0.000 000 002 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 002 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
0.000 000 002 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: 0.000 000 002 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) 0.000 000 002 3 × 2 = 0 + 0.000 000 004 6;
  • 2) 0.000 000 004 6 × 2 = 0 + 0.000 000 009 2;
  • 3) 0.000 000 009 2 × 2 = 0 + 0.000 000 018 4;
  • 4) 0.000 000 018 4 × 2 = 0 + 0.000 000 036 8;
  • 5) 0.000 000 036 8 × 2 = 0 + 0.000 000 073 6;
  • 6) 0.000 000 073 6 × 2 = 0 + 0.000 000 147 2;
  • 7) 0.000 000 147 2 × 2 = 0 + 0.000 000 294 4;
  • 8) 0.000 000 294 4 × 2 = 0 + 0.000 000 588 8;
  • 9) 0.000 000 588 8 × 2 = 0 + 0.000 001 177 6;
  • 10) 0.000 001 177 6 × 2 = 0 + 0.000 002 355 2;
  • 11) 0.000 002 355 2 × 2 = 0 + 0.000 004 710 4;
  • 12) 0.000 004 710 4 × 2 = 0 + 0.000 009 420 8;
  • 13) 0.000 009 420 8 × 2 = 0 + 0.000 018 841 6;
  • 14) 0.000 018 841 6 × 2 = 0 + 0.000 037 683 2;
  • 15) 0.000 037 683 2 × 2 = 0 + 0.000 075 366 4;
  • 16) 0.000 075 366 4 × 2 = 0 + 0.000 150 732 8;
  • 17) 0.000 150 732 8 × 2 = 0 + 0.000 301 465 6;
  • 18) 0.000 301 465 6 × 2 = 0 + 0.000 602 931 2;
  • 19) 0.000 602 931 2 × 2 = 0 + 0.001 205 862 4;
  • 20) 0.001 205 862 4 × 2 = 0 + 0.002 411 724 8;
  • 21) 0.002 411 724 8 × 2 = 0 + 0.004 823 449 6;
  • 22) 0.004 823 449 6 × 2 = 0 + 0.009 646 899 2;
  • 23) 0.009 646 899 2 × 2 = 0 + 0.019 293 798 4;
  • 24) 0.019 293 798 4 × 2 = 0 + 0.038 587 596 8;
  • 25) 0.038 587 596 8 × 2 = 0 + 0.077 175 193 6;
  • 26) 0.077 175 193 6 × 2 = 0 + 0.154 350 387 2;
  • 27) 0.154 350 387 2 × 2 = 0 + 0.308 700 774 4;
  • 28) 0.308 700 774 4 × 2 = 0 + 0.617 401 548 8;
  • 29) 0.617 401 548 8 × 2 = 1 + 0.234 803 097 6;
  • 30) 0.234 803 097 6 × 2 = 0 + 0.469 606 195 2;
  • 31) 0.469 606 195 2 × 2 = 0 + 0.939 212 390 4;
  • 32) 0.939 212 390 4 × 2 = 1 + 0.878 424 780 8;
  • 33) 0.878 424 780 8 × 2 = 1 + 0.756 849 561 6;
  • 34) 0.756 849 561 6 × 2 = 1 + 0.513 699 123 2;
  • 35) 0.513 699 123 2 × 2 = 1 + 0.027 398 246 4;
  • 36) 0.027 398 246 4 × 2 = 0 + 0.054 796 492 8;
  • 37) 0.054 796 492 8 × 2 = 0 + 0.109 592 985 6;
  • 38) 0.109 592 985 6 × 2 = 0 + 0.219 185 971 2;
  • 39) 0.219 185 971 2 × 2 = 0 + 0.438 371 942 4;
  • 40) 0.438 371 942 4 × 2 = 0 + 0.876 743 884 8;
  • 41) 0.876 743 884 8 × 2 = 1 + 0.753 487 769 6;
  • 42) 0.753 487 769 6 × 2 = 1 + 0.506 975 539 2;
  • 43) 0.506 975 539 2 × 2 = 1 + 0.013 951 078 4;
  • 44) 0.013 951 078 4 × 2 = 0 + 0.027 902 156 8;
  • 45) 0.027 902 156 8 × 2 = 0 + 0.055 804 313 6;
  • 46) 0.055 804 313 6 × 2 = 0 + 0.111 608 627 2;
  • 47) 0.111 608 627 2 × 2 = 0 + 0.223 217 254 4;
  • 48) 0.223 217 254 4 × 2 = 0 + 0.446 434 508 8;
  • 49) 0.446 434 508 8 × 2 = 0 + 0.892 869 017 6;
  • 50) 0.892 869 017 6 × 2 = 1 + 0.785 738 035 2;
  • 51) 0.785 738 035 2 × 2 = 1 + 0.571 476 070 4;
  • 52) 0.571 476 070 4 × 2 = 1 + 0.142 952 140 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:


0.000 000 002 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 1001 1110 0000 1110 0000 0111(2)

5. Positive number before normalization:

0.000 000 002 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 1001 1110 0000 1110 0000 0111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 29 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 002 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 1001 1110 0000 1110 0000 0111(2) =

0.0000 0000 0000 0000 0000 0000 0000 1001 1110 0000 1110 0000 0111(2) × 20 =

1.0011 1100 0001 1100 0000 111(2) × 2-29


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): -29

Mantissa (not normalized):
1.0011 1100 0001 1100 0000 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-29 + 2(8-1) - 1 =

(-29 + 127)(10) =

98(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;
  • 98 ÷ 2 = 49 + 0;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

98(10) =

0110 0010(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. 001 1110 0000 1110 0000 0111 =

001 1110 0000 1110 0000 0111


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 0010

Mantissa (23 bits) =
001 1110 0000 1110 0000 0111


Decimal number 0.000 000 002 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0010 - 001 1110 0000 1110 0000 0111

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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