Convert 10 011 101.011 010 1 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

10 011 101.011 010 1(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: 10 011 101.
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;
  • 10 011 101 ÷ 2 = 5 005 550 + 1;
  • 5 005 550 ÷ 2 = 2 502 775 + 0;
  • 2 502 775 ÷ 2 = 1 251 387 + 1;
  • 1 251 387 ÷ 2 = 625 693 + 1;
  • 625 693 ÷ 2 = 312 846 + 1;
  • 312 846 ÷ 2 = 156 423 + 0;
  • 156 423 ÷ 2 = 78 211 + 1;
  • 78 211 ÷ 2 = 39 105 + 1;
  • 39 105 ÷ 2 = 19 552 + 1;
  • 19 552 ÷ 2 = 9 776 + 0;
  • 9 776 ÷ 2 = 4 888 + 0;
  • 4 888 ÷ 2 = 2 444 + 0;
  • 2 444 ÷ 2 = 1 222 + 0;
  • 1 222 ÷ 2 = 611 + 0;
  • 611 ÷ 2 = 305 + 1;
  • 305 ÷ 2 = 152 + 1;
  • 152 ÷ 2 = 76 + 0;
  • 76 ÷ 2 = 38 + 0;
  • 38 ÷ 2 = 19 + 0;
  • 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.

10 011 101(10) =


1001 1000 1100 0001 1101 1101(2)


3. Convert to the binary (base 2) the fractional part: 0.011 010 1.

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) 0.011 010 1 × 2 = 0 + 0.022 020 2;
  • 2) 0.022 020 2 × 2 = 0 + 0.044 040 4;
  • 3) 0.044 040 4 × 2 = 0 + 0.088 080 8;
  • 4) 0.088 080 8 × 2 = 0 + 0.176 161 6;
  • 5) 0.176 161 6 × 2 = 0 + 0.352 323 2;
  • 6) 0.352 323 2 × 2 = 0 + 0.704 646 4;
  • 7) 0.704 646 4 × 2 = 1 + 0.409 292 8;
  • 8) 0.409 292 8 × 2 = 0 + 0.818 585 6;
  • 9) 0.818 585 6 × 2 = 1 + 0.637 171 2;
  • 10) 0.637 171 2 × 2 = 1 + 0.274 342 4;
  • 11) 0.274 342 4 × 2 = 0 + 0.548 684 8;
  • 12) 0.548 684 8 × 2 = 1 + 0.097 369 6;
  • 13) 0.097 369 6 × 2 = 0 + 0.194 739 2;
  • 14) 0.194 739 2 × 2 = 0 + 0.389 478 4;
  • 15) 0.389 478 4 × 2 = 0 + 0.778 956 8;
  • 16) 0.778 956 8 × 2 = 1 + 0.557 913 6;
  • 17) 0.557 913 6 × 2 = 1 + 0.115 827 2;
  • 18) 0.115 827 2 × 2 = 0 + 0.231 654 4;
  • 19) 0.231 654 4 × 2 = 0 + 0.463 308 8;
  • 20) 0.463 308 8 × 2 = 0 + 0.926 617 6;
  • 21) 0.926 617 6 × 2 = 1 + 0.853 235 2;
  • 22) 0.853 235 2 × 2 = 1 + 0.706 470 4;
  • 23) 0.706 470 4 × 2 = 1 + 0.412 940 8;
  • 24) 0.412 940 8 × 2 = 0 + 0.825 881 6;

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:

0.011 010 1(10) =


0.0000 0010 1101 0001 1000 1110(2)


5. Positive number before normalization:

10 011 101.011 010 1(10) =


1001 1000 1100 0001 1101 1101.0000 0010 1101 0001 1000 1110(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the left so that only one non zero digit remains to the left of it:

10 011 101.011 010 1(10) =


1001 1000 1100 0001 1101 1101.0000 0010 1101 0001 1000 1110(2) =


1001 1000 1100 0001 1101 1101.0000 0010 1101 0001 1000 1110(2) × 20 =


1.0011 0001 1000 0011 1011 1010 0000 0101 1010 0011 0001 110(2) × 223


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): 23


Mantissa (not normalized):
1.0011 0001 1000 0011 1011 1010 0000 0101 1010 0011 0001 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


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


23 + 2(8-1) - 1 =


(23 + 127)(10) =


150(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;
  • 150 ÷ 2 = 75 + 0;
  • 75 ÷ 2 = 37 + 1;
  • 37 ÷ 2 = 18 + 1;
  • 18 ÷ 2 = 9 + 0;
  • 9 ÷ 2 = 4 + 1;
  • 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) =


150(10) =


1001 0110(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 1000 1100 0001 1101 1101 0000 0010 1101 0001 1000 1110 =


001 1000 1100 0001 1101 1101


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) =
1001 0110


Mantissa (23 bits) =
001 1000 1100 0001 1101 1101


Number 10 011 101.011 010 1 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 1001 0110 - 001 1000 1100 0001 1101 1101

(32 bits IEEE 754)

More operations of this kind:

10 011 101.011 01 = ? ... 10 011 101.011 010 2 = ?


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

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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