1.324 234 234 246 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.324 234 234 246(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
1.324 234 234 246(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.324 234 234 246.

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) 0.324 234 234 246 × 2 = 0 + 0.648 468 468 492;
  • 2) 0.648 468 468 492 × 2 = 1 + 0.296 936 936 984;
  • 3) 0.296 936 936 984 × 2 = 0 + 0.593 873 873 968;
  • 4) 0.593 873 873 968 × 2 = 1 + 0.187 747 747 936;
  • 5) 0.187 747 747 936 × 2 = 0 + 0.375 495 495 872;
  • 6) 0.375 495 495 872 × 2 = 0 + 0.750 990 991 744;
  • 7) 0.750 990 991 744 × 2 = 1 + 0.501 981 983 488;
  • 8) 0.501 981 983 488 × 2 = 1 + 0.003 963 966 976;
  • 9) 0.003 963 966 976 × 2 = 0 + 0.007 927 933 952;
  • 10) 0.007 927 933 952 × 2 = 0 + 0.015 855 867 904;
  • 11) 0.015 855 867 904 × 2 = 0 + 0.031 711 735 808;
  • 12) 0.031 711 735 808 × 2 = 0 + 0.063 423 471 616;
  • 13) 0.063 423 471 616 × 2 = 0 + 0.126 846 943 232;
  • 14) 0.126 846 943 232 × 2 = 0 + 0.253 693 886 464;
  • 15) 0.253 693 886 464 × 2 = 0 + 0.507 387 772 928;
  • 16) 0.507 387 772 928 × 2 = 1 + 0.014 775 545 856;
  • 17) 0.014 775 545 856 × 2 = 0 + 0.029 551 091 712;
  • 18) 0.029 551 091 712 × 2 = 0 + 0.059 102 183 424;
  • 19) 0.059 102 183 424 × 2 = 0 + 0.118 204 366 848;
  • 20) 0.118 204 366 848 × 2 = 0 + 0.236 408 733 696;
  • 21) 0.236 408 733 696 × 2 = 0 + 0.472 817 467 392;
  • 22) 0.472 817 467 392 × 2 = 0 + 0.945 634 934 784;
  • 23) 0.945 634 934 784 × 2 = 1 + 0.891 269 869 568;
  • 24) 0.891 269 869 568 × 2 = 1 + 0.782 539 739 136;
  • 25) 0.782 539 739 136 × 2 = 1 + 0.565 079 478 272;
  • 26) 0.565 079 478 272 × 2 = 1 + 0.130 158 956 544;
  • 27) 0.130 158 956 544 × 2 = 0 + 0.260 317 913 088;
  • 28) 0.260 317 913 088 × 2 = 0 + 0.520 635 826 176;
  • 29) 0.520 635 826 176 × 2 = 1 + 0.041 271 652 352;
  • 30) 0.041 271 652 352 × 2 = 0 + 0.082 543 304 704;
  • 31) 0.082 543 304 704 × 2 = 0 + 0.165 086 609 408;
  • 32) 0.165 086 609 408 × 2 = 0 + 0.330 173 218 816;
  • 33) 0.330 173 218 816 × 2 = 0 + 0.660 346 437 632;
  • 34) 0.660 346 437 632 × 2 = 1 + 0.320 692 875 264;
  • 35) 0.320 692 875 264 × 2 = 0 + 0.641 385 750 528;
  • 36) 0.641 385 750 528 × 2 = 1 + 0.282 771 501 056;
  • 37) 0.282 771 501 056 × 2 = 0 + 0.565 543 002 112;
  • 38) 0.565 543 002 112 × 2 = 1 + 0.131 086 004 224;
  • 39) 0.131 086 004 224 × 2 = 0 + 0.262 172 008 448;
  • 40) 0.262 172 008 448 × 2 = 0 + 0.524 344 016 896;
  • 41) 0.524 344 016 896 × 2 = 1 + 0.048 688 033 792;
  • 42) 0.048 688 033 792 × 2 = 0 + 0.097 376 067 584;
  • 43) 0.097 376 067 584 × 2 = 0 + 0.194 752 135 168;
  • 44) 0.194 752 135 168 × 2 = 0 + 0.389 504 270 336;
  • 45) 0.389 504 270 336 × 2 = 0 + 0.779 008 540 672;
  • 46) 0.779 008 540 672 × 2 = 1 + 0.558 017 081 344;
  • 47) 0.558 017 081 344 × 2 = 1 + 0.116 034 162 688;
  • 48) 0.116 034 162 688 × 2 = 0 + 0.232 068 325 376;
  • 49) 0.232 068 325 376 × 2 = 0 + 0.464 136 650 752;
  • 50) 0.464 136 650 752 × 2 = 0 + 0.928 273 301 504;
  • 51) 0.928 273 301 504 × 2 = 1 + 0.856 546 603 008;
  • 52) 0.856 546 603 008 × 2 = 1 + 0.713 093 206 016;
  • 53) 0.713 093 206 016 × 2 = 1 + 0.426 186 412 032;

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:


0.324 234 234 246(10) =

0.0101 0011 0000 0001 0000 0011 1100 1000 0101 0100 1000 0110 0011 1(2)

5. Positive number before normalization:

1.324 234 234 246(10) =

1.0101 0011 0000 0001 0000 0011 1100 1000 0101 0100 1000 0110 0011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.324 234 234 246(10) =

1.0101 0011 0000 0001 0000 0011 1100 1000 0101 0100 1000 0110 0011 1(2) =

1.0101 0011 0000 0001 0000 0011 1100 1000 0101 0100 1000 0110 0011 1(2) × 20


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): 0

Mantissa (not normalized):
1.0101 0011 0000 0001 0000 0011 1100 1000 0101 0100 1000 0110 0011 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 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) =

1023(10) =

011 1111 1111(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. 0101 0011 0000 0001 0000 0011 1100 1000 0101 0100 1000 0110 0011 1 =

0101 0011 0000 0001 0000 0011 1100 1000 0101 0100 1000 0110 0011


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 1111

Mantissa (52 bits) =
0101 0011 0000 0001 0000 0011 1100 1000 0101 0100 1000 0110 0011


Decimal number 1.324 234 234 246 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0101 0011 0000 0001 0000 0011 1100 1000 0101 0100 1000 0110 0011

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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