0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 012 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 012(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
0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 012(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: 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 012.

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) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 012 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 024;
  • 2) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 024 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 048;
  • 3) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 048 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 096;
  • 4) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 096 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 776 192;
  • 5) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 776 192 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 552 384;
  • 6) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 552 384 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 104 768;
  • 7) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 104 768 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 209 536;
  • 8) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 209 536 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 419 072;
  • 9) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 419 072 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 838 144;
  • 10) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 838 144 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 676 288;
  • 11) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 676 288 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 352 576;
  • 12) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 352 576 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 110 705 152;
  • 13) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 110 705 152 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 221 410 304;
  • 14) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 221 410 304 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 442 820 608;
  • 15) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 442 820 608 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 885 641 216;
  • 16) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 885 641 216 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 771 282 432;
  • 17) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 771 282 432 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 542 564 864;
  • 18) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 542 564 864 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 085 129 728;
  • 19) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 085 129 728 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 170 259 456;
  • 20) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 170 259 456 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 340 518 912;
  • 21) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 340 518 912 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 681 037 824;
  • 22) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 681 037 824 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 362 075 648;
  • 23) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 362 075 648 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 554 724 151 296;
  • 24) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 554 724 151 296 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 109 448 302 592;
  • 25) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 109 448 302 592 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 218 896 605 184;
  • 26) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 218 896 605 184 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 437 793 210 368;
  • 27) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 437 793 210 368 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 875 586 420 736;

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:


0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 012(10) =

0.0001 1100 0111 0001 1100 0111 000(2)

5. Positive number before normalization:

0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 012(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:


0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 012(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 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 012 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1011 - 110 0011 1000 1110 0011 1000

Share

» Convert decimal 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 110 925 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions & Market Value. Gemini Notebook YouTube Video of 6.55 minutes

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