0.026 916 53 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.026 916 53(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
0.026 916 53(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: 0.026 916 53.

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) 0.026 916 53 × 2 = 0 + 0.053 833 06;
  • 2) 0.053 833 06 × 2 = 0 + 0.107 666 12;
  • 3) 0.107 666 12 × 2 = 0 + 0.215 332 24;
  • 4) 0.215 332 24 × 2 = 0 + 0.430 664 48;
  • 5) 0.430 664 48 × 2 = 0 + 0.861 328 96;
  • 6) 0.861 328 96 × 2 = 1 + 0.722 657 92;
  • 7) 0.722 657 92 × 2 = 1 + 0.445 315 84;
  • 8) 0.445 315 84 × 2 = 0 + 0.890 631 68;
  • 9) 0.890 631 68 × 2 = 1 + 0.781 263 36;
  • 10) 0.781 263 36 × 2 = 1 + 0.562 526 72;
  • 11) 0.562 526 72 × 2 = 1 + 0.125 053 44;
  • 12) 0.125 053 44 × 2 = 0 + 0.250 106 88;
  • 13) 0.250 106 88 × 2 = 0 + 0.500 213 76;
  • 14) 0.500 213 76 × 2 = 1 + 0.000 427 52;
  • 15) 0.000 427 52 × 2 = 0 + 0.000 855 04;
  • 16) 0.000 855 04 × 2 = 0 + 0.001 710 08;
  • 17) 0.001 710 08 × 2 = 0 + 0.003 420 16;
  • 18) 0.003 420 16 × 2 = 0 + 0.006 840 32;
  • 19) 0.006 840 32 × 2 = 0 + 0.013 680 64;
  • 20) 0.013 680 64 × 2 = 0 + 0.027 361 28;
  • 21) 0.027 361 28 × 2 = 0 + 0.054 722 56;
  • 22) 0.054 722 56 × 2 = 0 + 0.109 445 12;
  • 23) 0.109 445 12 × 2 = 0 + 0.218 890 24;
  • 24) 0.218 890 24 × 2 = 0 + 0.437 780 48;
  • 25) 0.437 780 48 × 2 = 0 + 0.875 560 96;
  • 26) 0.875 560 96 × 2 = 1 + 0.751 121 92;
  • 27) 0.751 121 92 × 2 = 1 + 0.502 243 84;
  • 28) 0.502 243 84 × 2 = 1 + 0.004 487 68;
  • 29) 0.004 487 68 × 2 = 0 + 0.008 975 36;
  • 30) 0.008 975 36 × 2 = 0 + 0.017 950 72;
  • 31) 0.017 950 72 × 2 = 0 + 0.035 901 44;
  • 32) 0.035 901 44 × 2 = 0 + 0.071 802 88;
  • 33) 0.071 802 88 × 2 = 0 + 0.143 605 76;
  • 34) 0.143 605 76 × 2 = 0 + 0.287 211 52;
  • 35) 0.287 211 52 × 2 = 0 + 0.574 423 04;
  • 36) 0.574 423 04 × 2 = 1 + 0.148 846 08;
  • 37) 0.148 846 08 × 2 = 0 + 0.297 692 16;
  • 38) 0.297 692 16 × 2 = 0 + 0.595 384 32;
  • 39) 0.595 384 32 × 2 = 1 + 0.190 768 64;
  • 40) 0.190 768 64 × 2 = 0 + 0.381 537 28;
  • 41) 0.381 537 28 × 2 = 0 + 0.763 074 56;
  • 42) 0.763 074 56 × 2 = 1 + 0.526 149 12;
  • 43) 0.526 149 12 × 2 = 1 + 0.052 298 24;
  • 44) 0.052 298 24 × 2 = 0 + 0.104 596 48;
  • 45) 0.104 596 48 × 2 = 0 + 0.209 192 96;
  • 46) 0.209 192 96 × 2 = 0 + 0.418 385 92;
  • 47) 0.418 385 92 × 2 = 0 + 0.836 771 84;
  • 48) 0.836 771 84 × 2 = 1 + 0.673 543 68;
  • 49) 0.673 543 68 × 2 = 1 + 0.347 087 36;
  • 50) 0.347 087 36 × 2 = 0 + 0.694 174 72;
  • 51) 0.694 174 72 × 2 = 1 + 0.388 349 44;
  • 52) 0.388 349 44 × 2 = 0 + 0.776 698 88;
  • 53) 0.776 698 88 × 2 = 1 + 0.553 397 76;
  • 54) 0.553 397 76 × 2 = 1 + 0.106 795 52;
  • 55) 0.106 795 52 × 2 = 0 + 0.213 591 04;
  • 56) 0.213 591 04 × 2 = 0 + 0.427 182 08;
  • 57) 0.427 182 08 × 2 = 0 + 0.854 364 16;
  • 58) 0.854 364 16 × 2 = 1 + 0.708 728 32;

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:


0.026 916 53(10) =

0.0000 0110 1110 0100 0000 0000 0111 0000 0001 0010 0110 0001 1010 1100 01(2)

5. Positive number before normalization:

0.026 916 53(10) =

0.0000 0110 1110 0100 0000 0000 0111 0000 0001 0010 0110 0001 1010 1100 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.026 916 53(10) =

0.0000 0110 1110 0100 0000 0000 0111 0000 0001 0010 0110 0001 1010 1100 01(2) =

0.0000 0110 1110 0100 0000 0000 0111 0000 0001 0010 0110 0001 1010 1100 01(2) × 20 =

1.1011 1001 0000 0000 0001 1100 0000 0100 1001 1000 0110 1011 0001(2) × 2-6


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): -6

Mantissa (not normalized):
1.1011 1001 0000 0000 0001 1100 0000 0100 1001 1000 0110 1011 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 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) =

1017(10) =

011 1111 1001(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. 1011 1001 0000 0000 0001 1100 0000 0100 1001 1000 0110 1011 0001 =

1011 1001 0000 0000 0001 1100 0000 0100 1001 1000 0110 1011 0001


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 1001

Mantissa (52 bits) =
1011 1001 0000 0000 0001 1100 0000 0100 1001 1000 0110 1011 0001


Decimal number 0.026 916 53 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1001 - 1011 1001 0000 0000 0001 1100 0000 0100 1001 1000 0110 1011 0001

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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