-0.000 000 000 000 000 000 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 000 15(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 15(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 15| = 0.000 000 000 000 000 000 15


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 15.

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 15 × 2 = 0 + 0.000 000 000 000 000 000 3;
  • 2) 0.000 000 000 000 000 000 3 × 2 = 0 + 0.000 000 000 000 000 000 6;
  • 3) 0.000 000 000 000 000 000 6 × 2 = 0 + 0.000 000 000 000 000 001 2;
  • 4) 0.000 000 000 000 000 001 2 × 2 = 0 + 0.000 000 000 000 000 002 4;
  • 5) 0.000 000 000 000 000 002 4 × 2 = 0 + 0.000 000 000 000 000 004 8;
  • 6) 0.000 000 000 000 000 004 8 × 2 = 0 + 0.000 000 000 000 000 009 6;
  • 7) 0.000 000 000 000 000 009 6 × 2 = 0 + 0.000 000 000 000 000 019 2;
  • 8) 0.000 000 000 000 000 019 2 × 2 = 0 + 0.000 000 000 000 000 038 4;
  • 9) 0.000 000 000 000 000 038 4 × 2 = 0 + 0.000 000 000 000 000 076 8;
  • 10) 0.000 000 000 000 000 076 8 × 2 = 0 + 0.000 000 000 000 000 153 6;
  • 11) 0.000 000 000 000 000 153 6 × 2 = 0 + 0.000 000 000 000 000 307 2;
  • 12) 0.000 000 000 000 000 307 2 × 2 = 0 + 0.000 000 000 000 000 614 4;
  • 13) 0.000 000 000 000 000 614 4 × 2 = 0 + 0.000 000 000 000 001 228 8;
  • 14) 0.000 000 000 000 001 228 8 × 2 = 0 + 0.000 000 000 000 002 457 6;
  • 15) 0.000 000 000 000 002 457 6 × 2 = 0 + 0.000 000 000 000 004 915 2;
  • 16) 0.000 000 000 000 004 915 2 × 2 = 0 + 0.000 000 000 000 009 830 4;
  • 17) 0.000 000 000 000 009 830 4 × 2 = 0 + 0.000 000 000 000 019 660 8;
  • 18) 0.000 000 000 000 019 660 8 × 2 = 0 + 0.000 000 000 000 039 321 6;
  • 19) 0.000 000 000 000 039 321 6 × 2 = 0 + 0.000 000 000 000 078 643 2;
  • 20) 0.000 000 000 000 078 643 2 × 2 = 0 + 0.000 000 000 000 157 286 4;
  • 21) 0.000 000 000 000 157 286 4 × 2 = 0 + 0.000 000 000 000 314 572 8;
  • 22) 0.000 000 000 000 314 572 8 × 2 = 0 + 0.000 000 000 000 629 145 6;
  • 23) 0.000 000 000 000 629 145 6 × 2 = 0 + 0.000 000 000 001 258 291 2;
  • 24) 0.000 000 000 001 258 291 2 × 2 = 0 + 0.000 000 000 002 516 582 4;
  • 25) 0.000 000 000 002 516 582 4 × 2 = 0 + 0.000 000 000 005 033 164 8;
  • 26) 0.000 000 000 005 033 164 8 × 2 = 0 + 0.000 000 000 010 066 329 6;
  • 27) 0.000 000 000 010 066 329 6 × 2 = 0 + 0.000 000 000 020 132 659 2;
  • 28) 0.000 000 000 020 132 659 2 × 2 = 0 + 0.000 000 000 040 265 318 4;
  • 29) 0.000 000 000 040 265 318 4 × 2 = 0 + 0.000 000 000 080 530 636 8;
  • 30) 0.000 000 000 080 530 636 8 × 2 = 0 + 0.000 000 000 161 061 273 6;
  • 31) 0.000 000 000 161 061 273 6 × 2 = 0 + 0.000 000 000 322 122 547 2;
  • 32) 0.000 000 000 322 122 547 2 × 2 = 0 + 0.000 000 000 644 245 094 4;
  • 33) 0.000 000 000 644 245 094 4 × 2 = 0 + 0.000 000 001 288 490 188 8;
  • 34) 0.000 000 001 288 490 188 8 × 2 = 0 + 0.000 000 002 576 980 377 6;
  • 35) 0.000 000 002 576 980 377 6 × 2 = 0 + 0.000 000 005 153 960 755 2;
  • 36) 0.000 000 005 153 960 755 2 × 2 = 0 + 0.000 000 010 307 921 510 4;
  • 37) 0.000 000 010 307 921 510 4 × 2 = 0 + 0.000 000 020 615 843 020 8;
  • 38) 0.000 000 020 615 843 020 8 × 2 = 0 + 0.000 000 041 231 686 041 6;
  • 39) 0.000 000 041 231 686 041 6 × 2 = 0 + 0.000 000 082 463 372 083 2;
  • 40) 0.000 000 082 463 372 083 2 × 2 = 0 + 0.000 000 164 926 744 166 4;
  • 41) 0.000 000 164 926 744 166 4 × 2 = 0 + 0.000 000 329 853 488 332 8;
  • 42) 0.000 000 329 853 488 332 8 × 2 = 0 + 0.000 000 659 706 976 665 6;
  • 43) 0.000 000 659 706 976 665 6 × 2 = 0 + 0.000 001 319 413 953 331 2;
  • 44) 0.000 001 319 413 953 331 2 × 2 = 0 + 0.000 002 638 827 906 662 4;
  • 45) 0.000 002 638 827 906 662 4 × 2 = 0 + 0.000 005 277 655 813 324 8;
  • 46) 0.000 005 277 655 813 324 8 × 2 = 0 + 0.000 010 555 311 626 649 6;
  • 47) 0.000 010 555 311 626 649 6 × 2 = 0 + 0.000 021 110 623 253 299 2;
  • 48) 0.000 021 110 623 253 299 2 × 2 = 0 + 0.000 042 221 246 506 598 4;
  • 49) 0.000 042 221 246 506 598 4 × 2 = 0 + 0.000 084 442 493 013 196 8;
  • 50) 0.000 084 442 493 013 196 8 × 2 = 0 + 0.000 168 884 986 026 393 6;
  • 51) 0.000 168 884 986 026 393 6 × 2 = 0 + 0.000 337 769 972 052 787 2;
  • 52) 0.000 337 769 972 052 787 2 × 2 = 0 + 0.000 675 539 944 105 574 4;
  • 53) 0.000 675 539 944 105 574 4 × 2 = 0 + 0.001 351 079 888 211 148 8;
  • 54) 0.001 351 079 888 211 148 8 × 2 = 0 + 0.002 702 159 776 422 297 6;
  • 55) 0.002 702 159 776 422 297 6 × 2 = 0 + 0.005 404 319 552 844 595 2;
  • 56) 0.005 404 319 552 844 595 2 × 2 = 0 + 0.010 808 639 105 689 190 4;
  • 57) 0.010 808 639 105 689 190 4 × 2 = 0 + 0.021 617 278 211 378 380 8;
  • 58) 0.021 617 278 211 378 380 8 × 2 = 0 + 0.043 234 556 422 756 761 6;
  • 59) 0.043 234 556 422 756 761 6 × 2 = 0 + 0.086 469 112 845 513 523 2;
  • 60) 0.086 469 112 845 513 523 2 × 2 = 0 + 0.172 938 225 691 027 046 4;
  • 61) 0.172 938 225 691 027 046 4 × 2 = 0 + 0.345 876 451 382 054 092 8;
  • 62) 0.345 876 451 382 054 092 8 × 2 = 0 + 0.691 752 902 764 108 185 6;
  • 63) 0.691 752 902 764 108 185 6 × 2 = 1 + 0.383 505 805 528 216 371 2;
  • 64) 0.383 505 805 528 216 371 2 × 2 = 0 + 0.767 011 611 056 432 742 4;
  • 65) 0.767 011 611 056 432 742 4 × 2 = 1 + 0.534 023 222 112 865 484 8;
  • 66) 0.534 023 222 112 865 484 8 × 2 = 1 + 0.068 046 444 225 730 969 6;
  • 67) 0.068 046 444 225 730 969 6 × 2 = 0 + 0.136 092 888 451 461 939 2;
  • 68) 0.136 092 888 451 461 939 2 × 2 = 0 + 0.272 185 776 902 923 878 4;
  • 69) 0.272 185 776 902 923 878 4 × 2 = 0 + 0.544 371 553 805 847 756 8;
  • 70) 0.544 371 553 805 847 756 8 × 2 = 1 + 0.088 743 107 611 695 513 6;
  • 71) 0.088 743 107 611 695 513 6 × 2 = 0 + 0.177 486 215 223 391 027 2;
  • 72) 0.177 486 215 223 391 027 2 × 2 = 0 + 0.354 972 430 446 782 054 4;
  • 73) 0.354 972 430 446 782 054 4 × 2 = 0 + 0.709 944 860 893 564 108 8;
  • 74) 0.709 944 860 893 564 108 8 × 2 = 1 + 0.419 889 721 787 128 217 6;
  • 75) 0.419 889 721 787 128 217 6 × 2 = 0 + 0.839 779 443 574 256 435 2;
  • 76) 0.839 779 443 574 256 435 2 × 2 = 1 + 0.679 558 887 148 512 870 4;
  • 77) 0.679 558 887 148 512 870 4 × 2 = 1 + 0.359 117 774 297 025 740 8;
  • 78) 0.359 117 774 297 025 740 8 × 2 = 0 + 0.718 235 548 594 051 481 6;
  • 79) 0.718 235 548 594 051 481 6 × 2 = 1 + 0.436 471 097 188 102 963 2;
  • 80) 0.436 471 097 188 102 963 2 × 2 = 0 + 0.872 942 194 376 205 926 4;
  • 81) 0.872 942 194 376 205 926 4 × 2 = 1 + 0.745 884 388 752 411 852 8;
  • 82) 0.745 884 388 752 411 852 8 × 2 = 1 + 0.491 768 777 504 823 705 6;
  • 83) 0.491 768 777 504 823 705 6 × 2 = 0 + 0.983 537 555 009 647 411 2;
  • 84) 0.983 537 555 009 647 411 2 × 2 = 1 + 0.967 075 110 019 294 822 4;
  • 85) 0.967 075 110 019 294 822 4 × 2 = 1 + 0.934 150 220 038 589 644 8;
  • 86) 0.934 150 220 038 589 644 8 × 2 = 1 + 0.868 300 440 077 179 289 6;
  • 87) 0.868 300 440 077 179 289 6 × 2 = 1 + 0.736 600 880 154 358 579 2;
  • 88) 0.736 600 880 154 358 579 2 × 2 = 1 + 0.473 201 760 308 717 158 4;
  • 89) 0.473 201 760 308 717 158 4 × 2 = 0 + 0.946 403 520 617 434 316 8;
  • 90) 0.946 403 520 617 434 316 8 × 2 = 1 + 0.892 807 041 234 868 633 6;
  • 91) 0.892 807 041 234 868 633 6 × 2 = 1 + 0.785 614 082 469 737 267 2;
  • 92) 0.785 614 082 469 737 267 2 × 2 = 1 + 0.571 228 164 939 474 534 4;
  • 93) 0.571 228 164 939 474 534 4 × 2 = 1 + 0.142 456 329 878 949 068 8;
  • 94) 0.142 456 329 878 949 068 8 × 2 = 0 + 0.284 912 659 757 898 137 6;
  • 95) 0.284 912 659 757 898 137 6 × 2 = 0 + 0.569 825 319 515 796 275 2;
  • 96) 0.569 825 319 515 796 275 2 × 2 = 1 + 0.139 650 639 031 592 550 4;
  • 97) 0.139 650 639 031 592 550 4 × 2 = 0 + 0.279 301 278 063 185 100 8;
  • 98) 0.279 301 278 063 185 100 8 × 2 = 0 + 0.558 602 556 126 370 201 6;
  • 99) 0.558 602 556 126 370 201 6 × 2 = 1 + 0.117 205 112 252 740 403 2;
  • 100) 0.117 205 112 252 740 403 2 × 2 = 0 + 0.234 410 224 505 480 806 4;
  • 101) 0.234 410 224 505 480 806 4 × 2 = 0 + 0.468 820 449 010 961 612 8;
  • 102) 0.468 820 449 010 961 612 8 × 2 = 0 + 0.937 640 898 021 923 225 6;
  • 103) 0.937 640 898 021 923 225 6 × 2 = 1 + 0.875 281 796 043 846 451 2;
  • 104) 0.875 281 796 043 846 451 2 × 2 = 1 + 0.750 563 592 087 692 902 4;
  • 105) 0.750 563 592 087 692 902 4 × 2 = 1 + 0.501 127 184 175 385 804 8;
  • 106) 0.501 127 184 175 385 804 8 × 2 = 1 + 0.002 254 368 350 771 609 6;
  • 107) 0.002 254 368 350 771 609 6 × 2 = 0 + 0.004 508 736 701 543 219 2;
  • 108) 0.004 508 736 701 543 219 2 × 2 = 0 + 0.009 017 473 403 086 438 4;
  • 109) 0.009 017 473 403 086 438 4 × 2 = 0 + 0.018 034 946 806 172 876 8;
  • 110) 0.018 034 946 806 172 876 8 × 2 = 0 + 0.036 069 893 612 345 753 6;
  • 111) 0.036 069 893 612 345 753 6 × 2 = 0 + 0.072 139 787 224 691 507 2;
  • 112) 0.072 139 787 224 691 507 2 × 2 = 0 + 0.144 279 574 449 383 014 4;
  • 113) 0.144 279 574 449 383 014 4 × 2 = 0 + 0.288 559 148 898 766 028 8;
  • 114) 0.288 559 148 898 766 028 8 × 2 = 0 + 0.577 118 297 797 532 057 6;
  • 115) 0.577 118 297 797 532 057 6 × 2 = 1 + 0.154 236 595 595 064 115 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:


0.000 000 000 000 000 000 15(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1100 0100 0101 1010 1101 1111 0111 1001 0010 0011 1100 0000 001(2)

6. Positive number before normalization:

0.000 000 000 000 000 000 15(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1100 0100 0101 1010 1101 1111 0111 1001 0010 0011 1100 0000 001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 63 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 15(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1100 0100 0101 1010 1101 1111 0111 1001 0010 0011 1100 0000 001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1100 0100 0101 1010 1101 1111 0111 1001 0010 0011 1100 0000 001(2) × 20 =

1.0110 0010 0010 1101 0110 1111 1011 1100 1001 0001 1110 0000 0001(2) × 2-63


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): -63

Mantissa (not normalized):
1.0110 0010 0010 1101 0110 1111 1011 1100 1001 0001 1110 0000 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-63 + 2(11-1) - 1 =

(-63 + 1 023)(10) =

960(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;
  • 960 ÷ 2 = 480 + 0;
  • 480 ÷ 2 = 240 + 0;
  • 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) =

960(10) =

011 1100 0000(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. 0110 0010 0010 1101 0110 1111 1011 1100 1001 0001 1110 0000 0001 =

0110 0010 0010 1101 0110 1111 1011 1100 1001 0001 1110 0000 0001


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 0000

Mantissa (52 bits) =
0110 0010 0010 1101 0110 1111 1011 1100 1001 0001 1110 0000 0001


Decimal number -0.000 000 000 000 000 000 15 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1100 0000 - 0110 0010 0010 1101 0110 1111 1011 1100 1001 0001 1110 0000 0001

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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