0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 102 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 102 4(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
0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 102 4(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. 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: 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 102 4.

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) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 102 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 204 8;
  • 2) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 204 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 409 6;
  • 3) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 409 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 819 2;
  • 4) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 819 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 638 4;
  • 5) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 638 4 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 276 8;
  • 6) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 276 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 110 553 6;
  • 7) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 110 553 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 221 107 2;
  • 8) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 221 107 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 442 214 4;
  • 9) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 442 214 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 884 428 8;
  • 10) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 884 428 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 768 857 6;
  • 11) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 768 857 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 537 715 2;
  • 12) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 537 715 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 075 430 4;
  • 13) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 075 430 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 150 860 8;
  • 14) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 150 860 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 301 721 6;
  • 15) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 301 721 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 603 443 2;
  • 16) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 603 443 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 206 886 4;
  • 17) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 206 886 4 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 554 413 772 8;
  • 18) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 554 413 772 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 108 827 545 6;
  • 19) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 108 827 545 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 217 655 091 2;
  • 20) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 217 655 091 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 435 310 182 4;
  • 21) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 435 310 182 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 870 620 364 8;
  • 22) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 870 620 364 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 741 240 729 6;
  • 23) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 741 240 729 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 482 481 459 2;
  • 24) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 482 481 459 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 110 964 962 918 4;
  • 25) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 110 964 962 918 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 221 929 925 836 8;
  • 26) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 221 929 925 836 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 443 859 851 673 6;
  • 27) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 443 859 851 673 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 887 719 703 347 2;

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:


0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 102 4(10) =

0.0001 1100 0111 0001 1100 0111 000(2)

5. Positive number before normalization:

0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 102 4(10) =

0.0001 1100 0111 0001 1100 0111 000(2)

6. Normalize the binary representation of the number.

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


0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 102 4(10) =

0.0001 1100 0111 0001 1100 0111 000(2) =

0.0001 1100 0111 0001 1100 0111 000(2) × 20 =

1.1100 0111 0001 1100 0111 000(2) × 2-4


7. 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 0 (a positive number)

Exponent (unadjusted): -4

Mantissa (not normalized):
1.1100 0111 0001 1100 0111 000


8. 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) =

123(10)


9. 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;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

123(10) =

0111 1011(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 110 0011 1000 1110 0011 1000 =

110 0011 1000 1110 0011 1000


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

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0111 1011

Mantissa (23 bits) =
110 0011 1000 1110 0011 1000


Decimal number 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 102 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1011 - 110 0011 1000 1110 0011 1000

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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