0.000 000 155 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 155(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
0.000 000 155(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: 0.000 000 155.

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) 0.000 000 155 × 2 = 0 + 0.000 000 31;
  • 2) 0.000 000 31 × 2 = 0 + 0.000 000 62;
  • 3) 0.000 000 62 × 2 = 0 + 0.000 001 24;
  • 4) 0.000 001 24 × 2 = 0 + 0.000 002 48;
  • 5) 0.000 002 48 × 2 = 0 + 0.000 004 96;
  • 6) 0.000 004 96 × 2 = 0 + 0.000 009 92;
  • 7) 0.000 009 92 × 2 = 0 + 0.000 019 84;
  • 8) 0.000 019 84 × 2 = 0 + 0.000 039 68;
  • 9) 0.000 039 68 × 2 = 0 + 0.000 079 36;
  • 10) 0.000 079 36 × 2 = 0 + 0.000 158 72;
  • 11) 0.000 158 72 × 2 = 0 + 0.000 317 44;
  • 12) 0.000 317 44 × 2 = 0 + 0.000 634 88;
  • 13) 0.000 634 88 × 2 = 0 + 0.001 269 76;
  • 14) 0.001 269 76 × 2 = 0 + 0.002 539 52;
  • 15) 0.002 539 52 × 2 = 0 + 0.005 079 04;
  • 16) 0.005 079 04 × 2 = 0 + 0.010 158 08;
  • 17) 0.010 158 08 × 2 = 0 + 0.020 316 16;
  • 18) 0.020 316 16 × 2 = 0 + 0.040 632 32;
  • 19) 0.040 632 32 × 2 = 0 + 0.081 264 64;
  • 20) 0.081 264 64 × 2 = 0 + 0.162 529 28;
  • 21) 0.162 529 28 × 2 = 0 + 0.325 058 56;
  • 22) 0.325 058 56 × 2 = 0 + 0.650 117 12;
  • 23) 0.650 117 12 × 2 = 1 + 0.300 234 24;
  • 24) 0.300 234 24 × 2 = 0 + 0.600 468 48;
  • 25) 0.600 468 48 × 2 = 1 + 0.200 936 96;
  • 26) 0.200 936 96 × 2 = 0 + 0.401 873 92;
  • 27) 0.401 873 92 × 2 = 0 + 0.803 747 84;
  • 28) 0.803 747 84 × 2 = 1 + 0.607 495 68;
  • 29) 0.607 495 68 × 2 = 1 + 0.214 991 36;
  • 30) 0.214 991 36 × 2 = 0 + 0.429 982 72;
  • 31) 0.429 982 72 × 2 = 0 + 0.859 965 44;
  • 32) 0.859 965 44 × 2 = 1 + 0.719 930 88;
  • 33) 0.719 930 88 × 2 = 1 + 0.439 861 76;
  • 34) 0.439 861 76 × 2 = 0 + 0.879 723 52;
  • 35) 0.879 723 52 × 2 = 1 + 0.759 447 04;
  • 36) 0.759 447 04 × 2 = 1 + 0.518 894 08;
  • 37) 0.518 894 08 × 2 = 1 + 0.037 788 16;
  • 38) 0.037 788 16 × 2 = 0 + 0.075 576 32;
  • 39) 0.075 576 32 × 2 = 0 + 0.151 152 64;
  • 40) 0.151 152 64 × 2 = 0 + 0.302 305 28;
  • 41) 0.302 305 28 × 2 = 0 + 0.604 610 56;
  • 42) 0.604 610 56 × 2 = 1 + 0.209 221 12;
  • 43) 0.209 221 12 × 2 = 0 + 0.418 442 24;
  • 44) 0.418 442 24 × 2 = 0 + 0.836 884 48;
  • 45) 0.836 884 48 × 2 = 1 + 0.673 768 96;
  • 46) 0.673 768 96 × 2 = 1 + 0.347 537 92;

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:


0.000 000 155(10) =

0.0000 0000 0000 0000 0000 0010 1001 1001 1011 1000 0100 11(2)

5. Positive number before normalization:

0.000 000 155(10) =

0.0000 0000 0000 0000 0000 0010 1001 1001 1011 1000 0100 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 155(10) =

0.0000 0000 0000 0000 0000 0010 1001 1001 1011 1000 0100 11(2) =

0.0000 0000 0000 0000 0000 0010 1001 1001 1011 1000 0100 11(2) × 20 =

1.0100 1100 1101 1100 0010 011(2) × 2-23


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): -23

Mantissa (not normalized):
1.0100 1100 1101 1100 0010 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(8-1) - 1 =

(-23 + 127)(10) =

104(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;
  • 104 ÷ 2 = 52 + 0;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 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) =

104(10) =

0110 1000(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. 010 0110 0110 1110 0001 0011 =

010 0110 0110 1110 0001 0011


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 1000

Mantissa (23 bits) =
010 0110 0110 1110 0001 0011


Decimal number 0.000 000 155 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1000 - 010 0110 0110 1110 0001 0011

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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