64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 384 747 294.484 849 294 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 384 747 294.484 849 294(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: 384 747 294.
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;
  • 384 747 294 ÷ 2 = 192 373 647 + 0;
  • 192 373 647 ÷ 2 = 96 186 823 + 1;
  • 96 186 823 ÷ 2 = 48 093 411 + 1;
  • 48 093 411 ÷ 2 = 24 046 705 + 1;
  • 24 046 705 ÷ 2 = 12 023 352 + 1;
  • 12 023 352 ÷ 2 = 6 011 676 + 0;
  • 6 011 676 ÷ 2 = 3 005 838 + 0;
  • 3 005 838 ÷ 2 = 1 502 919 + 0;
  • 1 502 919 ÷ 2 = 751 459 + 1;
  • 751 459 ÷ 2 = 375 729 + 1;
  • 375 729 ÷ 2 = 187 864 + 1;
  • 187 864 ÷ 2 = 93 932 + 0;
  • 93 932 ÷ 2 = 46 966 + 0;
  • 46 966 ÷ 2 = 23 483 + 0;
  • 23 483 ÷ 2 = 11 741 + 1;
  • 11 741 ÷ 2 = 5 870 + 1;
  • 5 870 ÷ 2 = 2 935 + 0;
  • 2 935 ÷ 2 = 1 467 + 1;
  • 1 467 ÷ 2 = 733 + 1;
  • 733 ÷ 2 = 366 + 1;
  • 366 ÷ 2 = 183 + 0;
  • 183 ÷ 2 = 91 + 1;
  • 91 ÷ 2 = 45 + 1;
  • 45 ÷ 2 = 22 + 1;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 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.


384 747 294(10) =


1 0110 1110 1110 1100 0111 0001 1110(2)


3. Convert to binary (base 2) the fractional part: 0.484 849 294.

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) 0.484 849 294 × 2 = 0 + 0.969 698 588;
  • 2) 0.969 698 588 × 2 = 1 + 0.939 397 176;
  • 3) 0.939 397 176 × 2 = 1 + 0.878 794 352;
  • 4) 0.878 794 352 × 2 = 1 + 0.757 588 704;
  • 5) 0.757 588 704 × 2 = 1 + 0.515 177 408;
  • 6) 0.515 177 408 × 2 = 1 + 0.030 354 816;
  • 7) 0.030 354 816 × 2 = 0 + 0.060 709 632;
  • 8) 0.060 709 632 × 2 = 0 + 0.121 419 264;
  • 9) 0.121 419 264 × 2 = 0 + 0.242 838 528;
  • 10) 0.242 838 528 × 2 = 0 + 0.485 677 056;
  • 11) 0.485 677 056 × 2 = 0 + 0.971 354 112;
  • 12) 0.971 354 112 × 2 = 1 + 0.942 708 224;
  • 13) 0.942 708 224 × 2 = 1 + 0.885 416 448;
  • 14) 0.885 416 448 × 2 = 1 + 0.770 832 896;
  • 15) 0.770 832 896 × 2 = 1 + 0.541 665 792;
  • 16) 0.541 665 792 × 2 = 1 + 0.083 331 584;
  • 17) 0.083 331 584 × 2 = 0 + 0.166 663 168;
  • 18) 0.166 663 168 × 2 = 0 + 0.333 326 336;
  • 19) 0.333 326 336 × 2 = 0 + 0.666 652 672;
  • 20) 0.666 652 672 × 2 = 1 + 0.333 305 344;
  • 21) 0.333 305 344 × 2 = 0 + 0.666 610 688;
  • 22) 0.666 610 688 × 2 = 1 + 0.333 221 376;
  • 23) 0.333 221 376 × 2 = 0 + 0.666 442 752;
  • 24) 0.666 442 752 × 2 = 1 + 0.332 885 504;
  • 25) 0.332 885 504 × 2 = 0 + 0.665 771 008;
  • 26) 0.665 771 008 × 2 = 1 + 0.331 542 016;
  • 27) 0.331 542 016 × 2 = 0 + 0.663 084 032;
  • 28) 0.663 084 032 × 2 = 1 + 0.326 168 064;
  • 29) 0.326 168 064 × 2 = 0 + 0.652 336 128;
  • 30) 0.652 336 128 × 2 = 1 + 0.304 672 256;
  • 31) 0.304 672 256 × 2 = 0 + 0.609 344 512;
  • 32) 0.609 344 512 × 2 = 1 + 0.218 689 024;
  • 33) 0.218 689 024 × 2 = 0 + 0.437 378 048;
  • 34) 0.437 378 048 × 2 = 0 + 0.874 756 096;
  • 35) 0.874 756 096 × 2 = 1 + 0.749 512 192;
  • 36) 0.749 512 192 × 2 = 1 + 0.499 024 384;
  • 37) 0.499 024 384 × 2 = 0 + 0.998 048 768;
  • 38) 0.998 048 768 × 2 = 1 + 0.996 097 536;
  • 39) 0.996 097 536 × 2 = 1 + 0.992 195 072;
  • 40) 0.992 195 072 × 2 = 1 + 0.984 390 144;
  • 41) 0.984 390 144 × 2 = 1 + 0.968 780 288;
  • 42) 0.968 780 288 × 2 = 1 + 0.937 560 576;
  • 43) 0.937 560 576 × 2 = 1 + 0.875 121 152;
  • 44) 0.875 121 152 × 2 = 1 + 0.750 242 304;
  • 45) 0.750 242 304 × 2 = 1 + 0.500 484 608;
  • 46) 0.500 484 608 × 2 = 1 + 0.000 969 216;
  • 47) 0.000 969 216 × 2 = 0 + 0.001 938 432;
  • 48) 0.001 938 432 × 2 = 0 + 0.003 876 864;
  • 49) 0.003 876 864 × 2 = 0 + 0.007 753 728;
  • 50) 0.007 753 728 × 2 = 0 + 0.015 507 456;
  • 51) 0.015 507 456 × 2 = 0 + 0.031 014 912;
  • 52) 0.031 014 912 × 2 = 0 + 0.062 029 824;
  • 53) 0.062 029 824 × 2 = 0 + 0.124 059 648;

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:


0.484 849 294(10) =


0.0111 1100 0001 1111 0001 0101 0101 0101 0011 0111 1111 1100 0000 0(2)


5. Positive number before normalization:

384 747 294.484 849 294(10) =


1 0110 1110 1110 1100 0111 0001 1110.0111 1100 0001 1111 0001 0101 0101 0101 0011 0111 1111 1100 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the left, so that only one non zero digit remains to the left of it:


384 747 294.484 849 294(10) =


1 0110 1110 1110 1100 0111 0001 1110.0111 1100 0001 1111 0001 0101 0101 0101 0011 0111 1111 1100 0000 0(2) =


1 0110 1110 1110 1100 0111 0001 1110.0111 1100 0001 1111 0001 0101 0101 0101 0011 0111 1111 1100 0000 0(2) × 20 =


1.0110 1110 1110 1100 0111 0001 1110 0111 1100 0001 1111 0001 0101 0101 0101 0011 0111 1111 1100 0000 0(2) × 228


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): 28


Mantissa (not normalized):
1.0110 1110 1110 1100 0111 0001 1110 0111 1100 0001 1111 0001 0101 0101 0101 0011 0111 1111 1100 0000 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


28 + 2(11-1) - 1 =


(28 + 1 023)(10) =


1 051(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;
  • 1 051 ÷ 2 = 525 + 1;
  • 525 ÷ 2 = 262 + 1;
  • 262 ÷ 2 = 131 + 0;
  • 131 ÷ 2 = 65 + 1;
  • 65 ÷ 2 = 32 + 1;
  • 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) =


1051(10) =


100 0001 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, 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. 0110 1110 1110 1100 0111 0001 1110 0111 1100 0001 1111 0001 0101 0 1010 1010 0110 1111 1111 1000 0000 =


0110 1110 1110 1100 0111 0001 1110 0111 1100 0001 1111 0001 0101


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 0001 1011


Mantissa (52 bits) =
0110 1110 1110 1100 0111 0001 1110 0111 1100 0001 1111 0001 0101


The base ten decimal number 384 747 294.484 849 294 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0001 1011 - 0110 1110 1110 1100 0111 0001 1110 0111 1100 0001 1111 0001 0101

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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