0.000 000 027 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 027 3(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
0.000 000 027 3(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: 0.000 000 027 3.

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) 0.000 000 027 3 × 2 = 0 + 0.000 000 054 6;
  • 2) 0.000 000 054 6 × 2 = 0 + 0.000 000 109 2;
  • 3) 0.000 000 109 2 × 2 = 0 + 0.000 000 218 4;
  • 4) 0.000 000 218 4 × 2 = 0 + 0.000 000 436 8;
  • 5) 0.000 000 436 8 × 2 = 0 + 0.000 000 873 6;
  • 6) 0.000 000 873 6 × 2 = 0 + 0.000 001 747 2;
  • 7) 0.000 001 747 2 × 2 = 0 + 0.000 003 494 4;
  • 8) 0.000 003 494 4 × 2 = 0 + 0.000 006 988 8;
  • 9) 0.000 006 988 8 × 2 = 0 + 0.000 013 977 6;
  • 10) 0.000 013 977 6 × 2 = 0 + 0.000 027 955 2;
  • 11) 0.000 027 955 2 × 2 = 0 + 0.000 055 910 4;
  • 12) 0.000 055 910 4 × 2 = 0 + 0.000 111 820 8;
  • 13) 0.000 111 820 8 × 2 = 0 + 0.000 223 641 6;
  • 14) 0.000 223 641 6 × 2 = 0 + 0.000 447 283 2;
  • 15) 0.000 447 283 2 × 2 = 0 + 0.000 894 566 4;
  • 16) 0.000 894 566 4 × 2 = 0 + 0.001 789 132 8;
  • 17) 0.001 789 132 8 × 2 = 0 + 0.003 578 265 6;
  • 18) 0.003 578 265 6 × 2 = 0 + 0.007 156 531 2;
  • 19) 0.007 156 531 2 × 2 = 0 + 0.014 313 062 4;
  • 20) 0.014 313 062 4 × 2 = 0 + 0.028 626 124 8;
  • 21) 0.028 626 124 8 × 2 = 0 + 0.057 252 249 6;
  • 22) 0.057 252 249 6 × 2 = 0 + 0.114 504 499 2;
  • 23) 0.114 504 499 2 × 2 = 0 + 0.229 008 998 4;
  • 24) 0.229 008 998 4 × 2 = 0 + 0.458 017 996 8;
  • 25) 0.458 017 996 8 × 2 = 0 + 0.916 035 993 6;
  • 26) 0.916 035 993 6 × 2 = 1 + 0.832 071 987 2;
  • 27) 0.832 071 987 2 × 2 = 1 + 0.664 143 974 4;
  • 28) 0.664 143 974 4 × 2 = 1 + 0.328 287 948 8;
  • 29) 0.328 287 948 8 × 2 = 0 + 0.656 575 897 6;
  • 30) 0.656 575 897 6 × 2 = 1 + 0.313 151 795 2;
  • 31) 0.313 151 795 2 × 2 = 0 + 0.626 303 590 4;
  • 32) 0.626 303 590 4 × 2 = 1 + 0.252 607 180 8;
  • 33) 0.252 607 180 8 × 2 = 0 + 0.505 214 361 6;
  • 34) 0.505 214 361 6 × 2 = 1 + 0.010 428 723 2;
  • 35) 0.010 428 723 2 × 2 = 0 + 0.020 857 446 4;
  • 36) 0.020 857 446 4 × 2 = 0 + 0.041 714 892 8;
  • 37) 0.041 714 892 8 × 2 = 0 + 0.083 429 785 6;
  • 38) 0.083 429 785 6 × 2 = 0 + 0.166 859 571 2;
  • 39) 0.166 859 571 2 × 2 = 0 + 0.333 719 142 4;
  • 40) 0.333 719 142 4 × 2 = 0 + 0.667 438 284 8;
  • 41) 0.667 438 284 8 × 2 = 1 + 0.334 876 569 6;
  • 42) 0.334 876 569 6 × 2 = 0 + 0.669 753 139 2;
  • 43) 0.669 753 139 2 × 2 = 1 + 0.339 506 278 4;
  • 44) 0.339 506 278 4 × 2 = 0 + 0.679 012 556 8;
  • 45) 0.679 012 556 8 × 2 = 1 + 0.358 025 113 6;
  • 46) 0.358 025 113 6 × 2 = 0 + 0.716 050 227 2;
  • 47) 0.716 050 227 2 × 2 = 1 + 0.432 100 454 4;
  • 48) 0.432 100 454 4 × 2 = 0 + 0.864 200 908 8;
  • 49) 0.864 200 908 8 × 2 = 1 + 0.728 401 817 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 - 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:


0.000 000 027 3(10) =

0.0000 0000 0000 0000 0000 0000 0111 0101 0100 0000 1010 1010 1(2)

5. Positive number before normalization:

0.000 000 027 3(10) =

0.0000 0000 0000 0000 0000 0000 0111 0101 0100 0000 1010 1010 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:


0.000 000 027 3(10) =

0.0000 0000 0000 0000 0000 0000 0111 0101 0100 0000 1010 1010 1(2) =

0.0000 0000 0000 0000 0000 0000 0111 0101 0100 0000 1010 1010 1(2) × 20 =

1.1101 0101 0000 0010 1010 101(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):
1.1101 0101 0000 0010 1010 101


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. 110 1010 1000 0001 0101 0101 =

110 1010 1000 0001 0101 0101


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) =
110 1010 1000 0001 0101 0101


Decimal number 0.000 000 027 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 110 1010 1000 0001 0101 0101

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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