0.000 001 73 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 001 73(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 73(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 73.

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 73 × 2 = 0 + 0.000 003 46;
  • 2) 0.000 003 46 × 2 = 0 + 0.000 006 92;
  • 3) 0.000 006 92 × 2 = 0 + 0.000 013 84;
  • 4) 0.000 013 84 × 2 = 0 + 0.000 027 68;
  • 5) 0.000 027 68 × 2 = 0 + 0.000 055 36;
  • 6) 0.000 055 36 × 2 = 0 + 0.000 110 72;
  • 7) 0.000 110 72 × 2 = 0 + 0.000 221 44;
  • 8) 0.000 221 44 × 2 = 0 + 0.000 442 88;
  • 9) 0.000 442 88 × 2 = 0 + 0.000 885 76;
  • 10) 0.000 885 76 × 2 = 0 + 0.001 771 52;
  • 11) 0.001 771 52 × 2 = 0 + 0.003 543 04;
  • 12) 0.003 543 04 × 2 = 0 + 0.007 086 08;
  • 13) 0.007 086 08 × 2 = 0 + 0.014 172 16;
  • 14) 0.014 172 16 × 2 = 0 + 0.028 344 32;
  • 15) 0.028 344 32 × 2 = 0 + 0.056 688 64;
  • 16) 0.056 688 64 × 2 = 0 + 0.113 377 28;
  • 17) 0.113 377 28 × 2 = 0 + 0.226 754 56;
  • 18) 0.226 754 56 × 2 = 0 + 0.453 509 12;
  • 19) 0.453 509 12 × 2 = 0 + 0.907 018 24;
  • 20) 0.907 018 24 × 2 = 1 + 0.814 036 48;
  • 21) 0.814 036 48 × 2 = 1 + 0.628 072 96;
  • 22) 0.628 072 96 × 2 = 1 + 0.256 145 92;
  • 23) 0.256 145 92 × 2 = 0 + 0.512 291 84;
  • 24) 0.512 291 84 × 2 = 1 + 0.024 583 68;
  • 25) 0.024 583 68 × 2 = 0 + 0.049 167 36;
  • 26) 0.049 167 36 × 2 = 0 + 0.098 334 72;
  • 27) 0.098 334 72 × 2 = 0 + 0.196 669 44;
  • 28) 0.196 669 44 × 2 = 0 + 0.393 338 88;
  • 29) 0.393 338 88 × 2 = 0 + 0.786 677 76;
  • 30) 0.786 677 76 × 2 = 1 + 0.573 355 52;
  • 31) 0.573 355 52 × 2 = 1 + 0.146 711 04;
  • 32) 0.146 711 04 × 2 = 0 + 0.293 422 08;
  • 33) 0.293 422 08 × 2 = 0 + 0.586 844 16;
  • 34) 0.586 844 16 × 2 = 1 + 0.173 688 32;
  • 35) 0.173 688 32 × 2 = 0 + 0.347 376 64;
  • 36) 0.347 376 64 × 2 = 0 + 0.694 753 28;
  • 37) 0.694 753 28 × 2 = 1 + 0.389 506 56;
  • 38) 0.389 506 56 × 2 = 0 + 0.779 013 12;
  • 39) 0.779 013 12 × 2 = 1 + 0.558 026 24;
  • 40) 0.558 026 24 × 2 = 1 + 0.116 052 48;
  • 41) 0.116 052 48 × 2 = 0 + 0.232 104 96;
  • 42) 0.232 104 96 × 2 = 0 + 0.464 209 92;
  • 43) 0.464 209 92 × 2 = 0 + 0.928 419 84;

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 73(10) =

0.0000 0000 0000 0000 0001 1101 0000 0110 0100 1011 000(2)

5. Positive number before normalization:

0.000 001 73(10) =

0.0000 0000 0000 0000 0001 1101 0000 0110 0100 1011 000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 73(10) =

0.0000 0000 0000 0000 0001 1101 0000 0110 0100 1011 000(2) =

0.0000 0000 0000 0000 0001 1101 0000 0110 0100 1011 000(2) × 20 =

1.1101 0000 0110 0100 1011 000(2) × 2-20


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): -20

Mantissa (not normalized):
1.1101 0000 0110 0100 1011 000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(8-1) - 1 =

(-20 + 127)(10) =

107(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;
  • 107 ÷ 2 = 53 + 1;
  • 53 ÷ 2 = 26 + 1;
  • 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) =

107(10) =

0110 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 1000 0011 0010 0101 1000 =

110 1000 0011 0010 0101 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) =
0110 1011

Mantissa (23 bits) =
110 1000 0011 0010 0101 1000


Decimal number 0.000 001 73 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1011 - 110 1000 0011 0010 0101 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