-17.222 222 222 222 222 222 222 57 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -17.222 222 222 222 222 222 222 57(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-17.222 222 222 222 222 222 222 57(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-17.222 222 222 222 222 222 222 57| = 17.222 222 222 222 222 222 222 57


2. First, convert to binary (in base 2) the integer part: 17.
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;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. 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.

17(10) =

1 0001(2)


4. Convert to binary (base 2) the fractional part: 0.222 222 222 222 222 222 222 57.

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) 0.222 222 222 222 222 222 222 57 × 2 = 0 + 0.444 444 444 444 444 444 445 14;
  • 2) 0.444 444 444 444 444 444 445 14 × 2 = 0 + 0.888 888 888 888 888 888 890 28;
  • 3) 0.888 888 888 888 888 888 890 28 × 2 = 1 + 0.777 777 777 777 777 777 780 56;
  • 4) 0.777 777 777 777 777 777 780 56 × 2 = 1 + 0.555 555 555 555 555 555 561 12;
  • 5) 0.555 555 555 555 555 555 561 12 × 2 = 1 + 0.111 111 111 111 111 111 122 24;
  • 6) 0.111 111 111 111 111 111 122 24 × 2 = 0 + 0.222 222 222 222 222 222 244 48;
  • 7) 0.222 222 222 222 222 222 244 48 × 2 = 0 + 0.444 444 444 444 444 444 488 96;
  • 8) 0.444 444 444 444 444 444 488 96 × 2 = 0 + 0.888 888 888 888 888 888 977 92;
  • 9) 0.888 888 888 888 888 888 977 92 × 2 = 1 + 0.777 777 777 777 777 777 955 84;
  • 10) 0.777 777 777 777 777 777 955 84 × 2 = 1 + 0.555 555 555 555 555 555 911 68;
  • 11) 0.555 555 555 555 555 555 911 68 × 2 = 1 + 0.111 111 111 111 111 111 823 36;
  • 12) 0.111 111 111 111 111 111 823 36 × 2 = 0 + 0.222 222 222 222 222 223 646 72;
  • 13) 0.222 222 222 222 222 223 646 72 × 2 = 0 + 0.444 444 444 444 444 447 293 44;
  • 14) 0.444 444 444 444 444 447 293 44 × 2 = 0 + 0.888 888 888 888 888 894 586 88;
  • 15) 0.888 888 888 888 888 894 586 88 × 2 = 1 + 0.777 777 777 777 777 789 173 76;
  • 16) 0.777 777 777 777 777 789 173 76 × 2 = 1 + 0.555 555 555 555 555 578 347 52;
  • 17) 0.555 555 555 555 555 578 347 52 × 2 = 1 + 0.111 111 111 111 111 156 695 04;
  • 18) 0.111 111 111 111 111 156 695 04 × 2 = 0 + 0.222 222 222 222 222 313 390 08;
  • 19) 0.222 222 222 222 222 313 390 08 × 2 = 0 + 0.444 444 444 444 444 626 780 16;
  • 20) 0.444 444 444 444 444 626 780 16 × 2 = 0 + 0.888 888 888 888 889 253 560 32;
  • 21) 0.888 888 888 888 889 253 560 32 × 2 = 1 + 0.777 777 777 777 778 507 120 64;
  • 22) 0.777 777 777 777 778 507 120 64 × 2 = 1 + 0.555 555 555 555 557 014 241 28;
  • 23) 0.555 555 555 555 557 014 241 28 × 2 = 1 + 0.111 111 111 111 114 028 482 56;
  • 24) 0.111 111 111 111 114 028 482 56 × 2 = 0 + 0.222 222 222 222 228 056 965 12;

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).


5. 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:


0.222 222 222 222 222 222 222 57(10) =

0.0011 1000 1110 0011 1000 1110(2)

6. Positive number before normalization:

17.222 222 222 222 222 222 222 57(10) =

1 0001.0011 1000 1110 0011 1000 1110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


17.222 222 222 222 222 222 222 57(10) =

1 0001.0011 1000 1110 0011 1000 1110(2) =

1 0001.0011 1000 1110 0011 1000 1110(2) × 20 =

1.0001 0011 1000 1110 0011 1000 1110(2) × 24


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): 4

Mantissa (not normalized):
1.0001 0011 1000 1110 0011 1000 1110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

4 + 2(8-1) - 1 =

(4 + 127)(10) =

131(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

131(10) =

1000 0011(2)


12. 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 23 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. 000 1001 1100 0111 0001 1100 0 1110 =

000 1001 1100 0111 0001 1100


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
1000 0011

Mantissa (23 bits) =
000 1001 1100 0111 0001 1100


Decimal number -17.222 222 222 222 222 222 222 57 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 1000 0011 - 000 1001 1100 0111 0001 1100

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111