0.126 380 952 43 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.126 380 952 43(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
0.126 380 952 43(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: 0.126 380 952 43.

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) 0.126 380 952 43 × 2 = 0 + 0.252 761 904 86;
  • 2) 0.252 761 904 86 × 2 = 0 + 0.505 523 809 72;
  • 3) 0.505 523 809 72 × 2 = 1 + 0.011 047 619 44;
  • 4) 0.011 047 619 44 × 2 = 0 + 0.022 095 238 88;
  • 5) 0.022 095 238 88 × 2 = 0 + 0.044 190 477 76;
  • 6) 0.044 190 477 76 × 2 = 0 + 0.088 380 955 52;
  • 7) 0.088 380 955 52 × 2 = 0 + 0.176 761 911 04;
  • 8) 0.176 761 911 04 × 2 = 0 + 0.353 523 822 08;
  • 9) 0.353 523 822 08 × 2 = 0 + 0.707 047 644 16;
  • 10) 0.707 047 644 16 × 2 = 1 + 0.414 095 288 32;
  • 11) 0.414 095 288 32 × 2 = 0 + 0.828 190 576 64;
  • 12) 0.828 190 576 64 × 2 = 1 + 0.656 381 153 28;
  • 13) 0.656 381 153 28 × 2 = 1 + 0.312 762 306 56;
  • 14) 0.312 762 306 56 × 2 = 0 + 0.625 524 613 12;
  • 15) 0.625 524 613 12 × 2 = 1 + 0.251 049 226 24;
  • 16) 0.251 049 226 24 × 2 = 0 + 0.502 098 452 48;
  • 17) 0.502 098 452 48 × 2 = 1 + 0.004 196 904 96;
  • 18) 0.004 196 904 96 × 2 = 0 + 0.008 393 809 92;
  • 19) 0.008 393 809 92 × 2 = 0 + 0.016 787 619 84;
  • 20) 0.016 787 619 84 × 2 = 0 + 0.033 575 239 68;
  • 21) 0.033 575 239 68 × 2 = 0 + 0.067 150 479 36;
  • 22) 0.067 150 479 36 × 2 = 0 + 0.134 300 958 72;
  • 23) 0.134 300 958 72 × 2 = 0 + 0.268 601 917 44;
  • 24) 0.268 601 917 44 × 2 = 0 + 0.537 203 834 88;
  • 25) 0.537 203 834 88 × 2 = 1 + 0.074 407 669 76;
  • 26) 0.074 407 669 76 × 2 = 0 + 0.148 815 339 52;
  • 27) 0.148 815 339 52 × 2 = 0 + 0.297 630 679 04;
  • 28) 0.297 630 679 04 × 2 = 0 + 0.595 261 358 08;
  • 29) 0.595 261 358 08 × 2 = 1 + 0.190 522 716 16;
  • 30) 0.190 522 716 16 × 2 = 0 + 0.381 045 432 32;
  • 31) 0.381 045 432 32 × 2 = 0 + 0.762 090 864 64;
  • 32) 0.762 090 864 64 × 2 = 1 + 0.524 181 729 28;
  • 33) 0.524 181 729 28 × 2 = 1 + 0.048 363 458 56;
  • 34) 0.048 363 458 56 × 2 = 0 + 0.096 726 917 12;
  • 35) 0.096 726 917 12 × 2 = 0 + 0.193 453 834 24;
  • 36) 0.193 453 834 24 × 2 = 0 + 0.386 907 668 48;
  • 37) 0.386 907 668 48 × 2 = 0 + 0.773 815 336 96;
  • 38) 0.773 815 336 96 × 2 = 1 + 0.547 630 673 92;
  • 39) 0.547 630 673 92 × 2 = 1 + 0.095 261 347 84;
  • 40) 0.095 261 347 84 × 2 = 0 + 0.190 522 695 68;
  • 41) 0.190 522 695 68 × 2 = 0 + 0.381 045 391 36;
  • 42) 0.381 045 391 36 × 2 = 0 + 0.762 090 782 72;
  • 43) 0.762 090 782 72 × 2 = 1 + 0.524 181 565 44;
  • 44) 0.524 181 565 44 × 2 = 1 + 0.048 363 130 88;
  • 45) 0.048 363 130 88 × 2 = 0 + 0.096 726 261 76;
  • 46) 0.096 726 261 76 × 2 = 0 + 0.193 452 523 52;
  • 47) 0.193 452 523 52 × 2 = 0 + 0.386 905 047 04;
  • 48) 0.386 905 047 04 × 2 = 0 + 0.773 810 094 08;
  • 49) 0.773 810 094 08 × 2 = 1 + 0.547 620 188 16;
  • 50) 0.547 620 188 16 × 2 = 1 + 0.095 240 376 32;
  • 51) 0.095 240 376 32 × 2 = 0 + 0.190 480 752 64;
  • 52) 0.190 480 752 64 × 2 = 0 + 0.380 961 505 28;
  • 53) 0.380 961 505 28 × 2 = 0 + 0.761 923 010 56;
  • 54) 0.761 923 010 56 × 2 = 1 + 0.523 846 021 12;
  • 55) 0.523 846 021 12 × 2 = 1 + 0.047 692 042 24;

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:


0.126 380 952 43(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1000 0110 0011 0000 1100 011(2)

5. Positive number before normalization:

0.126 380 952 43(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1000 0110 0011 0000 1100 011(2)

6. Normalize the binary representation of the number.

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


0.126 380 952 43(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1000 0110 0011 0000 1100 011(2) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1000 0110 0011 0000 1100 011(2) × 20 =

1.0000 0010 1101 0100 0000 0100 0100 1100 0011 0001 1000 0110 0011(2) × 2-3


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 0010 1101 0100 0000 0100 0100 1100 0011 0001 1000 0110 0011


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) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 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) =

1020(10) =

011 1111 1100(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. 0000 0010 1101 0100 0000 0100 0100 1100 0011 0001 1000 0110 0011 =

0000 0010 1101 0100 0000 0100 0100 1100 0011 0001 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 1100

Mantissa (52 bits) =
0000 0010 1101 0100 0000 0100 0100 1100 0011 0001 1000 0110 0011


Decimal number 0.126 380 952 43 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1100 - 0000 0010 1101 0100 0000 0100 0100 1100 0011 0001 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