32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 2.236 812 375 42 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 2.236 812 375 42(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: 2.
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;
  • 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.


2(10) =


10(2)


3. Convert to binary (base 2) the fractional part: 0.236 812 375 42.

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) 0.236 812 375 42 × 2 = 0 + 0.473 624 750 84;
  • 2) 0.473 624 750 84 × 2 = 0 + 0.947 249 501 68;
  • 3) 0.947 249 501 68 × 2 = 1 + 0.894 499 003 36;
  • 4) 0.894 499 003 36 × 2 = 1 + 0.788 998 006 72;
  • 5) 0.788 998 006 72 × 2 = 1 + 0.577 996 013 44;
  • 6) 0.577 996 013 44 × 2 = 1 + 0.155 992 026 88;
  • 7) 0.155 992 026 88 × 2 = 0 + 0.311 984 053 76;
  • 8) 0.311 984 053 76 × 2 = 0 + 0.623 968 107 52;
  • 9) 0.623 968 107 52 × 2 = 1 + 0.247 936 215 04;
  • 10) 0.247 936 215 04 × 2 = 0 + 0.495 872 430 08;
  • 11) 0.495 872 430 08 × 2 = 0 + 0.991 744 860 16;
  • 12) 0.991 744 860 16 × 2 = 1 + 0.983 489 720 32;
  • 13) 0.983 489 720 32 × 2 = 1 + 0.966 979 440 64;
  • 14) 0.966 979 440 64 × 2 = 1 + 0.933 958 881 28;
  • 15) 0.933 958 881 28 × 2 = 1 + 0.867 917 762 56;
  • 16) 0.867 917 762 56 × 2 = 1 + 0.735 835 525 12;
  • 17) 0.735 835 525 12 × 2 = 1 + 0.471 671 050 24;
  • 18) 0.471 671 050 24 × 2 = 0 + 0.943 342 100 48;
  • 19) 0.943 342 100 48 × 2 = 1 + 0.886 684 200 96;
  • 20) 0.886 684 200 96 × 2 = 1 + 0.773 368 401 92;
  • 21) 0.773 368 401 92 × 2 = 1 + 0.546 736 803 84;
  • 22) 0.546 736 803 84 × 2 = 1 + 0.093 473 607 68;
  • 23) 0.093 473 607 68 × 2 = 0 + 0.186 947 215 36;
  • 24) 0.186 947 215 36 × 2 = 0 + 0.373 894 430 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...)


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:


0.236 812 375 42(10) =


0.0011 1100 1001 1111 1011 1100(2)


5. Positive number before normalization:

2.236 812 375 42(10) =


10.0011 1100 1001 1111 1011 1100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.236 812 375 42(10) =


10.0011 1100 1001 1111 1011 1100(2) =


10.0011 1100 1001 1111 1011 1100(2) × 20 =


1.0001 1110 0100 1111 1101 1110 0(2) × 21


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): 1


Mantissa (not normalized):
1.0001 1110 0100 1111 1101 1110 0


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


1 + 2(8-1) - 1 =


(1 + 127)(10) =


128(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;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 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) =


128(10) =


1000 0000(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. 000 1111 0010 0111 1110 1111 00 =


000 1111 0010 0111 1110 1111


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 0000


Mantissa (23 bits) =
000 1111 0010 0111 1110 1111


The base ten decimal number 2.236 812 375 42 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0000 - 000 1111 0010 0111 1110 1111

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