0.022 120 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.022 120 7(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
0.022 120 7(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: 0.022 120 7.

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) 0.022 120 7 × 2 = 0 + 0.044 241 4;
  • 2) 0.044 241 4 × 2 = 0 + 0.088 482 8;
  • 3) 0.088 482 8 × 2 = 0 + 0.176 965 6;
  • 4) 0.176 965 6 × 2 = 0 + 0.353 931 2;
  • 5) 0.353 931 2 × 2 = 0 + 0.707 862 4;
  • 6) 0.707 862 4 × 2 = 1 + 0.415 724 8;
  • 7) 0.415 724 8 × 2 = 0 + 0.831 449 6;
  • 8) 0.831 449 6 × 2 = 1 + 0.662 899 2;
  • 9) 0.662 899 2 × 2 = 1 + 0.325 798 4;
  • 10) 0.325 798 4 × 2 = 0 + 0.651 596 8;
  • 11) 0.651 596 8 × 2 = 1 + 0.303 193 6;
  • 12) 0.303 193 6 × 2 = 0 + 0.606 387 2;
  • 13) 0.606 387 2 × 2 = 1 + 0.212 774 4;
  • 14) 0.212 774 4 × 2 = 0 + 0.425 548 8;
  • 15) 0.425 548 8 × 2 = 0 + 0.851 097 6;
  • 16) 0.851 097 6 × 2 = 1 + 0.702 195 2;
  • 17) 0.702 195 2 × 2 = 1 + 0.404 390 4;
  • 18) 0.404 390 4 × 2 = 0 + 0.808 780 8;
  • 19) 0.808 780 8 × 2 = 1 + 0.617 561 6;
  • 20) 0.617 561 6 × 2 = 1 + 0.235 123 2;
  • 21) 0.235 123 2 × 2 = 0 + 0.470 246 4;
  • 22) 0.470 246 4 × 2 = 0 + 0.940 492 8;
  • 23) 0.940 492 8 × 2 = 1 + 0.880 985 6;
  • 24) 0.880 985 6 × 2 = 1 + 0.761 971 2;
  • 25) 0.761 971 2 × 2 = 1 + 0.523 942 4;
  • 26) 0.523 942 4 × 2 = 1 + 0.047 884 8;
  • 27) 0.047 884 8 × 2 = 0 + 0.095 769 6;
  • 28) 0.095 769 6 × 2 = 0 + 0.191 539 2;
  • 29) 0.191 539 2 × 2 = 0 + 0.383 078 4;
  • 30) 0.383 078 4 × 2 = 0 + 0.766 156 8;
  • 31) 0.766 156 8 × 2 = 1 + 0.532 313 6;
  • 32) 0.532 313 6 × 2 = 1 + 0.064 627 2;
  • 33) 0.064 627 2 × 2 = 0 + 0.129 254 4;
  • 34) 0.129 254 4 × 2 = 0 + 0.258 508 8;
  • 35) 0.258 508 8 × 2 = 0 + 0.517 017 6;
  • 36) 0.517 017 6 × 2 = 1 + 0.034 035 2;
  • 37) 0.034 035 2 × 2 = 0 + 0.068 070 4;
  • 38) 0.068 070 4 × 2 = 0 + 0.136 140 8;
  • 39) 0.136 140 8 × 2 = 0 + 0.272 281 6;
  • 40) 0.272 281 6 × 2 = 0 + 0.544 563 2;
  • 41) 0.544 563 2 × 2 = 1 + 0.089 126 4;
  • 42) 0.089 126 4 × 2 = 0 + 0.178 252 8;
  • 43) 0.178 252 8 × 2 = 0 + 0.356 505 6;
  • 44) 0.356 505 6 × 2 = 0 + 0.713 011 2;
  • 45) 0.713 011 2 × 2 = 1 + 0.426 022 4;
  • 46) 0.426 022 4 × 2 = 0 + 0.852 044 8;
  • 47) 0.852 044 8 × 2 = 1 + 0.704 089 6;
  • 48) 0.704 089 6 × 2 = 1 + 0.408 179 2;
  • 49) 0.408 179 2 × 2 = 0 + 0.816 358 4;
  • 50) 0.816 358 4 × 2 = 1 + 0.632 716 8;
  • 51) 0.632 716 8 × 2 = 1 + 0.265 433 6;
  • 52) 0.265 433 6 × 2 = 0 + 0.530 867 2;
  • 53) 0.530 867 2 × 2 = 1 + 0.061 734 4;
  • 54) 0.061 734 4 × 2 = 0 + 0.123 468 8;
  • 55) 0.123 468 8 × 2 = 0 + 0.246 937 6;
  • 56) 0.246 937 6 × 2 = 0 + 0.493 875 2;
  • 57) 0.493 875 2 × 2 = 0 + 0.987 750 4;
  • 58) 0.987 750 4 × 2 = 1 + 0.975 500 8;

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:


0.022 120 7(10) =

0.0000 0101 1010 1001 1011 0011 1100 0011 0001 0000 1000 1011 0110 1000 01(2)

5. Positive number before normalization:

0.022 120 7(10) =

0.0000 0101 1010 1001 1011 0011 1100 0011 0001 0000 1000 1011 0110 1000 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:


0.022 120 7(10) =

0.0000 0101 1010 1001 1011 0011 1100 0011 0001 0000 1000 1011 0110 1000 01(2) =

0.0000 0101 1010 1001 1011 0011 1100 0011 0001 0000 1000 1011 0110 1000 01(2) × 20 =

1.0110 1010 0110 1100 1111 0000 1100 0100 0010 0010 1101 1010 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):
1.0110 1010 0110 1100 1111 0000 1100 0100 0010 0010 1101 1010 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. 0110 1010 0110 1100 1111 0000 1100 0100 0010 0010 1101 1010 0001 =

0110 1010 0110 1100 1111 0000 1100 0100 0010 0010 1101 1010 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) =
0110 1010 0110 1100 1111 0000 1100 0100 0010 0010 1101 1010 0001


Decimal number 0.022 120 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1001 - 0110 1010 0110 1100 1111 0000 1100 0100 0010 0010 1101 1010 0001

Share

» Convert decimal 0.022 112 6 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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