0.878 126 080 186 649 741 556 081 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.878 126 080 186 649 741 556 081 62(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
0.878 126 080 186 649 741 556 081 62(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: 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;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.878 126 080 186 649 741 556 081 62.

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) 0.878 126 080 186 649 741 556 081 62 × 2 = 1 + 0.756 252 160 373 299 483 112 163 24;
  • 2) 0.756 252 160 373 299 483 112 163 24 × 2 = 1 + 0.512 504 320 746 598 966 224 326 48;
  • 3) 0.512 504 320 746 598 966 224 326 48 × 2 = 1 + 0.025 008 641 493 197 932 448 652 96;
  • 4) 0.025 008 641 493 197 932 448 652 96 × 2 = 0 + 0.050 017 282 986 395 864 897 305 92;
  • 5) 0.050 017 282 986 395 864 897 305 92 × 2 = 0 + 0.100 034 565 972 791 729 794 611 84;
  • 6) 0.100 034 565 972 791 729 794 611 84 × 2 = 0 + 0.200 069 131 945 583 459 589 223 68;
  • 7) 0.200 069 131 945 583 459 589 223 68 × 2 = 0 + 0.400 138 263 891 166 919 178 447 36;
  • 8) 0.400 138 263 891 166 919 178 447 36 × 2 = 0 + 0.800 276 527 782 333 838 356 894 72;
  • 9) 0.800 276 527 782 333 838 356 894 72 × 2 = 1 + 0.600 553 055 564 667 676 713 789 44;
  • 10) 0.600 553 055 564 667 676 713 789 44 × 2 = 1 + 0.201 106 111 129 335 353 427 578 88;
  • 11) 0.201 106 111 129 335 353 427 578 88 × 2 = 0 + 0.402 212 222 258 670 706 855 157 76;
  • 12) 0.402 212 222 258 670 706 855 157 76 × 2 = 0 + 0.804 424 444 517 341 413 710 315 52;
  • 13) 0.804 424 444 517 341 413 710 315 52 × 2 = 1 + 0.608 848 889 034 682 827 420 631 04;
  • 14) 0.608 848 889 034 682 827 420 631 04 × 2 = 1 + 0.217 697 778 069 365 654 841 262 08;
  • 15) 0.217 697 778 069 365 654 841 262 08 × 2 = 0 + 0.435 395 556 138 731 309 682 524 16;
  • 16) 0.435 395 556 138 731 309 682 524 16 × 2 = 0 + 0.870 791 112 277 462 619 365 048 32;
  • 17) 0.870 791 112 277 462 619 365 048 32 × 2 = 1 + 0.741 582 224 554 925 238 730 096 64;
  • 18) 0.741 582 224 554 925 238 730 096 64 × 2 = 1 + 0.483 164 449 109 850 477 460 193 28;
  • 19) 0.483 164 449 109 850 477 460 193 28 × 2 = 0 + 0.966 328 898 219 700 954 920 386 56;
  • 20) 0.966 328 898 219 700 954 920 386 56 × 2 = 1 + 0.932 657 796 439 401 909 840 773 12;
  • 21) 0.932 657 796 439 401 909 840 773 12 × 2 = 1 + 0.865 315 592 878 803 819 681 546 24;
  • 22) 0.865 315 592 878 803 819 681 546 24 × 2 = 1 + 0.730 631 185 757 607 639 363 092 48;
  • 23) 0.730 631 185 757 607 639 363 092 48 × 2 = 1 + 0.461 262 371 515 215 278 726 184 96;
  • 24) 0.461 262 371 515 215 278 726 184 96 × 2 = 0 + 0.922 524 743 030 430 557 452 369 92;
  • 25) 0.922 524 743 030 430 557 452 369 92 × 2 = 1 + 0.845 049 486 060 861 114 904 739 84;
  • 26) 0.845 049 486 060 861 114 904 739 84 × 2 = 1 + 0.690 098 972 121 722 229 809 479 68;
  • 27) 0.690 098 972 121 722 229 809 479 68 × 2 = 1 + 0.380 197 944 243 444 459 618 959 36;
  • 28) 0.380 197 944 243 444 459 618 959 36 × 2 = 0 + 0.760 395 888 486 888 919 237 918 72;
  • 29) 0.760 395 888 486 888 919 237 918 72 × 2 = 1 + 0.520 791 776 973 777 838 475 837 44;
  • 30) 0.520 791 776 973 777 838 475 837 44 × 2 = 1 + 0.041 583 553 947 555 676 951 674 88;
  • 31) 0.041 583 553 947 555 676 951 674 88 × 2 = 0 + 0.083 167 107 895 111 353 903 349 76;
  • 32) 0.083 167 107 895 111 353 903 349 76 × 2 = 0 + 0.166 334 215 790 222 707 806 699 52;
  • 33) 0.166 334 215 790 222 707 806 699 52 × 2 = 0 + 0.332 668 431 580 445 415 613 399 04;
  • 34) 0.332 668 431 580 445 415 613 399 04 × 2 = 0 + 0.665 336 863 160 890 831 226 798 08;
  • 35) 0.665 336 863 160 890 831 226 798 08 × 2 = 1 + 0.330 673 726 321 781 662 453 596 16;
  • 36) 0.330 673 726 321 781 662 453 596 16 × 2 = 0 + 0.661 347 452 643 563 324 907 192 32;
  • 37) 0.661 347 452 643 563 324 907 192 32 × 2 = 1 + 0.322 694 905 287 126 649 814 384 64;
  • 38) 0.322 694 905 287 126 649 814 384 64 × 2 = 0 + 0.645 389 810 574 253 299 628 769 28;
  • 39) 0.645 389 810 574 253 299 628 769 28 × 2 = 1 + 0.290 779 621 148 506 599 257 538 56;
  • 40) 0.290 779 621 148 506 599 257 538 56 × 2 = 0 + 0.581 559 242 297 013 198 515 077 12;
  • 41) 0.581 559 242 297 013 198 515 077 12 × 2 = 1 + 0.163 118 484 594 026 397 030 154 24;
  • 42) 0.163 118 484 594 026 397 030 154 24 × 2 = 0 + 0.326 236 969 188 052 794 060 308 48;
  • 43) 0.326 236 969 188 052 794 060 308 48 × 2 = 0 + 0.652 473 938 376 105 588 120 616 96;
  • 44) 0.652 473 938 376 105 588 120 616 96 × 2 = 1 + 0.304 947 876 752 211 176 241 233 92;
  • 45) 0.304 947 876 752 211 176 241 233 92 × 2 = 0 + 0.609 895 753 504 422 352 482 467 84;
  • 46) 0.609 895 753 504 422 352 482 467 84 × 2 = 1 + 0.219 791 507 008 844 704 964 935 68;
  • 47) 0.219 791 507 008 844 704 964 935 68 × 2 = 0 + 0.439 583 014 017 689 409 929 871 36;
  • 48) 0.439 583 014 017 689 409 929 871 36 × 2 = 0 + 0.879 166 028 035 378 819 859 742 72;
  • 49) 0.879 166 028 035 378 819 859 742 72 × 2 = 1 + 0.758 332 056 070 757 639 719 485 44;
  • 50) 0.758 332 056 070 757 639 719 485 44 × 2 = 1 + 0.516 664 112 141 515 279 438 970 88;
  • 51) 0.516 664 112 141 515 279 438 970 88 × 2 = 1 + 0.033 328 224 283 030 558 877 941 76;
  • 52) 0.033 328 224 283 030 558 877 941 76 × 2 = 0 + 0.066 656 448 566 061 117 755 883 52;
  • 53) 0.066 656 448 566 061 117 755 883 52 × 2 = 0 + 0.133 312 897 132 122 235 511 767 04;

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:


0.878 126 080 186 649 741 556 081 62(10) =

0.1110 0000 1100 1100 1101 1110 1110 1100 0010 1010 1001 0100 1110 0(2)

5. Positive number before normalization:

0.878 126 080 186 649 741 556 081 62(10) =

0.1110 0000 1100 1100 1101 1110 1110 1100 0010 1010 1001 0100 1110 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.878 126 080 186 649 741 556 081 62(10) =

0.1110 0000 1100 1100 1101 1110 1110 1100 0010 1010 1001 0100 1110 0(2) =

0.1110 0000 1100 1100 1101 1110 1110 1100 0010 1010 1001 0100 1110 0(2) × 20 =

1.1100 0001 1001 1001 1011 1101 1101 1000 0101 0101 0010 1001 1100(2) × 2-1


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): -1

Mantissa (not normalized):
1.1100 0001 1001 1001 1011 1101 1101 1000 0101 0101 0010 1001 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1022(10) =

011 1111 1110(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1100 0001 1001 1001 1011 1101 1101 1000 0101 0101 0010 1001 1100 =

1100 0001 1001 1001 1011 1101 1101 1000 0101 0101 0010 1001 1100


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) =
011 1111 1110

Mantissa (52 bits) =
1100 0001 1001 1001 1011 1101 1101 1000 0101 0101 0010 1001 1100


Decimal number 0.878 126 080 186 649 741 556 081 62 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1100 0001 1001 1001 1011 1101 1101 1000 0101 0101 0010 1001 1100

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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