0.000 000 023 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 023 1(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 000 023 1(10) to 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: 0.000 000 023 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) 0.000 000 023 1 × 2 = 0 + 0.000 000 046 2;
  • 2) 0.000 000 046 2 × 2 = 0 + 0.000 000 092 4;
  • 3) 0.000 000 092 4 × 2 = 0 + 0.000 000 184 8;
  • 4) 0.000 000 184 8 × 2 = 0 + 0.000 000 369 6;
  • 5) 0.000 000 369 6 × 2 = 0 + 0.000 000 739 2;
  • 6) 0.000 000 739 2 × 2 = 0 + 0.000 001 478 4;
  • 7) 0.000 001 478 4 × 2 = 0 + 0.000 002 956 8;
  • 8) 0.000 002 956 8 × 2 = 0 + 0.000 005 913 6;
  • 9) 0.000 005 913 6 × 2 = 0 + 0.000 011 827 2;
  • 10) 0.000 011 827 2 × 2 = 0 + 0.000 023 654 4;
  • 11) 0.000 023 654 4 × 2 = 0 + 0.000 047 308 8;
  • 12) 0.000 047 308 8 × 2 = 0 + 0.000 094 617 6;
  • 13) 0.000 094 617 6 × 2 = 0 + 0.000 189 235 2;
  • 14) 0.000 189 235 2 × 2 = 0 + 0.000 378 470 4;
  • 15) 0.000 378 470 4 × 2 = 0 + 0.000 756 940 8;
  • 16) 0.000 756 940 8 × 2 = 0 + 0.001 513 881 6;
  • 17) 0.001 513 881 6 × 2 = 0 + 0.003 027 763 2;
  • 18) 0.003 027 763 2 × 2 = 0 + 0.006 055 526 4;
  • 19) 0.006 055 526 4 × 2 = 0 + 0.012 111 052 8;
  • 20) 0.012 111 052 8 × 2 = 0 + 0.024 222 105 6;
  • 21) 0.024 222 105 6 × 2 = 0 + 0.048 444 211 2;
  • 22) 0.048 444 211 2 × 2 = 0 + 0.096 888 422 4;
  • 23) 0.096 888 422 4 × 2 = 0 + 0.193 776 844 8;
  • 24) 0.193 776 844 8 × 2 = 0 + 0.387 553 689 6;
  • 25) 0.387 553 689 6 × 2 = 0 + 0.775 107 379 2;
  • 26) 0.775 107 379 2 × 2 = 1 + 0.550 214 758 4;
  • 27) 0.550 214 758 4 × 2 = 1 + 0.100 429 516 8;
  • 28) 0.100 429 516 8 × 2 = 0 + 0.200 859 033 6;
  • 29) 0.200 859 033 6 × 2 = 0 + 0.401 718 067 2;
  • 30) 0.401 718 067 2 × 2 = 0 + 0.803 436 134 4;
  • 31) 0.803 436 134 4 × 2 = 1 + 0.606 872 268 8;
  • 32) 0.606 872 268 8 × 2 = 1 + 0.213 744 537 6;
  • 33) 0.213 744 537 6 × 2 = 0 + 0.427 489 075 2;
  • 34) 0.427 489 075 2 × 2 = 0 + 0.854 978 150 4;
  • 35) 0.854 978 150 4 × 2 = 1 + 0.709 956 300 8;
  • 36) 0.709 956 300 8 × 2 = 1 + 0.419 912 601 6;
  • 37) 0.419 912 601 6 × 2 = 0 + 0.839 825 203 2;
  • 38) 0.839 825 203 2 × 2 = 1 + 0.679 650 406 4;
  • 39) 0.679 650 406 4 × 2 = 1 + 0.359 300 812 8;
  • 40) 0.359 300 812 8 × 2 = 0 + 0.718 601 625 6;
  • 41) 0.718 601 625 6 × 2 = 1 + 0.437 203 251 2;
  • 42) 0.437 203 251 2 × 2 = 0 + 0.874 406 502 4;
  • 43) 0.874 406 502 4 × 2 = 1 + 0.748 813 004 8;
  • 44) 0.748 813 004 8 × 2 = 1 + 0.497 626 009 6;
  • 45) 0.497 626 009 6 × 2 = 0 + 0.995 252 019 2;
  • 46) 0.995 252 019 2 × 2 = 1 + 0.990 504 038 4;
  • 47) 0.990 504 038 4 × 2 = 1 + 0.981 008 076 8;
  • 48) 0.981 008 076 8 × 2 = 1 + 0.962 016 153 6;
  • 49) 0.962 016 153 6 × 2 = 1 + 0.924 032 307 2;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.000 000 023 1(10) =

0.0000 0000 0000 0000 0000 0000 0110 0011 0011 0110 1011 0111 1(2)

5. Positive number before normalization:

0.000 000 023 1(10) =

0.0000 0000 0000 0000 0000 0000 0110 0011 0011 0110 1011 0111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 023 1(10) =

0.0000 0000 0000 0000 0000 0000 0110 0011 0011 0110 1011 0111 1(2) =

0.0000 0000 0000 0000 0000 0000 0110 0011 0011 0110 1011 0111 1(2) × 20 =

1.1000 1100 1101 1010 1101 111(2) × 2-26


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): -26

Mantissa (not normalized):
1.1000 1100 1101 1010 1101 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(8-1) - 1 =

(-26 + 127)(10) =

101(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;
  • 101 ÷ 2 = 50 + 1;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

101(10) =

0110 0101(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. 100 0110 0110 1101 0110 1111 =

100 0110 0110 1101 0110 1111


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) =
0110 0101

Mantissa (23 bits) =
100 0110 0110 1101 0110 1111


Decimal number 0.000 000 023 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 100 0110 0110 1101 0110 1111

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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