0.000 000 021 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 9(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
0.000 000 021 9(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: 0.000 000 021 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) 0.000 000 021 9 × 2 = 0 + 0.000 000 043 8;
  • 2) 0.000 000 043 8 × 2 = 0 + 0.000 000 087 6;
  • 3) 0.000 000 087 6 × 2 = 0 + 0.000 000 175 2;
  • 4) 0.000 000 175 2 × 2 = 0 + 0.000 000 350 4;
  • 5) 0.000 000 350 4 × 2 = 0 + 0.000 000 700 8;
  • 6) 0.000 000 700 8 × 2 = 0 + 0.000 001 401 6;
  • 7) 0.000 001 401 6 × 2 = 0 + 0.000 002 803 2;
  • 8) 0.000 002 803 2 × 2 = 0 + 0.000 005 606 4;
  • 9) 0.000 005 606 4 × 2 = 0 + 0.000 011 212 8;
  • 10) 0.000 011 212 8 × 2 = 0 + 0.000 022 425 6;
  • 11) 0.000 022 425 6 × 2 = 0 + 0.000 044 851 2;
  • 12) 0.000 044 851 2 × 2 = 0 + 0.000 089 702 4;
  • 13) 0.000 089 702 4 × 2 = 0 + 0.000 179 404 8;
  • 14) 0.000 179 404 8 × 2 = 0 + 0.000 358 809 6;
  • 15) 0.000 358 809 6 × 2 = 0 + 0.000 717 619 2;
  • 16) 0.000 717 619 2 × 2 = 0 + 0.001 435 238 4;
  • 17) 0.001 435 238 4 × 2 = 0 + 0.002 870 476 8;
  • 18) 0.002 870 476 8 × 2 = 0 + 0.005 740 953 6;
  • 19) 0.005 740 953 6 × 2 = 0 + 0.011 481 907 2;
  • 20) 0.011 481 907 2 × 2 = 0 + 0.022 963 814 4;
  • 21) 0.022 963 814 4 × 2 = 0 + 0.045 927 628 8;
  • 22) 0.045 927 628 8 × 2 = 0 + 0.091 855 257 6;
  • 23) 0.091 855 257 6 × 2 = 0 + 0.183 710 515 2;
  • 24) 0.183 710 515 2 × 2 = 0 + 0.367 421 030 4;
  • 25) 0.367 421 030 4 × 2 = 0 + 0.734 842 060 8;
  • 26) 0.734 842 060 8 × 2 = 1 + 0.469 684 121 6;
  • 27) 0.469 684 121 6 × 2 = 0 + 0.939 368 243 2;
  • 28) 0.939 368 243 2 × 2 = 1 + 0.878 736 486 4;
  • 29) 0.878 736 486 4 × 2 = 1 + 0.757 472 972 8;
  • 30) 0.757 472 972 8 × 2 = 1 + 0.514 945 945 6;
  • 31) 0.514 945 945 6 × 2 = 1 + 0.029 891 891 2;
  • 32) 0.029 891 891 2 × 2 = 0 + 0.059 783 782 4;
  • 33) 0.059 783 782 4 × 2 = 0 + 0.119 567 564 8;
  • 34) 0.119 567 564 8 × 2 = 0 + 0.239 135 129 6;
  • 35) 0.239 135 129 6 × 2 = 0 + 0.478 270 259 2;
  • 36) 0.478 270 259 2 × 2 = 0 + 0.956 540 518 4;
  • 37) 0.956 540 518 4 × 2 = 1 + 0.913 081 036 8;
  • 38) 0.913 081 036 8 × 2 = 1 + 0.826 162 073 6;
  • 39) 0.826 162 073 6 × 2 = 1 + 0.652 324 147 2;
  • 40) 0.652 324 147 2 × 2 = 1 + 0.304 648 294 4;
  • 41) 0.304 648 294 4 × 2 = 0 + 0.609 296 588 8;
  • 42) 0.609 296 588 8 × 2 = 1 + 0.218 593 177 6;
  • 43) 0.218 593 177 6 × 2 = 0 + 0.437 186 355 2;
  • 44) 0.437 186 355 2 × 2 = 0 + 0.874 372 710 4;
  • 45) 0.874 372 710 4 × 2 = 1 + 0.748 745 420 8;
  • 46) 0.748 745 420 8 × 2 = 1 + 0.497 490 841 6;
  • 47) 0.497 490 841 6 × 2 = 0 + 0.994 981 683 2;
  • 48) 0.994 981 683 2 × 2 = 1 + 0.989 963 366 4;
  • 49) 0.989 963 366 4 × 2 = 1 + 0.979 926 732 8;

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:


0.000 000 021 9(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0000 1111 0100 1101 1(2)

5. Positive number before normalization:

0.000 000 021 9(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0000 1111 0100 1101 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:


0.000 000 021 9(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0000 1111 0100 1101 1(2) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0000 1111 0100 1101 1(2) × 20 =

1.0111 1000 0011 1101 0011 011(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):
1.0111 1000 0011 1101 0011 011


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. 011 1100 0001 1110 1001 1011 =

011 1100 0001 1110 1001 1011


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) =
011 1100 0001 1110 1001 1011


Decimal number 0.000 000 021 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 011 1100 0001 1110 1001 1011

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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