32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.110 011 001 11 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.110 011 001 11(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: 0.110 011 001 11.

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) 0.110 011 001 11 × 2 = 0 + 0.220 022 002 22;
  • 2) 0.220 022 002 22 × 2 = 0 + 0.440 044 004 44;
  • 3) 0.440 044 004 44 × 2 = 0 + 0.880 088 008 88;
  • 4) 0.880 088 008 88 × 2 = 1 + 0.760 176 017 76;
  • 5) 0.760 176 017 76 × 2 = 1 + 0.520 352 035 52;
  • 6) 0.520 352 035 52 × 2 = 1 + 0.040 704 071 04;
  • 7) 0.040 704 071 04 × 2 = 0 + 0.081 408 142 08;
  • 8) 0.081 408 142 08 × 2 = 0 + 0.162 816 284 16;
  • 9) 0.162 816 284 16 × 2 = 0 + 0.325 632 568 32;
  • 10) 0.325 632 568 32 × 2 = 0 + 0.651 265 136 64;
  • 11) 0.651 265 136 64 × 2 = 1 + 0.302 530 273 28;
  • 12) 0.302 530 273 28 × 2 = 0 + 0.605 060 546 56;
  • 13) 0.605 060 546 56 × 2 = 1 + 0.210 121 093 12;
  • 14) 0.210 121 093 12 × 2 = 0 + 0.420 242 186 24;
  • 15) 0.420 242 186 24 × 2 = 0 + 0.840 484 372 48;
  • 16) 0.840 484 372 48 × 2 = 1 + 0.680 968 744 96;
  • 17) 0.680 968 744 96 × 2 = 1 + 0.361 937 489 92;
  • 18) 0.361 937 489 92 × 2 = 0 + 0.723 874 979 84;
  • 19) 0.723 874 979 84 × 2 = 1 + 0.447 749 959 68;
  • 20) 0.447 749 959 68 × 2 = 0 + 0.895 499 919 36;
  • 21) 0.895 499 919 36 × 2 = 1 + 0.790 999 838 72;
  • 22) 0.790 999 838 72 × 2 = 1 + 0.581 999 677 44;
  • 23) 0.581 999 677 44 × 2 = 1 + 0.163 999 354 88;
  • 24) 0.163 999 354 88 × 2 = 0 + 0.327 998 709 76;
  • 25) 0.327 998 709 76 × 2 = 0 + 0.655 997 419 52;
  • 26) 0.655 997 419 52 × 2 = 1 + 0.311 994 839 04;
  • 27) 0.311 994 839 04 × 2 = 0 + 0.623 989 678 08;

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:


0.110 011 001 11(10) =

0.0001 1100 0010 1001 1010 1110 010(2)


5. Positive number before normalization:

0.110 011 001 11(10) =

0.0001 1100 0010 1001 1010 1110 010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.110 011 001 11(10) =

0.0001 1100 0010 1001 1010 1110 010(2) =

0.0001 1100 0010 1001 1010 1110 010(2) × 20 =

1.1100 0010 1001 1010 1110 010(2) × 2-4


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): -4

Mantissa (not normalized):
1.1100 0010 1001 1010 1110 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(8-1) - 1 =

(-4 + 127)(10) =

123(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;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

123(10) =

0111 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 0001 0100 1101 0111 0010 =

110 0001 0100 1101 0111 0010


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 1011

Mantissa (23 bits) =
110 0001 0100 1101 0111 0010


The base ten decimal number 0.110 011 001 11 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 1011 - 110 0001 0100 1101 0111 0010

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

» Number 0.110 011 001 1 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

» Number 0.110 011 001 12 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

The latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

Number 7.68 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 23:40 UTC (GMT)
Number -68.623 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 23:40 UTC (GMT)
Number 4 592 620 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 23:40 UTC (GMT)
Number 8.331 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 23:40 UTC (GMT)
Number 8.331 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 23:40 UTC (GMT)
Number 125 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 23:40 UTC (GMT)
Number 10 100 111 111 000 011 010 001 009 999 992 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 23:40 UTC (GMT)
Number 2 029.1 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 23:40 UTC (GMT)
Number -1.504 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 23:40 UTC (GMT)
Number 19.270 751 955 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard Apr 18 23:40 UTC (GMT)
All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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