32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -8.451 239 585 876 464 843 9 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -8.451 239 585 876 464 843 9(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. Start with the positive version of the number:

|-8.451 239 585 876 464 843 9| = 8.451 239 585 876 464 843 9

2. First, convert to binary (in base 2) the integer part: 8.
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;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. 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.


8(10) =

1000(2)


4. Convert to binary (base 2) the fractional part: 0.451 239 585 876 464 843 9.

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) 0.451 239 585 876 464 843 9 × 2 = 0 + 0.902 479 171 752 929 687 8;
  • 2) 0.902 479 171 752 929 687 8 × 2 = 1 + 0.804 958 343 505 859 375 6;
  • 3) 0.804 958 343 505 859 375 6 × 2 = 1 + 0.609 916 687 011 718 751 2;
  • 4) 0.609 916 687 011 718 751 2 × 2 = 1 + 0.219 833 374 023 437 502 4;
  • 5) 0.219 833 374 023 437 502 4 × 2 = 0 + 0.439 666 748 046 875 004 8;
  • 6) 0.439 666 748 046 875 004 8 × 2 = 0 + 0.879 333 496 093 750 009 6;
  • 7) 0.879 333 496 093 750 009 6 × 2 = 1 + 0.758 666 992 187 500 019 2;
  • 8) 0.758 666 992 187 500 019 2 × 2 = 1 + 0.517 333 984 375 000 038 4;
  • 9) 0.517 333 984 375 000 038 4 × 2 = 1 + 0.034 667 968 750 000 076 8;
  • 10) 0.034 667 968 750 000 076 8 × 2 = 0 + 0.069 335 937 500 000 153 6;
  • 11) 0.069 335 937 500 000 153 6 × 2 = 0 + 0.138 671 875 000 000 307 2;
  • 12) 0.138 671 875 000 000 307 2 × 2 = 0 + 0.277 343 750 000 000 614 4;
  • 13) 0.277 343 750 000 000 614 4 × 2 = 0 + 0.554 687 500 000 001 228 8;
  • 14) 0.554 687 500 000 001 228 8 × 2 = 1 + 0.109 375 000 000 002 457 6;
  • 15) 0.109 375 000 000 002 457 6 × 2 = 0 + 0.218 750 000 000 004 915 2;
  • 16) 0.218 750 000 000 004 915 2 × 2 = 0 + 0.437 500 000 000 009 830 4;
  • 17) 0.437 500 000 000 009 830 4 × 2 = 0 + 0.875 000 000 000 019 660 8;
  • 18) 0.875 000 000 000 019 660 8 × 2 = 1 + 0.750 000 000 000 039 321 6;
  • 19) 0.750 000 000 000 039 321 6 × 2 = 1 + 0.500 000 000 000 078 643 2;
  • 20) 0.500 000 000 000 078 643 2 × 2 = 1 + 0.000 000 000 000 157 286 4;
  • 21) 0.000 000 000 000 157 286 4 × 2 = 0 + 0.000 000 000 000 314 572 8;
  • 22) 0.000 000 000 000 314 572 8 × 2 = 0 + 0.000 000 000 000 629 145 6;
  • 23) 0.000 000 000 000 629 145 6 × 2 = 0 + 0.000 000 000 001 258 291 2;
  • 24) 0.000 000 000 001 258 291 2 × 2 = 0 + 0.000 000 000 002 516 582 4;

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...)


5. 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:


0.451 239 585 876 464 843 9(10) =

0.0111 0011 1000 0100 0111 0000(2)


6. Positive number before normalization:

8.451 239 585 876 464 843 9(10) =

1000.0111 0011 1000 0100 0111 0000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


8.451 239 585 876 464 843 9(10) =

1000.0111 0011 1000 0100 0111 0000(2) =

1000.0111 0011 1000 0100 0111 0000(2) × 20 =

1.0000 1110 0111 0000 1000 1110 000(2) × 23


8. 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 1 (a negative number)

Exponent (unadjusted): 3

Mantissa (not normalized):
1.0000 1110 0111 0000 1000 1110 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

3 + 2(8-1) - 1 =

(3 + 127)(10) =

130(10)


10. 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;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

130(10) =

1000 0010(2)


12. 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. 000 0111 0011 1000 0100 0111 0000 =

000 0111 0011 1000 0100 0111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
1000 0010

Mantissa (23 bits) =
000 0111 0011 1000 0100 0111


The base ten decimal number -8.451 239 585 876 464 843 9 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 1000 0010 - 000 0111 0011 1000 0100 0111

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

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» Number -8.451 239 585 876 464 843 8 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