8.108 695 652 173 912 193 916 294 199 880 210 89 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 8.108 695 652 173 912 193 916 294 199 880 210 89(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
8.108 695 652 173 912 193 916 294 199 880 210 89(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. First, convert to binary (in base 2) the integer part: 8.
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;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

8(10) =

1000(2)


3. Convert to binary (base 2) the fractional part: 0.108 695 652 173 912 193 916 294 199 880 210 89.

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) 0.108 695 652 173 912 193 916 294 199 880 210 89 × 2 = 0 + 0.217 391 304 347 824 387 832 588 399 760 421 78;
  • 2) 0.217 391 304 347 824 387 832 588 399 760 421 78 × 2 = 0 + 0.434 782 608 695 648 775 665 176 799 520 843 56;
  • 3) 0.434 782 608 695 648 775 665 176 799 520 843 56 × 2 = 0 + 0.869 565 217 391 297 551 330 353 599 041 687 12;
  • 4) 0.869 565 217 391 297 551 330 353 599 041 687 12 × 2 = 1 + 0.739 130 434 782 595 102 660 707 198 083 374 24;
  • 5) 0.739 130 434 782 595 102 660 707 198 083 374 24 × 2 = 1 + 0.478 260 869 565 190 205 321 414 396 166 748 48;
  • 6) 0.478 260 869 565 190 205 321 414 396 166 748 48 × 2 = 0 + 0.956 521 739 130 380 410 642 828 792 333 496 96;
  • 7) 0.956 521 739 130 380 410 642 828 792 333 496 96 × 2 = 1 + 0.913 043 478 260 760 821 285 657 584 666 993 92;
  • 8) 0.913 043 478 260 760 821 285 657 584 666 993 92 × 2 = 1 + 0.826 086 956 521 521 642 571 315 169 333 987 84;
  • 9) 0.826 086 956 521 521 642 571 315 169 333 987 84 × 2 = 1 + 0.652 173 913 043 043 285 142 630 338 667 975 68;
  • 10) 0.652 173 913 043 043 285 142 630 338 667 975 68 × 2 = 1 + 0.304 347 826 086 086 570 285 260 677 335 951 36;
  • 11) 0.304 347 826 086 086 570 285 260 677 335 951 36 × 2 = 0 + 0.608 695 652 172 173 140 570 521 354 671 902 72;
  • 12) 0.608 695 652 172 173 140 570 521 354 671 902 72 × 2 = 1 + 0.217 391 304 344 346 281 141 042 709 343 805 44;
  • 13) 0.217 391 304 344 346 281 141 042 709 343 805 44 × 2 = 0 + 0.434 782 608 688 692 562 282 085 418 687 610 88;
  • 14) 0.434 782 608 688 692 562 282 085 418 687 610 88 × 2 = 0 + 0.869 565 217 377 385 124 564 170 837 375 221 76;
  • 15) 0.869 565 217 377 385 124 564 170 837 375 221 76 × 2 = 1 + 0.739 130 434 754 770 249 128 341 674 750 443 52;
  • 16) 0.739 130 434 754 770 249 128 341 674 750 443 52 × 2 = 1 + 0.478 260 869 509 540 498 256 683 349 500 887 04;
  • 17) 0.478 260 869 509 540 498 256 683 349 500 887 04 × 2 = 0 + 0.956 521 739 019 080 996 513 366 699 001 774 08;
  • 18) 0.956 521 739 019 080 996 513 366 699 001 774 08 × 2 = 1 + 0.913 043 478 038 161 993 026 733 398 003 548 16;
  • 19) 0.913 043 478 038 161 993 026 733 398 003 548 16 × 2 = 1 + 0.826 086 956 076 323 986 053 466 796 007 096 32;
  • 20) 0.826 086 956 076 323 986 053 466 796 007 096 32 × 2 = 1 + 0.652 173 912 152 647 972 106 933 592 014 192 64;
  • 21) 0.652 173 912 152 647 972 106 933 592 014 192 64 × 2 = 1 + 0.304 347 824 305 295 944 213 867 184 028 385 28;
  • 22) 0.304 347 824 305 295 944 213 867 184 028 385 28 × 2 = 0 + 0.608 695 648 610 591 888 427 734 368 056 770 56;
  • 23) 0.608 695 648 610 591 888 427 734 368 056 770 56 × 2 = 1 + 0.217 391 297 221 183 776 855 468 736 113 541 12;
  • 24) 0.217 391 297 221 183 776 855 468 736 113 541 12 × 2 = 0 + 0.434 782 594 442 367 553 710 937 472 227 082 24;
  • 25) 0.434 782 594 442 367 553 710 937 472 227 082 24 × 2 = 0 + 0.869 565 188 884 735 107 421 874 944 454 164 48;
  • 26) 0.869 565 188 884 735 107 421 874 944 454 164 48 × 2 = 1 + 0.739 130 377 769 470 214 843 749 888 908 328 96;
  • 27) 0.739 130 377 769 470 214 843 749 888 908 328 96 × 2 = 1 + 0.478 260 755 538 940 429 687 499 777 816 657 92;
  • 28) 0.478 260 755 538 940 429 687 499 777 816 657 92 × 2 = 0 + 0.956 521 511 077 880 859 374 999 555 633 315 84;
  • 29) 0.956 521 511 077 880 859 374 999 555 633 315 84 × 2 = 1 + 0.913 043 022 155 761 718 749 999 111 266 631 68;
  • 30) 0.913 043 022 155 761 718 749 999 111 266 631 68 × 2 = 1 + 0.826 086 044 311 523 437 499 998 222 533 263 36;
  • 31) 0.826 086 044 311 523 437 499 998 222 533 263 36 × 2 = 1 + 0.652 172 088 623 046 874 999 996 445 066 526 72;
  • 32) 0.652 172 088 623 046 874 999 996 445 066 526 72 × 2 = 1 + 0.304 344 177 246 093 749 999 992 890 133 053 44;
  • 33) 0.304 344 177 246 093 749 999 992 890 133 053 44 × 2 = 0 + 0.608 688 354 492 187 499 999 985 780 266 106 88;
  • 34) 0.608 688 354 492 187 499 999 985 780 266 106 88 × 2 = 1 + 0.217 376 708 984 374 999 999 971 560 532 213 76;
  • 35) 0.217 376 708 984 374 999 999 971 560 532 213 76 × 2 = 0 + 0.434 753 417 968 749 999 999 943 121 064 427 52;
  • 36) 0.434 753 417 968 749 999 999 943 121 064 427 52 × 2 = 0 + 0.869 506 835 937 499 999 999 886 242 128 855 04;
  • 37) 0.869 506 835 937 499 999 999 886 242 128 855 04 × 2 = 1 + 0.739 013 671 874 999 999 999 772 484 257 710 08;
  • 38) 0.739 013 671 874 999 999 999 772 484 257 710 08 × 2 = 1 + 0.478 027 343 749 999 999 999 544 968 515 420 16;
  • 39) 0.478 027 343 749 999 999 999 544 968 515 420 16 × 2 = 0 + 0.956 054 687 499 999 999 999 089 937 030 840 32;
  • 40) 0.956 054 687 499 999 999 999 089 937 030 840 32 × 2 = 1 + 0.912 109 374 999 999 999 998 179 874 061 680 64;
  • 41) 0.912 109 374 999 999 999 998 179 874 061 680 64 × 2 = 1 + 0.824 218 749 999 999 999 996 359 748 123 361 28;
  • 42) 0.824 218 749 999 999 999 996 359 748 123 361 28 × 2 = 1 + 0.648 437 499 999 999 999 992 719 496 246 722 56;
  • 43) 0.648 437 499 999 999 999 992 719 496 246 722 56 × 2 = 1 + 0.296 874 999 999 999 999 985 438 992 493 445 12;
  • 44) 0.296 874 999 999 999 999 985 438 992 493 445 12 × 2 = 0 + 0.593 749 999 999 999 999 970 877 984 986 890 24;
  • 45) 0.593 749 999 999 999 999 970 877 984 986 890 24 × 2 = 1 + 0.187 499 999 999 999 999 941 755 969 973 780 48;
  • 46) 0.187 499 999 999 999 999 941 755 969 973 780 48 × 2 = 0 + 0.374 999 999 999 999 999 883 511 939 947 560 96;
  • 47) 0.374 999 999 999 999 999 883 511 939 947 560 96 × 2 = 0 + 0.749 999 999 999 999 999 767 023 879 895 121 92;
  • 48) 0.749 999 999 999 999 999 767 023 879 895 121 92 × 2 = 1 + 0.499 999 999 999 999 999 534 047 759 790 243 84;
  • 49) 0.499 999 999 999 999 999 534 047 759 790 243 84 × 2 = 0 + 0.999 999 999 999 999 999 068 095 519 580 487 68;
  • 50) 0.999 999 999 999 999 999 068 095 519 580 487 68 × 2 = 1 + 0.999 999 999 999 999 998 136 191 039 160 975 36;
  • 51) 0.999 999 999 999 999 998 136 191 039 160 975 36 × 2 = 1 + 0.999 999 999 999 999 996 272 382 078 321 950 72;
  • 52) 0.999 999 999 999 999 996 272 382 078 321 950 72 × 2 = 1 + 0.999 999 999 999 999 992 544 764 156 643 901 44;
  • 53) 0.999 999 999 999 999 992 544 764 156 643 901 44 × 2 = 1 + 0.999 999 999 999 999 985 089 528 313 287 802 88;

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).


4. 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:


0.108 695 652 173 912 193 916 294 199 880 210 89(10) =

0.0001 1011 1101 0011 0111 1010 0110 1111 0100 1101 1110 1001 0111 1(2)

5. Positive number before normalization:

8.108 695 652 173 912 193 916 294 199 880 210 89(10) =

1000.0001 1011 1101 0011 0111 1010 0110 1111 0100 1101 1110 1001 0111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


8.108 695 652 173 912 193 916 294 199 880 210 89(10) =

1000.0001 1011 1101 0011 0111 1010 0110 1111 0100 1101 1110 1001 0111 1(2) =

1000.0001 1011 1101 0011 0111 1010 0110 1111 0100 1101 1110 1001 0111 1(2) × 20 =

1.0000 0011 0111 1010 0110 1111 0100 1101 1110 1001 1011 1101 0010 1111(2) × 23


7. 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 0 (a positive number)

Exponent (unadjusted): 3

Mantissa (not normalized):
1.0000 0011 0111 1010 0110 1111 0100 1101 1110 1001 1011 1101 0010 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

3 + 2(11-1) - 1 =

(3 + 1 023)(10) =

1 026(10)


9. 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;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1026(10) =

100 0000 0010(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0000 0011 0111 1010 0110 1111 0100 1101 1110 1001 1011 1101 0010 1111 =

0000 0011 0111 1010 0110 1111 0100 1101 1110 1001 1011 1101 0010


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0010

Mantissa (52 bits) =
0000 0011 0111 1010 0110 1111 0100 1101 1110 1001 1011 1101 0010


Decimal number 8.108 695 652 173 912 193 916 294 199 880 210 89 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0010 - 0000 0011 0111 1010 0110 1111 0100 1101 1110 1001 1011 1101 0010

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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