32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.333 333 313 465 118 408 203 4 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.333 333 313 465 118 408 203 4(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: 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: 0.333 333 313 465 118 408 203 4.

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) 0.333 333 313 465 118 408 203 4 × 2 = 0 + 0.666 666 626 930 236 816 406 8;
  • 2) 0.666 666 626 930 236 816 406 8 × 2 = 1 + 0.333 333 253 860 473 632 813 6;
  • 3) 0.333 333 253 860 473 632 813 6 × 2 = 0 + 0.666 666 507 720 947 265 627 2;
  • 4) 0.666 666 507 720 947 265 627 2 × 2 = 1 + 0.333 333 015 441 894 531 254 4;
  • 5) 0.333 333 015 441 894 531 254 4 × 2 = 0 + 0.666 666 030 883 789 062 508 8;
  • 6) 0.666 666 030 883 789 062 508 8 × 2 = 1 + 0.333 332 061 767 578 125 017 6;
  • 7) 0.333 332 061 767 578 125 017 6 × 2 = 0 + 0.666 664 123 535 156 250 035 2;
  • 8) 0.666 664 123 535 156 250 035 2 × 2 = 1 + 0.333 328 247 070 312 500 070 4;
  • 9) 0.333 328 247 070 312 500 070 4 × 2 = 0 + 0.666 656 494 140 625 000 140 8;
  • 10) 0.666 656 494 140 625 000 140 8 × 2 = 1 + 0.333 312 988 281 250 000 281 6;
  • 11) 0.333 312 988 281 250 000 281 6 × 2 = 0 + 0.666 625 976 562 500 000 563 2;
  • 12) 0.666 625 976 562 500 000 563 2 × 2 = 1 + 0.333 251 953 125 000 001 126 4;
  • 13) 0.333 251 953 125 000 001 126 4 × 2 = 0 + 0.666 503 906 250 000 002 252 8;
  • 14) 0.666 503 906 250 000 002 252 8 × 2 = 1 + 0.333 007 812 500 000 004 505 6;
  • 15) 0.333 007 812 500 000 004 505 6 × 2 = 0 + 0.666 015 625 000 000 009 011 2;
  • 16) 0.666 015 625 000 000 009 011 2 × 2 = 1 + 0.332 031 250 000 000 018 022 4;
  • 17) 0.332 031 250 000 000 018 022 4 × 2 = 0 + 0.664 062 500 000 000 036 044 8;
  • 18) 0.664 062 500 000 000 036 044 8 × 2 = 1 + 0.328 125 000 000 000 072 089 6;
  • 19) 0.328 125 000 000 000 072 089 6 × 2 = 0 + 0.656 250 000 000 000 144 179 2;
  • 20) 0.656 250 000 000 000 144 179 2 × 2 = 1 + 0.312 500 000 000 000 288 358 4;
  • 21) 0.312 500 000 000 000 288 358 4 × 2 = 0 + 0.625 000 000 000 000 576 716 8;
  • 22) 0.625 000 000 000 000 576 716 8 × 2 = 1 + 0.250 000 000 000 001 153 433 6;
  • 23) 0.250 000 000 000 001 153 433 6 × 2 = 0 + 0.500 000 000 000 002 306 867 2;
  • 24) 0.500 000 000 000 002 306 867 2 × 2 = 1 + 0.000 000 000 000 004 613 734 4;
  • 25) 0.000 000 000 000 004 613 734 4 × 2 = 0 + 0.000 000 000 000 009 227 468 8;

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:


0.333 333 313 465 118 408 203 4(10) =

0.0101 0101 0101 0101 0101 0101 0(2)


5. Positive number before normalization:

0.333 333 313 465 118 408 203 4(10) =

0.0101 0101 0101 0101 0101 0101 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.333 333 313 465 118 408 203 4(10) =

0.0101 0101 0101 0101 0101 0101 0(2) =

0.0101 0101 0101 0101 0101 0101 0(2) × 20 =

1.0101 0101 0101 0101 0101 010(2) × 2-2


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): -2

Mantissa (not normalized):
1.0101 0101 0101 0101 0101 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-2 + 127)(10) =

125(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;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

125(10) =

0111 1101(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. 010 1010 1010 1010 1010 1010 =

010 1010 1010 1010 1010 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) =
0111 1101

Mantissa (23 bits) =
010 1010 1010 1010 1010 1010


The base ten decimal number 0.333 333 313 465 118 408 203 4 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 1101 - 010 1010 1010 1010 1010 1010

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» Number 0.333 333 313 465 118 408 203 3 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

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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