0.000 001 685 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 001 685(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
0.000 001 685(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: 0.000 001 685.

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) 0.000 001 685 × 2 = 0 + 0.000 003 37;
  • 2) 0.000 003 37 × 2 = 0 + 0.000 006 74;
  • 3) 0.000 006 74 × 2 = 0 + 0.000 013 48;
  • 4) 0.000 013 48 × 2 = 0 + 0.000 026 96;
  • 5) 0.000 026 96 × 2 = 0 + 0.000 053 92;
  • 6) 0.000 053 92 × 2 = 0 + 0.000 107 84;
  • 7) 0.000 107 84 × 2 = 0 + 0.000 215 68;
  • 8) 0.000 215 68 × 2 = 0 + 0.000 431 36;
  • 9) 0.000 431 36 × 2 = 0 + 0.000 862 72;
  • 10) 0.000 862 72 × 2 = 0 + 0.001 725 44;
  • 11) 0.001 725 44 × 2 = 0 + 0.003 450 88;
  • 12) 0.003 450 88 × 2 = 0 + 0.006 901 76;
  • 13) 0.006 901 76 × 2 = 0 + 0.013 803 52;
  • 14) 0.013 803 52 × 2 = 0 + 0.027 607 04;
  • 15) 0.027 607 04 × 2 = 0 + 0.055 214 08;
  • 16) 0.055 214 08 × 2 = 0 + 0.110 428 16;
  • 17) 0.110 428 16 × 2 = 0 + 0.220 856 32;
  • 18) 0.220 856 32 × 2 = 0 + 0.441 712 64;
  • 19) 0.441 712 64 × 2 = 0 + 0.883 425 28;
  • 20) 0.883 425 28 × 2 = 1 + 0.766 850 56;
  • 21) 0.766 850 56 × 2 = 1 + 0.533 701 12;
  • 22) 0.533 701 12 × 2 = 1 + 0.067 402 24;
  • 23) 0.067 402 24 × 2 = 0 + 0.134 804 48;
  • 24) 0.134 804 48 × 2 = 0 + 0.269 608 96;
  • 25) 0.269 608 96 × 2 = 0 + 0.539 217 92;
  • 26) 0.539 217 92 × 2 = 1 + 0.078 435 84;
  • 27) 0.078 435 84 × 2 = 0 + 0.156 871 68;
  • 28) 0.156 871 68 × 2 = 0 + 0.313 743 36;
  • 29) 0.313 743 36 × 2 = 0 + 0.627 486 72;
  • 30) 0.627 486 72 × 2 = 1 + 0.254 973 44;
  • 31) 0.254 973 44 × 2 = 0 + 0.509 946 88;
  • 32) 0.509 946 88 × 2 = 1 + 0.019 893 76;
  • 33) 0.019 893 76 × 2 = 0 + 0.039 787 52;
  • 34) 0.039 787 52 × 2 = 0 + 0.079 575 04;
  • 35) 0.079 575 04 × 2 = 0 + 0.159 150 08;
  • 36) 0.159 150 08 × 2 = 0 + 0.318 300 16;
  • 37) 0.318 300 16 × 2 = 0 + 0.636 600 32;
  • 38) 0.636 600 32 × 2 = 1 + 0.273 200 64;
  • 39) 0.273 200 64 × 2 = 0 + 0.546 401 28;
  • 40) 0.546 401 28 × 2 = 1 + 0.092 802 56;
  • 41) 0.092 802 56 × 2 = 0 + 0.185 605 12;
  • 42) 0.185 605 12 × 2 = 0 + 0.371 210 24;
  • 43) 0.371 210 24 × 2 = 0 + 0.742 420 48;

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:


0.000 001 685(10) =

0.0000 0000 0000 0000 0001 1100 0100 0101 0000 0101 000(2)

5. Positive number before normalization:

0.000 001 685(10) =

0.0000 0000 0000 0000 0001 1100 0100 0101 0000 0101 000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 685(10) =

0.0000 0000 0000 0000 0001 1100 0100 0101 0000 0101 000(2) =

0.0000 0000 0000 0000 0001 1100 0100 0101 0000 0101 000(2) × 20 =

1.1100 0100 0101 0000 0101 000(2) × 2-20


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): -20

Mantissa (not normalized):
1.1100 0100 0101 0000 0101 000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(8-1) - 1 =

(-20 + 127)(10) =

107(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;
  • 107 ÷ 2 = 53 + 1;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

107(10) =

0110 1011(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 0010 0010 1000 0010 1000 =

110 0010 0010 1000 0010 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 1011

Mantissa (23 bits) =
110 0010 0010 1000 0010 1000


Decimal number 0.000 001 685 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1011 - 110 0010 0010 1000 0010 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