0.000 012 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 012 6(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
0.000 012 6(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: 0.000 012 6.

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) 0.000 012 6 × 2 = 0 + 0.000 025 2;
  • 2) 0.000 025 2 × 2 = 0 + 0.000 050 4;
  • 3) 0.000 050 4 × 2 = 0 + 0.000 100 8;
  • 4) 0.000 100 8 × 2 = 0 + 0.000 201 6;
  • 5) 0.000 201 6 × 2 = 0 + 0.000 403 2;
  • 6) 0.000 403 2 × 2 = 0 + 0.000 806 4;
  • 7) 0.000 806 4 × 2 = 0 + 0.001 612 8;
  • 8) 0.001 612 8 × 2 = 0 + 0.003 225 6;
  • 9) 0.003 225 6 × 2 = 0 + 0.006 451 2;
  • 10) 0.006 451 2 × 2 = 0 + 0.012 902 4;
  • 11) 0.012 902 4 × 2 = 0 + 0.025 804 8;
  • 12) 0.025 804 8 × 2 = 0 + 0.051 609 6;
  • 13) 0.051 609 6 × 2 = 0 + 0.103 219 2;
  • 14) 0.103 219 2 × 2 = 0 + 0.206 438 4;
  • 15) 0.206 438 4 × 2 = 0 + 0.412 876 8;
  • 16) 0.412 876 8 × 2 = 0 + 0.825 753 6;
  • 17) 0.825 753 6 × 2 = 1 + 0.651 507 2;
  • 18) 0.651 507 2 × 2 = 1 + 0.303 014 4;
  • 19) 0.303 014 4 × 2 = 0 + 0.606 028 8;
  • 20) 0.606 028 8 × 2 = 1 + 0.212 057 6;
  • 21) 0.212 057 6 × 2 = 0 + 0.424 115 2;
  • 22) 0.424 115 2 × 2 = 0 + 0.848 230 4;
  • 23) 0.848 230 4 × 2 = 1 + 0.696 460 8;
  • 24) 0.696 460 8 × 2 = 1 + 0.392 921 6;
  • 25) 0.392 921 6 × 2 = 0 + 0.785 843 2;
  • 26) 0.785 843 2 × 2 = 1 + 0.571 686 4;
  • 27) 0.571 686 4 × 2 = 1 + 0.143 372 8;
  • 28) 0.143 372 8 × 2 = 0 + 0.286 745 6;
  • 29) 0.286 745 6 × 2 = 0 + 0.573 491 2;
  • 30) 0.573 491 2 × 2 = 1 + 0.146 982 4;
  • 31) 0.146 982 4 × 2 = 0 + 0.293 964 8;
  • 32) 0.293 964 8 × 2 = 0 + 0.587 929 6;
  • 33) 0.587 929 6 × 2 = 1 + 0.175 859 2;
  • 34) 0.175 859 2 × 2 = 0 + 0.351 718 4;
  • 35) 0.351 718 4 × 2 = 0 + 0.703 436 8;
  • 36) 0.703 436 8 × 2 = 1 + 0.406 873 6;
  • 37) 0.406 873 6 × 2 = 0 + 0.813 747 2;
  • 38) 0.813 747 2 × 2 = 1 + 0.627 494 4;
  • 39) 0.627 494 4 × 2 = 1 + 0.254 988 8;
  • 40) 0.254 988 8 × 2 = 0 + 0.509 977 6;

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:


0.000 012 6(10) =

0.0000 0000 0000 0000 1101 0011 0110 0100 1001 0110(2)

5. Positive number before normalization:

0.000 012 6(10) =

0.0000 0000 0000 0000 1101 0011 0110 0100 1001 0110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 012 6(10) =

0.0000 0000 0000 0000 1101 0011 0110 0100 1001 0110(2) =

0.0000 0000 0000 0000 1101 0011 0110 0100 1001 0110(2) × 20 =

1.1010 0110 1100 1001 0010 110(2) × 2-17


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): -17

Mantissa (not normalized):
1.1010 0110 1100 1001 0010 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(8-1) - 1 =

(-17 + 127)(10) =

110(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;
  • 110 ÷ 2 = 55 + 0;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

110(10) =

0110 1110(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. 101 0011 0110 0100 1001 0110 =

101 0011 0110 0100 1001 0110


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) =
0110 1110

Mantissa (23 bits) =
101 0011 0110 0100 1001 0110


Decimal number 0.000 012 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1110 - 101 0011 0110 0100 1001 0110

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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