0.000 000 017 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 017 5(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
0.000 000 017 5(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: 0.000 000 017 5.

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) 0.000 000 017 5 × 2 = 0 + 0.000 000 035;
  • 2) 0.000 000 035 × 2 = 0 + 0.000 000 07;
  • 3) 0.000 000 07 × 2 = 0 + 0.000 000 14;
  • 4) 0.000 000 14 × 2 = 0 + 0.000 000 28;
  • 5) 0.000 000 28 × 2 = 0 + 0.000 000 56;
  • 6) 0.000 000 56 × 2 = 0 + 0.000 001 12;
  • 7) 0.000 001 12 × 2 = 0 + 0.000 002 24;
  • 8) 0.000 002 24 × 2 = 0 + 0.000 004 48;
  • 9) 0.000 004 48 × 2 = 0 + 0.000 008 96;
  • 10) 0.000 008 96 × 2 = 0 + 0.000 017 92;
  • 11) 0.000 017 92 × 2 = 0 + 0.000 035 84;
  • 12) 0.000 035 84 × 2 = 0 + 0.000 071 68;
  • 13) 0.000 071 68 × 2 = 0 + 0.000 143 36;
  • 14) 0.000 143 36 × 2 = 0 + 0.000 286 72;
  • 15) 0.000 286 72 × 2 = 0 + 0.000 573 44;
  • 16) 0.000 573 44 × 2 = 0 + 0.001 146 88;
  • 17) 0.001 146 88 × 2 = 0 + 0.002 293 76;
  • 18) 0.002 293 76 × 2 = 0 + 0.004 587 52;
  • 19) 0.004 587 52 × 2 = 0 + 0.009 175 04;
  • 20) 0.009 175 04 × 2 = 0 + 0.018 350 08;
  • 21) 0.018 350 08 × 2 = 0 + 0.036 700 16;
  • 22) 0.036 700 16 × 2 = 0 + 0.073 400 32;
  • 23) 0.073 400 32 × 2 = 0 + 0.146 800 64;
  • 24) 0.146 800 64 × 2 = 0 + 0.293 601 28;
  • 25) 0.293 601 28 × 2 = 0 + 0.587 202 56;
  • 26) 0.587 202 56 × 2 = 1 + 0.174 405 12;
  • 27) 0.174 405 12 × 2 = 0 + 0.348 810 24;
  • 28) 0.348 810 24 × 2 = 0 + 0.697 620 48;
  • 29) 0.697 620 48 × 2 = 1 + 0.395 240 96;
  • 30) 0.395 240 96 × 2 = 0 + 0.790 481 92;
  • 31) 0.790 481 92 × 2 = 1 + 0.580 963 84;
  • 32) 0.580 963 84 × 2 = 1 + 0.161 927 68;
  • 33) 0.161 927 68 × 2 = 0 + 0.323 855 36;
  • 34) 0.323 855 36 × 2 = 0 + 0.647 710 72;
  • 35) 0.647 710 72 × 2 = 1 + 0.295 421 44;
  • 36) 0.295 421 44 × 2 = 0 + 0.590 842 88;
  • 37) 0.590 842 88 × 2 = 1 + 0.181 685 76;
  • 38) 0.181 685 76 × 2 = 0 + 0.363 371 52;
  • 39) 0.363 371 52 × 2 = 0 + 0.726 743 04;
  • 40) 0.726 743 04 × 2 = 1 + 0.453 486 08;
  • 41) 0.453 486 08 × 2 = 0 + 0.906 972 16;
  • 42) 0.906 972 16 × 2 = 1 + 0.813 944 32;
  • 43) 0.813 944 32 × 2 = 1 + 0.627 888 64;
  • 44) 0.627 888 64 × 2 = 1 + 0.255 777 28;
  • 45) 0.255 777 28 × 2 = 0 + 0.511 554 56;
  • 46) 0.511 554 56 × 2 = 1 + 0.023 109 12;
  • 47) 0.023 109 12 × 2 = 0 + 0.046 218 24;
  • 48) 0.046 218 24 × 2 = 0 + 0.092 436 48;
  • 49) 0.092 436 48 × 2 = 0 + 0.184 872 96;

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:


0.000 000 017 5(10) =

0.0000 0000 0000 0000 0000 0000 0100 1011 0010 1001 0111 0100 0(2)

5. Positive number before normalization:

0.000 000 017 5(10) =

0.0000 0000 0000 0000 0000 0000 0100 1011 0010 1001 0111 0100 0(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:


0.000 000 017 5(10) =

0.0000 0000 0000 0000 0000 0000 0100 1011 0010 1001 0111 0100 0(2) =

0.0000 0000 0000 0000 0000 0000 0100 1011 0010 1001 0111 0100 0(2) × 20 =

1.0010 1100 1010 0101 1101 000(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):
1.0010 1100 1010 0101 1101 000


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. 001 0110 0101 0010 1110 1000 =

001 0110 0101 0010 1110 1000


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) =
001 0110 0101 0010 1110 1000


Decimal number 0.000 000 017 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 001 0110 0101 0010 1110 1000

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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