0.000 001 93 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 001 93(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
0.000 001 93(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: 0.000 001 93.

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) 0.000 001 93 × 2 = 0 + 0.000 003 86;
  • 2) 0.000 003 86 × 2 = 0 + 0.000 007 72;
  • 3) 0.000 007 72 × 2 = 0 + 0.000 015 44;
  • 4) 0.000 015 44 × 2 = 0 + 0.000 030 88;
  • 5) 0.000 030 88 × 2 = 0 + 0.000 061 76;
  • 6) 0.000 061 76 × 2 = 0 + 0.000 123 52;
  • 7) 0.000 123 52 × 2 = 0 + 0.000 247 04;
  • 8) 0.000 247 04 × 2 = 0 + 0.000 494 08;
  • 9) 0.000 494 08 × 2 = 0 + 0.000 988 16;
  • 10) 0.000 988 16 × 2 = 0 + 0.001 976 32;
  • 11) 0.001 976 32 × 2 = 0 + 0.003 952 64;
  • 12) 0.003 952 64 × 2 = 0 + 0.007 905 28;
  • 13) 0.007 905 28 × 2 = 0 + 0.015 810 56;
  • 14) 0.015 810 56 × 2 = 0 + 0.031 621 12;
  • 15) 0.031 621 12 × 2 = 0 + 0.063 242 24;
  • 16) 0.063 242 24 × 2 = 0 + 0.126 484 48;
  • 17) 0.126 484 48 × 2 = 0 + 0.252 968 96;
  • 18) 0.252 968 96 × 2 = 0 + 0.505 937 92;
  • 19) 0.505 937 92 × 2 = 1 + 0.011 875 84;
  • 20) 0.011 875 84 × 2 = 0 + 0.023 751 68;
  • 21) 0.023 751 68 × 2 = 0 + 0.047 503 36;
  • 22) 0.047 503 36 × 2 = 0 + 0.095 006 72;
  • 23) 0.095 006 72 × 2 = 0 + 0.190 013 44;
  • 24) 0.190 013 44 × 2 = 0 + 0.380 026 88;
  • 25) 0.380 026 88 × 2 = 0 + 0.760 053 76;
  • 26) 0.760 053 76 × 2 = 1 + 0.520 107 52;
  • 27) 0.520 107 52 × 2 = 1 + 0.040 215 04;
  • 28) 0.040 215 04 × 2 = 0 + 0.080 430 08;
  • 29) 0.080 430 08 × 2 = 0 + 0.160 860 16;
  • 30) 0.160 860 16 × 2 = 0 + 0.321 720 32;
  • 31) 0.321 720 32 × 2 = 0 + 0.643 440 64;
  • 32) 0.643 440 64 × 2 = 1 + 0.286 881 28;
  • 33) 0.286 881 28 × 2 = 0 + 0.573 762 56;
  • 34) 0.573 762 56 × 2 = 1 + 0.147 525 12;
  • 35) 0.147 525 12 × 2 = 0 + 0.295 050 24;
  • 36) 0.295 050 24 × 2 = 0 + 0.590 100 48;
  • 37) 0.590 100 48 × 2 = 1 + 0.180 200 96;
  • 38) 0.180 200 96 × 2 = 0 + 0.360 401 92;
  • 39) 0.360 401 92 × 2 = 0 + 0.720 803 84;
  • 40) 0.720 803 84 × 2 = 1 + 0.441 607 68;
  • 41) 0.441 607 68 × 2 = 0 + 0.883 215 36;
  • 42) 0.883 215 36 × 2 = 1 + 0.766 430 72;

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:


0.000 001 93(10) =

0.0000 0000 0000 0000 0010 0000 0110 0001 0100 1001 01(2)

5. Positive number before normalization:

0.000 001 93(10) =

0.0000 0000 0000 0000 0010 0000 0110 0001 0100 1001 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 93(10) =

0.0000 0000 0000 0000 0010 0000 0110 0001 0100 1001 01(2) =

0.0000 0000 0000 0000 0010 0000 0110 0001 0100 1001 01(2) × 20 =

1.0000 0011 0000 1010 0100 101(2) × 2-19


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): -19

Mantissa (not normalized):
1.0000 0011 0000 1010 0100 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-19 + 2(8-1) - 1 =

(-19 + 127)(10) =

108(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;
  • 108 ÷ 2 = 54 + 0;
  • 54 ÷ 2 = 27 + 0;
  • 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) =

108(10) =

0110 1100(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. 000 0001 1000 0101 0010 0101 =

000 0001 1000 0101 0010 0101


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 1100

Mantissa (23 bits) =
000 0001 1000 0101 0010 0101


Decimal number 0.000 001 93 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1100 - 000 0001 1000 0101 0010 0101

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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