000 076 965 813 944 32 × 2 = 0 + 0.000 000 153 931 627 888 64;
  • 38) 0.000 000 153 931 627 888 64 × 2 = 0 + 0.000 000 307 863 255 777 28;
  • 39) 0.000 000 307 863 255 777 28 × 2 = 0 + 0.000 000 615 726 511 554 56;
  • 40) 0.000 000 615 726 511 554 56 × 2 = 0 + 0.000 001 231 453 023 109 12;
  • 41) 0.000 001 231 453 023 109 12 × 2 = 0 + 0.000 002 462 906 046 218 24;
  • 42) 0.000 002 462 906 046 218 24 × 2 = 0 + 0.000 004 925 812 092 436 48;
  • 43) 0.000 004 925 812 092 436 48 × 2 = 0 + 0.000 009 851 624 184 872 96;
  • 44) 0.000 009 851 624 184 872 96 × 2 = 0 + 0.000 019 703 248 369 745 92;
  • 45) 0.000 019 703 248 369 745 92 × 2 = 0 + 0.000 039 406 496 739 491 84;
  • 46) 0.000 039 406 496 739 491 84 × 2 = 0 + 0.000 078 812 993 478 983 68;
  • 47) 0.000 078 812 993 478 983 68 × 2 = 0 + 0.000 157 625 986 957 967 36;
  • 48) 0.000 157 625 986 957 967 36 × 2 = 0 + 0.000 315 251 973 915 934 72;
  • 49) 0.000 315 251 973 915 934 72 × 2 = 0 + 0.000 630 503 947 831 869 44;
  • 50) 0.000 630 503 947 831 869 44 × 2 = 0 + 0.001 261 007 895 663 738 88;
  • 51) 0.001 261 007 895 663 738 88 × 2 = 0 + 0.002 522 015 791 327 477 76;
  • 52) 0.002 522 015 791 327 477 76 × 2 = 0 + 0.005 044 031 582 654 955 52;
  • 53) 0.005 044 031 582 654 955 52 × 2 = 0 + 0.010 088 063 165 309 911 04;
  • 54) 0.010 088 063 165 309 911 04 × 2 = 0 + 0.020 176 126 330 619 822 08;
  • 55) 0.020 176 126 330 619 822 08 × 2 = 0 + 0.040 352 252 661 239 644 16;
  • 56) 0.040 352 252 661 239 644 16 × 2 = 0 + 0.080 704 505 322 479 288 32;
  • 57) 0.080 704 505 322 479 288 32 × 2 = 0 + 0.161 409 010 644 958 576 64;
  • 58) 0.161 409 010 644 958 576 64 × 2 = 0 + 0.322 818 021 289 917 153 28;
  • 59) 0.322 818 021 289 917 153 28 × 2 = 0 + 0.645 636 042 579 834 306 56;
  • 60) 0.645 636 042 579 834 306 56 × 2 = 1 + 0.291 272 085 159 668 613 12;
  • 61) 0.291 272 085 159 668 613 12 × 2 = 0 + 0.582 544 170 319 337 226 24;
  • 62) 0.582 544 170 319 337 226 24 × 2 = 1 + 0.165 088 340 638 674 452 48;
  • 63) 0.165 088 340 638 674 452 48 × 2 = 0 + 0.330 176 681 277 348 904 96;
  • 64) 0.330 176 681 277 348 904 96 × 2 = 0 + 0.660 353 362 554 697 809 92;
  • 65) 0.660 353 362 554 697 809 92 × 2 = 1 + 0.320 706 725 109 395 619 84;
  • 66) 0.320 706 725 109 395 619 84 × 2 = 0 + 0.641 413 450 218 791 239 68;
  • 67) 0.641 413 450 218 791 239 68 × 2 = 1 + 0.282 826 900 437 582 479 36;
  • 68) 0.282 826 900 437 582 479 36 × 2 = 0 + 0.565 653 800 875 164 958 72;
  • 69) 0.565 653 800 875 164 958 72 × 2 = 1 + 0.131 307 601 750 329 917 44;
  • 70) 0.131 307 601 750 329 917 44 × 2 = 0 + 0.262 615 203 500 659 834 88;
  • 71) 0.262 615 203 500 659 834 88 × 2 = 0 + 0.525 230 407 001 319 669 76;
  • 72) 0.525 230 407 001 319 669 76 × 2 = 1 + 0.050 460 814 002 639 339 52;
  • 73) 0.050 460 814 002 639 339 52 × 2 = 0 + 0.100 921 628 005 278 679 04;
  • 74) 0.100 921 628 005 278 679 04 × 2 = 0 + 0.201 843 256 010 557 358 08;
  • 75) 0.201 843 256 010 557 358 08 × 2 = 0 + 0.403 686 512 021 114 716 16;
  • 76) 0.403 686 512 021 114 716 16 × 2 = 0 + 0.807 373 024 042 229 432 32;
  • 77) 0.807 373 024 042 229 432 32 × 2 = 1 + 0.614 746 048 084 458 864 64;
  • 78) 0.614 746 048 084 458 864 64 × 2 = 1 + 0.229 492 096 168 917 729 28;
  • 79) 0.229 492 096 168 917 729 28 × 2 = 0 + 0.458 984 192 337 835 458 56;
  • 80) 0.458 984 192 337 835 458 56 × 2 = 0 + 0.917 968 384 675 670 917 12;
  • 81) 0.917 968 384 675 670 917 12 × 2 = 1 + 0.835 936 769 351 341 834 24;
  • 82) 0.835 936 769 351 341 834 24 × 2 = 1 + 0.671 873 538 702 683 668 48;
  • 83) 0.671 873 538 702 683 668 48 × 2 = 1 + 0.343 747 077 405 367 336 96;
  • 84) 0.343 747 077 405 367 336 96 × 2 = 0 + 0.687 494 154 810 734 673 92;
  • 85) 0.687 494 154 810 734 673 92 × 2 = 1 + 0.374 988 309 621 469 347 84;
  • 86) 0.374 988 309 621 469 347 84 × 2 = 0 + 0.749 976 619 242 938 695 68;
  • 87) 0.749 976 619 242 938 695 68 × 2 = 1 + 0.499 953 238 485 877 391 36;
  • 88) 0.499 953 238 485 877 391 36 × 2 = 0 + 0.999 906 476 971 754 782 72;
  • 89) 0.999 906 476 971 754 782 72 × 2 = 1 + 0.999 812 953 943 509 565 44;
  • 90) 0.999 812 953 943 509 565 44 × 2 = 1 + 0.999 625 907 887 019 130 88;
  • 91) 0.999 625 907 887 019 130 88 × 2 = 1 + 0.999 251 815 774 038 261 76;
  • 92) 0.999 251 815 774 038 261 76 × 2 = 1 + 0.998 503 631 548 076 523 52;
  • 93) 0.998 503 631 548 076 523 52 × 2 = 1 + 0.997 007 263 096 153 047 04;
  • 94) 0.997 007 263 096 153 047 04 × 2 = 1 + 0.994 014 526 192 306 094 08;
  • 95) 0.994 014 526 192 306 094 08 × 2 = 1 + 0.988 029 052 384 612 188 16;
  • 96) 0.988 029 052 384 612 188 16 × 2 = 1 + 0.976 058 104 769 224 376 32;
  • 97) 0.976 058 104 769 224 376 32 × 2 = 1 + 0.952 116 209 538 448 752 64;
  • 98) 0.952 116 209 538 448 752 64 × 2 = 1 + 0.904 232 419 076 897 505 28;
  • 99) 0.904 232 419 076 897 505 28 × 2 = 1 + 0.808 464 838 153 795 010 56;
  • 100) 0.808 464 838 153 795 010 56 × 2 = 1 + 0.616 929 676 307 590 021 12;
  • 101) 0.616 929 676 307 590 021 12 × 2 = 1 + 0.233 859 352 615 180 042 24;
  • 102) 0.233 859 352 615 180 042 24 × 2 = 0 + 0.467 718 705 230 360 084 48;
  • 103) 0.467 718 705 230 360 084 48 × 2 = 0 + 0.935 437 410 460 720 168 96;
  • 104) 0.935 437 410 460 720 168 96 × 2 = 1 + 0.870 874 820 921 440 337 92;
  • 105) 0.870 874 820 921 440 337 92 × 2 = 1 + 0.741 749 641 842 880 675 84;
  • 106) 0.741 749 641 842 880 675 84 × 2 = 1 + 0.483 499 283 685 761 351 68;
  • 107) 0.483 499 283 685 761 351 68 × 2 = 0 + 0.966 998 567 371 522 703 36;
  • 108) 0.966 998 567 371 522 703 36 × 2 = 1 + 0.933 997 134 743 045 406 72;
  • 109) 0.933 997 134 743 045 406 72 × 2 = 1 + 0.867 994 269 486 090 813 44;
  • 110) 0.867 994 269 486 090 813 44 × 2 = 1 + 0.735 988 538 972 181 626 88;
  • 111) 0.735 988 538 972 181 626 88 × 2 = 1 + 0.471 977 077 944 363 253 76;
  • 112) 0.471 977 077 944 363 253 76 × 2 = 0 + 0.943 954 155 888 726 507 52;

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:


0.000 000 000 000 000 001 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0100 1010 1001 0000 1100 1110 1010 1111 1111 1111 1001 1101 1110(2)

6. Positive number before normalization:

0.000 000 000 000 000 001 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0100 1010 1001 0000 1100 1110 1010 1111 1111 1111 1001 1101 1110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 60 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 001 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0100 1010 1001 0000 1100 1110 1010 1111 1111 1111 1001 1101 1110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0100 1010 1001 0000 1100 1110 1010 1111 1111 1111 1001 1101 1110(2) × 20 =

1.0100 1010 1001 0000 1100 1110 1010 1111 1111 1111 1001 1101 1110(2) × 2-60


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): -60

Mantissa (not normalized):
1.0100 1010 1001 0000 1100 1110 1010 1111 1111 1111 1001 1101 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-60 + 2(11-1) - 1 =

(-60 + 1 023)(10) =

963(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;
  • 963 ÷ 2 = 481 + 1;
  • 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) =

963(10) =

011 1100 0011(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. 0100 1010 1001 0000 1100 1110 1010 1111 1111 1111 1001 1101 1110 =

0100 1010 1001 0000 1100 1110 1010 1111 1111 1111 1001 1101 1110


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 0011

Mantissa (52 bits) =
0100 1010 1001 0000 1100 1110 1010 1111 1111 1111 1001 1101 1110


Decimal number -0.000 000 000 000 000 001 12 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1100 0011 - 0100 1010 1001 0000 1100 1110 1010 1111 1111 1111 1001 1101 1110

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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