0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 106 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 106 8(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
0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 106 8(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: 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 106 8.

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) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 106 8 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 213 6;
  • 2) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 213 6 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 427 2;
  • 3) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 427 2 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 854 4;
  • 4) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 854 4 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 708 8;
  • 5) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 708 8 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 417 6;
  • 6) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 417 6 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 110 835 2;
  • 7) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 110 835 2 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 221 670 4;
  • 8) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 221 670 4 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 443 340 8;
  • 9) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 443 340 8 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 886 681 6;
  • 10) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 886 681 6 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 773 363 2;
  • 11) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 773 363 2 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 546 726 4;
  • 12) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 546 726 4 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 093 452 8;
  • 13) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 093 452 8 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 186 905 6;
  • 14) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 186 905 6 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 373 811 2;
  • 15) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 373 811 2 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 747 622 4;
  • 16) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 747 622 4 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 495 244 8;
  • 17) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 495 244 8 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 554 990 489 6;
  • 18) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 554 990 489 6 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 109 980 979 2;
  • 19) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 109 980 979 2 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 219 961 958 4;
  • 20) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 219 961 958 4 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 439 923 916 8;
  • 21) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 439 923 916 8 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 879 847 833 6;
  • 22) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 879 847 833 6 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 759 695 667 2;
  • 23) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 759 695 667 2 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 519 391 334 4;
  • 24) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 519 391 334 4 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 038 782 668 8;
  • 25) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 038 782 668 8 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 077 565 337 6;
  • 26) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 077 565 337 6 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 155 130 675 2;
  • 27) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 155 130 675 2 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 310 261 350 4;

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:


0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 106 8(10) =

0.0001 1100 0111 0001 1100 0111 000(2)

5. Positive number before normalization:

0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 106 8(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:


0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 106 8(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 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 106 8 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