0.000 000 285 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 285(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
0.000 000 285(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: 0.000 000 285.

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) 0.000 000 285 × 2 = 0 + 0.000 000 57;
  • 2) 0.000 000 57 × 2 = 0 + 0.000 001 14;
  • 3) 0.000 001 14 × 2 = 0 + 0.000 002 28;
  • 4) 0.000 002 28 × 2 = 0 + 0.000 004 56;
  • 5) 0.000 004 56 × 2 = 0 + 0.000 009 12;
  • 6) 0.000 009 12 × 2 = 0 + 0.000 018 24;
  • 7) 0.000 018 24 × 2 = 0 + 0.000 036 48;
  • 8) 0.000 036 48 × 2 = 0 + 0.000 072 96;
  • 9) 0.000 072 96 × 2 = 0 + 0.000 145 92;
  • 10) 0.000 145 92 × 2 = 0 + 0.000 291 84;
  • 11) 0.000 291 84 × 2 = 0 + 0.000 583 68;
  • 12) 0.000 583 68 × 2 = 0 + 0.001 167 36;
  • 13) 0.001 167 36 × 2 = 0 + 0.002 334 72;
  • 14) 0.002 334 72 × 2 = 0 + 0.004 669 44;
  • 15) 0.004 669 44 × 2 = 0 + 0.009 338 88;
  • 16) 0.009 338 88 × 2 = 0 + 0.018 677 76;
  • 17) 0.018 677 76 × 2 = 0 + 0.037 355 52;
  • 18) 0.037 355 52 × 2 = 0 + 0.074 711 04;
  • 19) 0.074 711 04 × 2 = 0 + 0.149 422 08;
  • 20) 0.149 422 08 × 2 = 0 + 0.298 844 16;
  • 21) 0.298 844 16 × 2 = 0 + 0.597 688 32;
  • 22) 0.597 688 32 × 2 = 1 + 0.195 376 64;
  • 23) 0.195 376 64 × 2 = 0 + 0.390 753 28;
  • 24) 0.390 753 28 × 2 = 0 + 0.781 506 56;
  • 25) 0.781 506 56 × 2 = 1 + 0.563 013 12;
  • 26) 0.563 013 12 × 2 = 1 + 0.126 026 24;
  • 27) 0.126 026 24 × 2 = 0 + 0.252 052 48;
  • 28) 0.252 052 48 × 2 = 0 + 0.504 104 96;
  • 29) 0.504 104 96 × 2 = 1 + 0.008 209 92;
  • 30) 0.008 209 92 × 2 = 0 + 0.016 419 84;
  • 31) 0.016 419 84 × 2 = 0 + 0.032 839 68;
  • 32) 0.032 839 68 × 2 = 0 + 0.065 679 36;
  • 33) 0.065 679 36 × 2 = 0 + 0.131 358 72;
  • 34) 0.131 358 72 × 2 = 0 + 0.262 717 44;
  • 35) 0.262 717 44 × 2 = 0 + 0.525 434 88;
  • 36) 0.525 434 88 × 2 = 1 + 0.050 869 76;
  • 37) 0.050 869 76 × 2 = 0 + 0.101 739 52;
  • 38) 0.101 739 52 × 2 = 0 + 0.203 479 04;
  • 39) 0.203 479 04 × 2 = 0 + 0.406 958 08;
  • 40) 0.406 958 08 × 2 = 0 + 0.813 916 16;
  • 41) 0.813 916 16 × 2 = 1 + 0.627 832 32;
  • 42) 0.627 832 32 × 2 = 1 + 0.255 664 64;
  • 43) 0.255 664 64 × 2 = 0 + 0.511 329 28;
  • 44) 0.511 329 28 × 2 = 1 + 0.022 658 56;
  • 45) 0.022 658 56 × 2 = 0 + 0.045 317 12;

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:


0.000 000 285(10) =

0.0000 0000 0000 0000 0000 0100 1100 1000 0001 0000 1101 0(2)

5. Positive number before normalization:

0.000 000 285(10) =

0.0000 0000 0000 0000 0000 0100 1100 1000 0001 0000 1101 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 285(10) =

0.0000 0000 0000 0000 0000 0100 1100 1000 0001 0000 1101 0(2) =

0.0000 0000 0000 0000 0000 0100 1100 1000 0001 0000 1101 0(2) × 20 =

1.0011 0010 0000 0100 0011 010(2) × 2-22


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): -22

Mantissa (not normalized):
1.0011 0010 0000 0100 0011 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(8-1) - 1 =

(-22 + 127)(10) =

105(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;
  • 105 ÷ 2 = 52 + 1;
  • 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) =

105(10) =

0110 1001(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. 001 1001 0000 0010 0001 1010 =

001 1001 0000 0010 0001 1010


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 1001

Mantissa (23 bits) =
001 1001 0000 0010 0001 1010


Decimal number 0.000 000 285 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 001 1001 0000 0010 0001 1010

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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