64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.088 993 507 098 078 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.088 993 507 098 078(10) converted and written in 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: 0.088 993 507 098 078.

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) 0.088 993 507 098 078 × 2 = 0 + 0.177 987 014 196 156;
  • 2) 0.177 987 014 196 156 × 2 = 0 + 0.355 974 028 392 312;
  • 3) 0.355 974 028 392 312 × 2 = 0 + 0.711 948 056 784 624;
  • 4) 0.711 948 056 784 624 × 2 = 1 + 0.423 896 113 569 248;
  • 5) 0.423 896 113 569 248 × 2 = 0 + 0.847 792 227 138 496;
  • 6) 0.847 792 227 138 496 × 2 = 1 + 0.695 584 454 276 992;
  • 7) 0.695 584 454 276 992 × 2 = 1 + 0.391 168 908 553 984;
  • 8) 0.391 168 908 553 984 × 2 = 0 + 0.782 337 817 107 968;
  • 9) 0.782 337 817 107 968 × 2 = 1 + 0.564 675 634 215 936;
  • 10) 0.564 675 634 215 936 × 2 = 1 + 0.129 351 268 431 872;
  • 11) 0.129 351 268 431 872 × 2 = 0 + 0.258 702 536 863 744;
  • 12) 0.258 702 536 863 744 × 2 = 0 + 0.517 405 073 727 488;
  • 13) 0.517 405 073 727 488 × 2 = 1 + 0.034 810 147 454 976;
  • 14) 0.034 810 147 454 976 × 2 = 0 + 0.069 620 294 909 952;
  • 15) 0.069 620 294 909 952 × 2 = 0 + 0.139 240 589 819 904;
  • 16) 0.139 240 589 819 904 × 2 = 0 + 0.278 481 179 639 808;
  • 17) 0.278 481 179 639 808 × 2 = 0 + 0.556 962 359 279 616;
  • 18) 0.556 962 359 279 616 × 2 = 1 + 0.113 924 718 559 232;
  • 19) 0.113 924 718 559 232 × 2 = 0 + 0.227 849 437 118 464;
  • 20) 0.227 849 437 118 464 × 2 = 0 + 0.455 698 874 236 928;
  • 21) 0.455 698 874 236 928 × 2 = 0 + 0.911 397 748 473 856;
  • 22) 0.911 397 748 473 856 × 2 = 1 + 0.822 795 496 947 712;
  • 23) 0.822 795 496 947 712 × 2 = 1 + 0.645 590 993 895 424;
  • 24) 0.645 590 993 895 424 × 2 = 1 + 0.291 181 987 790 848;
  • 25) 0.291 181 987 790 848 × 2 = 0 + 0.582 363 975 581 696;
  • 26) 0.582 363 975 581 696 × 2 = 1 + 0.164 727 951 163 392;
  • 27) 0.164 727 951 163 392 × 2 = 0 + 0.329 455 902 326 784;
  • 28) 0.329 455 902 326 784 × 2 = 0 + 0.658 911 804 653 568;
  • 29) 0.658 911 804 653 568 × 2 = 1 + 0.317 823 609 307 136;
  • 30) 0.317 823 609 307 136 × 2 = 0 + 0.635 647 218 614 272;
  • 31) 0.635 647 218 614 272 × 2 = 1 + 0.271 294 437 228 544;
  • 32) 0.271 294 437 228 544 × 2 = 0 + 0.542 588 874 457 088;
  • 33) 0.542 588 874 457 088 × 2 = 1 + 0.085 177 748 914 176;
  • 34) 0.085 177 748 914 176 × 2 = 0 + 0.170 355 497 828 352;
  • 35) 0.170 355 497 828 352 × 2 = 0 + 0.340 710 995 656 704;
  • 36) 0.340 710 995 656 704 × 2 = 0 + 0.681 421 991 313 408;
  • 37) 0.681 421 991 313 408 × 2 = 1 + 0.362 843 982 626 816;
  • 38) 0.362 843 982 626 816 × 2 = 0 + 0.725 687 965 253 632;
  • 39) 0.725 687 965 253 632 × 2 = 1 + 0.451 375 930 507 264;
  • 40) 0.451 375 930 507 264 × 2 = 0 + 0.902 751 861 014 528;
  • 41) 0.902 751 861 014 528 × 2 = 1 + 0.805 503 722 029 056;
  • 42) 0.805 503 722 029 056 × 2 = 1 + 0.611 007 444 058 112;
  • 43) 0.611 007 444 058 112 × 2 = 1 + 0.222 014 888 116 224;
  • 44) 0.222 014 888 116 224 × 2 = 0 + 0.444 029 776 232 448;
  • 45) 0.444 029 776 232 448 × 2 = 0 + 0.888 059 552 464 896;
  • 46) 0.888 059 552 464 896 × 2 = 1 + 0.776 119 104 929 792;
  • 47) 0.776 119 104 929 792 × 2 = 1 + 0.552 238 209 859 584;
  • 48) 0.552 238 209 859 584 × 2 = 1 + 0.104 476 419 719 168;
  • 49) 0.104 476 419 719 168 × 2 = 0 + 0.208 952 839 438 336;
  • 50) 0.208 952 839 438 336 × 2 = 0 + 0.417 905 678 876 672;
  • 51) 0.417 905 678 876 672 × 2 = 0 + 0.835 811 357 753 344;
  • 52) 0.835 811 357 753 344 × 2 = 1 + 0.671 622 715 506 688;
  • 53) 0.671 622 715 506 688 × 2 = 1 + 0.343 245 431 013 376;
  • 54) 0.343 245 431 013 376 × 2 = 0 + 0.686 490 862 026 752;
  • 55) 0.686 490 862 026 752 × 2 = 1 + 0.372 981 724 053 504;
  • 56) 0.372 981 724 053 504 × 2 = 0 + 0.745 963 448 107 008;

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...)


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:


0.088 993 507 098 078(10) =


0.0001 0110 1100 1000 0100 0111 0100 1010 1000 1010 1110 0111 0001 1010(2)


5. Positive number before normalization:

0.088 993 507 098 078(10) =


0.0001 0110 1100 1000 0100 0111 0100 1010 1000 1010 1110 0111 0001 1010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.088 993 507 098 078(10) =


0.0001 0110 1100 1000 0100 0111 0100 1010 1000 1010 1110 0111 0001 1010(2) =


0.0001 0110 1100 1000 0100 0111 0100 1010 1000 1010 1110 0111 0001 1010(2) × 20 =


1.0110 1100 1000 0100 0111 0100 1010 1000 1010 1110 0111 0001 1010(2) × 2-4


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): -4


Mantissa (not normalized):
1.0110 1100 1000 0100 0111 0100 1010 1000 1010 1110 0111 0001 1010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


-4 + 2(11-1) - 1 =


(-4 + 1 023)(10) =


1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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) =


1019(10) =


011 1111 1011(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. 0110 1100 1000 0100 0111 0100 1010 1000 1010 1110 0111 0001 1010 =


0110 1100 1000 0100 0111 0100 1010 1000 1010 1110 0111 0001 1010


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 1011


Mantissa (52 bits) =
0110 1100 1000 0100 0111 0100 1010 1000 1010 1110 0111 0001 1010


The base ten decimal number 0.088 993 507 098 078 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1011 - 0110 1100 1000 0100 0111 0100 1010 1000 1010 1110 0111 0001 1010

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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