32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 47.367 187 7 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 47.367 187 7(10) converted and written in 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: 47.
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;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.


47(10) =


10 1111(2)


3. Convert to binary (base 2) the fractional part: 0.367 187 7.

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) 0.367 187 7 × 2 = 0 + 0.734 375 4;
  • 2) 0.734 375 4 × 2 = 1 + 0.468 750 8;
  • 3) 0.468 750 8 × 2 = 0 + 0.937 501 6;
  • 4) 0.937 501 6 × 2 = 1 + 0.875 003 2;
  • 5) 0.875 003 2 × 2 = 1 + 0.750 006 4;
  • 6) 0.750 006 4 × 2 = 1 + 0.500 012 8;
  • 7) 0.500 012 8 × 2 = 1 + 0.000 025 6;
  • 8) 0.000 025 6 × 2 = 0 + 0.000 051 2;
  • 9) 0.000 051 2 × 2 = 0 + 0.000 102 4;
  • 10) 0.000 102 4 × 2 = 0 + 0.000 204 8;
  • 11) 0.000 204 8 × 2 = 0 + 0.000 409 6;
  • 12) 0.000 409 6 × 2 = 0 + 0.000 819 2;
  • 13) 0.000 819 2 × 2 = 0 + 0.001 638 4;
  • 14) 0.001 638 4 × 2 = 0 + 0.003 276 8;
  • 15) 0.003 276 8 × 2 = 0 + 0.006 553 6;
  • 16) 0.006 553 6 × 2 = 0 + 0.013 107 2;
  • 17) 0.013 107 2 × 2 = 0 + 0.026 214 4;
  • 18) 0.026 214 4 × 2 = 0 + 0.052 428 8;
  • 19) 0.052 428 8 × 2 = 0 + 0.104 857 6;
  • 20) 0.104 857 6 × 2 = 0 + 0.209 715 2;
  • 21) 0.209 715 2 × 2 = 0 + 0.419 430 4;
  • 22) 0.419 430 4 × 2 = 0 + 0.838 860 8;
  • 23) 0.838 860 8 × 2 = 1 + 0.677 721 6;
  • 24) 0.677 721 6 × 2 = 1 + 0.355 443 2;

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...)


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:


0.367 187 7(10) =


0.0101 1110 0000 0000 0000 0011(2)


5. Positive number before normalization:

47.367 187 7(10) =


10 1111.0101 1110 0000 0000 0000 0011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the left, so that only one non zero digit remains to the left of it:


47.367 187 7(10) =


10 1111.0101 1110 0000 0000 0000 0011(2) =


10 1111.0101 1110 0000 0000 0000 0011(2) × 20 =


1.0111 1010 1111 0000 0000 0000 0001 1(2) × 25


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): 5


Mantissa (not normalized):
1.0111 1010 1111 0000 0000 0000 0001 1


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


5 + 2(8-1) - 1 =


(5 + 127)(10) =


132(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;
  • 132 ÷ 2 = 66 + 0;
  • 66 ÷ 2 = 33 + 0;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


132(10) =


1000 0100(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 011 1101 0111 1000 0000 0000 00 0011 =


011 1101 0111 1000 0000 0000


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) =
1000 0100


Mantissa (23 bits) =
011 1101 0111 1000 0000 0000


The base ten decimal number 47.367 187 7 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0100 - 011 1101 0111 1000 0000 0000

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