Convert 2 532.129 83 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

2 532.129 83(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

1. First, convert to the binary (base 2) the integer part: 2 532.
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;
  • 2 532 ÷ 2 = 1 266 + 0;
  • 1 266 ÷ 2 = 633 + 0;
  • 633 ÷ 2 = 316 + 1;
  • 316 ÷ 2 = 158 + 0;
  • 158 ÷ 2 = 79 + 0;
  • 79 ÷ 2 = 39 + 1;
  • 39 ÷ 2 = 19 + 1;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 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.


2 532(10) =

1001 1110 0100(2)


3. Convert to the binary (base 2) the fractional part: 0.129 83.

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) 0.129 83 × 2 = 0 + 0.259 66;
  • 2) 0.259 66 × 2 = 0 + 0.519 32;
  • 3) 0.519 32 × 2 = 1 + 0.038 64;
  • 4) 0.038 64 × 2 = 0 + 0.077 28;
  • 5) 0.077 28 × 2 = 0 + 0.154 56;
  • 6) 0.154 56 × 2 = 0 + 0.309 12;
  • 7) 0.309 12 × 2 = 0 + 0.618 24;
  • 8) 0.618 24 × 2 = 1 + 0.236 48;
  • 9) 0.236 48 × 2 = 0 + 0.472 96;
  • 10) 0.472 96 × 2 = 0 + 0.945 92;
  • 11) 0.945 92 × 2 = 1 + 0.891 84;
  • 12) 0.891 84 × 2 = 1 + 0.783 68;
  • 13) 0.783 68 × 2 = 1 + 0.567 36;
  • 14) 0.567 36 × 2 = 1 + 0.134 72;
  • 15) 0.134 72 × 2 = 0 + 0.269 44;
  • 16) 0.269 44 × 2 = 0 + 0.538 88;
  • 17) 0.538 88 × 2 = 1 + 0.077 76;
  • 18) 0.077 76 × 2 = 0 + 0.155 52;
  • 19) 0.155 52 × 2 = 0 + 0.311 04;
  • 20) 0.311 04 × 2 = 0 + 0.622 08;
  • 21) 0.622 08 × 2 = 1 + 0.244 16;
  • 22) 0.244 16 × 2 = 0 + 0.488 32;
  • 23) 0.488 32 × 2 = 0 + 0.976 64;
  • 24) 0.976 64 × 2 = 1 + 0.953 28;

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:


0.129 83(10) =

0.0010 0001 0011 1100 1000 1001(2)


5. Positive number before normalization:

2 532.129 83(10) =

1001 1110 0100.0010 0001 0011 1100 1000 1001(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the left so that only one non zero digit remains to the left of it:


2 532.129 83(10) =

1001 1110 0100.0010 0001 0011 1100 1000 1001(2) =

1001 1110 0100.0010 0001 0011 1100 1000 1001(2) × 20 =

1.0011 1100 1000 0100 0010 0111 1001 0001 001(2) × 211


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): 11

Mantissa (not normalized):
1.0011 1100 1000 0100 0010 0111 1001 0001 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

11 + 2(8-1) - 1 =

(11 + 127)(10) =

138(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;
  • 138 ÷ 2 = 69 + 0;
  • 69 ÷ 2 = 34 + 1;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 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) =

138(10) =

1000 1010(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. 001 1110 0100 0010 0001 0011 1100 1000 1001 =

001 1110 0100 0010 0001 0011


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 1010

Mantissa (23 bits) =
001 1110 0100 0010 0001 0011


Number 2 532.129 83 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 1000 1010 - 001 1110 0100 0010 0001 0011

(32 bits IEEE 754)

More operations of this kind:

2 532.129 82 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ?

2 532.129 84 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

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

2 532.129 83 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:34 UTC (GMT)
8 192.003 417 961 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:34 UTC (GMT)
406.054 687 7 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:34 UTC (GMT)
1 137 176 603 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:34 UTC (GMT)
0.132 812 3 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:34 UTC (GMT)
440.2 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:34 UTC (GMT)
1.234 567 890 112 342 342 342 323 532 4 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:34 UTC (GMT)
11 000 001 110 101 109 999 999 999 999 978 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:33 UTC (GMT)
639 493 665 029 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:33 UTC (GMT)
16 014 072 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:33 UTC (GMT)
99.026 9 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:33 UTC (GMT)
10 111 110 111 000 999 999 999 999 999 982 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:33 UTC (GMT)
1.111 6 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ? May 29 14:33 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