32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 110 010 100 011.011 101 02 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 110 010 100 011.011 101 02(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: 110 010 100 011.
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;
  • 110 010 100 011 ÷ 2 = 55 005 050 005 + 1;
  • 55 005 050 005 ÷ 2 = 27 502 525 002 + 1;
  • 27 502 525 002 ÷ 2 = 13 751 262 501 + 0;
  • 13 751 262 501 ÷ 2 = 6 875 631 250 + 1;
  • 6 875 631 250 ÷ 2 = 3 437 815 625 + 0;
  • 3 437 815 625 ÷ 2 = 1 718 907 812 + 1;
  • 1 718 907 812 ÷ 2 = 859 453 906 + 0;
  • 859 453 906 ÷ 2 = 429 726 953 + 0;
  • 429 726 953 ÷ 2 = 214 863 476 + 1;
  • 214 863 476 ÷ 2 = 107 431 738 + 0;
  • 107 431 738 ÷ 2 = 53 715 869 + 0;
  • 53 715 869 ÷ 2 = 26 857 934 + 1;
  • 26 857 934 ÷ 2 = 13 428 967 + 0;
  • 13 428 967 ÷ 2 = 6 714 483 + 1;
  • 6 714 483 ÷ 2 = 3 357 241 + 1;
  • 3 357 241 ÷ 2 = 1 678 620 + 1;
  • 1 678 620 ÷ 2 = 839 310 + 0;
  • 839 310 ÷ 2 = 419 655 + 0;
  • 419 655 ÷ 2 = 209 827 + 1;
  • 209 827 ÷ 2 = 104 913 + 1;
  • 104 913 ÷ 2 = 52 456 + 1;
  • 52 456 ÷ 2 = 26 228 + 0;
  • 26 228 ÷ 2 = 13 114 + 0;
  • 13 114 ÷ 2 = 6 557 + 0;
  • 6 557 ÷ 2 = 3 278 + 1;
  • 3 278 ÷ 2 = 1 639 + 0;
  • 1 639 ÷ 2 = 819 + 1;
  • 819 ÷ 2 = 409 + 1;
  • 409 ÷ 2 = 204 + 1;
  • 204 ÷ 2 = 102 + 0;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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.


110 010 100 011(10) =

1 1001 1001 1101 0001 1100 1110 1001 0010 1011(2)


3. Convert to binary (base 2) the fractional part: 0.011 101 02.

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) 0.011 101 02 × 2 = 0 + 0.022 202 04;
  • 2) 0.022 202 04 × 2 = 0 + 0.044 404 08;
  • 3) 0.044 404 08 × 2 = 0 + 0.088 808 16;
  • 4) 0.088 808 16 × 2 = 0 + 0.177 616 32;
  • 5) 0.177 616 32 × 2 = 0 + 0.355 232 64;
  • 6) 0.355 232 64 × 2 = 0 + 0.710 465 28;
  • 7) 0.710 465 28 × 2 = 1 + 0.420 930 56;
  • 8) 0.420 930 56 × 2 = 0 + 0.841 861 12;
  • 9) 0.841 861 12 × 2 = 1 + 0.683 722 24;
  • 10) 0.683 722 24 × 2 = 1 + 0.367 444 48;
  • 11) 0.367 444 48 × 2 = 0 + 0.734 888 96;
  • 12) 0.734 888 96 × 2 = 1 + 0.469 777 92;
  • 13) 0.469 777 92 × 2 = 0 + 0.939 555 84;
  • 14) 0.939 555 84 × 2 = 1 + 0.879 111 68;
  • 15) 0.879 111 68 × 2 = 1 + 0.758 223 36;
  • 16) 0.758 223 36 × 2 = 1 + 0.516 446 72;
  • 17) 0.516 446 72 × 2 = 1 + 0.032 893 44;
  • 18) 0.032 893 44 × 2 = 0 + 0.065 786 88;
  • 19) 0.065 786 88 × 2 = 0 + 0.131 573 76;
  • 20) 0.131 573 76 × 2 = 0 + 0.263 147 52;
  • 21) 0.263 147 52 × 2 = 0 + 0.526 295 04;
  • 22) 0.526 295 04 × 2 = 1 + 0.052 590 08;
  • 23) 0.052 590 08 × 2 = 0 + 0.105 180 16;
  • 24) 0.105 180 16 × 2 = 0 + 0.210 360 32;

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:


0.011 101 02(10) =

0.0000 0010 1101 0111 1000 0100(2)


5. Positive number before normalization:

110 010 100 011.011 101 02(10) =

1 1001 1001 1101 0001 1100 1110 1001 0010 1011.0000 0010 1101 0111 1000 0100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 36 positions to the left, so that only one non zero digit remains to the left of it:


110 010 100 011.011 101 02(10) =

1 1001 1001 1101 0001 1100 1110 1001 0010 1011.0000 0010 1101 0111 1000 0100(2) =

1 1001 1001 1101 0001 1100 1110 1001 0010 1011.0000 0010 1101 0111 1000 0100(2) × 20 =

1.1001 1001 1101 0001 1100 1110 1001 0010 1011 0000 0010 1101 0111 1000 0100(2) × 236


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): 36

Mantissa (not normalized):
1.1001 1001 1101 0001 1100 1110 1001 0010 1011 0000 0010 1101 0111 1000 0100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

36 + 2(8-1) - 1 =

(36 + 127)(10) =

163(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;
  • 163 ÷ 2 = 81 + 1;
  • 81 ÷ 2 = 40 + 1;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 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) =

163(10) =

1010 0011(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. 100 1100 1110 1000 1110 0111 0 1001 0010 1011 0000 0010 1101 0111 1000 0100 =

100 1100 1110 1000 1110 0111


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) =
1010 0011

Mantissa (23 bits) =
100 1100 1110 1000 1110 0111


The base ten decimal number 110 010 100 011.011 101 02 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1010 0011 - 100 1100 1110 1000 1110 0111

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

» Number 110 010 100 011.011 101 01 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