-0.000 000 13 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 13(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 13(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. Start with the positive version of the number:

|-0.000 000 13| = 0.000 000 13


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 13.

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 13 × 2 = 0 + 0.000 000 26;
  • 2) 0.000 000 26 × 2 = 0 + 0.000 000 52;
  • 3) 0.000 000 52 × 2 = 0 + 0.000 001 04;
  • 4) 0.000 001 04 × 2 = 0 + 0.000 002 08;
  • 5) 0.000 002 08 × 2 = 0 + 0.000 004 16;
  • 6) 0.000 004 16 × 2 = 0 + 0.000 008 32;
  • 7) 0.000 008 32 × 2 = 0 + 0.000 016 64;
  • 8) 0.000 016 64 × 2 = 0 + 0.000 033 28;
  • 9) 0.000 033 28 × 2 = 0 + 0.000 066 56;
  • 10) 0.000 066 56 × 2 = 0 + 0.000 133 12;
  • 11) 0.000 133 12 × 2 = 0 + 0.000 266 24;
  • 12) 0.000 266 24 × 2 = 0 + 0.000 532 48;
  • 13) 0.000 532 48 × 2 = 0 + 0.001 064 96;
  • 14) 0.001 064 96 × 2 = 0 + 0.002 129 92;
  • 15) 0.002 129 92 × 2 = 0 + 0.004 259 84;
  • 16) 0.004 259 84 × 2 = 0 + 0.008 519 68;
  • 17) 0.008 519 68 × 2 = 0 + 0.017 039 36;
  • 18) 0.017 039 36 × 2 = 0 + 0.034 078 72;
  • 19) 0.034 078 72 × 2 = 0 + 0.068 157 44;
  • 20) 0.068 157 44 × 2 = 0 + 0.136 314 88;
  • 21) 0.136 314 88 × 2 = 0 + 0.272 629 76;
  • 22) 0.272 629 76 × 2 = 0 + 0.545 259 52;
  • 23) 0.545 259 52 × 2 = 1 + 0.090 519 04;
  • 24) 0.090 519 04 × 2 = 0 + 0.181 038 08;
  • 25) 0.181 038 08 × 2 = 0 + 0.362 076 16;
  • 26) 0.362 076 16 × 2 = 0 + 0.724 152 32;
  • 27) 0.724 152 32 × 2 = 1 + 0.448 304 64;
  • 28) 0.448 304 64 × 2 = 0 + 0.896 609 28;
  • 29) 0.896 609 28 × 2 = 1 + 0.793 218 56;
  • 30) 0.793 218 56 × 2 = 1 + 0.586 437 12;
  • 31) 0.586 437 12 × 2 = 1 + 0.172 874 24;
  • 32) 0.172 874 24 × 2 = 0 + 0.345 748 48;
  • 33) 0.345 748 48 × 2 = 0 + 0.691 496 96;
  • 34) 0.691 496 96 × 2 = 1 + 0.382 993 92;
  • 35) 0.382 993 92 × 2 = 0 + 0.765 987 84;
  • 36) 0.765 987 84 × 2 = 1 + 0.531 975 68;
  • 37) 0.531 975 68 × 2 = 1 + 0.063 951 36;
  • 38) 0.063 951 36 × 2 = 0 + 0.127 902 72;
  • 39) 0.127 902 72 × 2 = 0 + 0.255 805 44;
  • 40) 0.255 805 44 × 2 = 0 + 0.511 610 88;
  • 41) 0.511 610 88 × 2 = 1 + 0.023 221 76;
  • 42) 0.023 221 76 × 2 = 0 + 0.046 443 52;
  • 43) 0.046 443 52 × 2 = 0 + 0.092 887 04;
  • 44) 0.092 887 04 × 2 = 0 + 0.185 774 08;
  • 45) 0.185 774 08 × 2 = 0 + 0.371 548 16;
  • 46) 0.371 548 16 × 2 = 0 + 0.743 096 32;

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).


5. 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 13(10) =

0.0000 0000 0000 0000 0000 0010 0010 1110 0101 1000 1000 00(2)

6. Positive number before normalization:

0.000 000 13(10) =

0.0000 0000 0000 0000 0000 0010 0010 1110 0101 1000 1000 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 13(10) =

0.0000 0000 0000 0000 0000 0010 0010 1110 0101 1000 1000 00(2) =

0.0000 0000 0000 0000 0000 0010 0010 1110 0101 1000 1000 00(2) × 20 =

1.0001 0111 0010 1100 0100 000(2) × 2-23


8. 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 1 (a negative number)

Exponent (unadjusted): -23

Mantissa (not normalized):
1.0001 0111 0010 1100 0100 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(8-1) - 1 =

(-23 + 127)(10) =

104(10)


10. 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;
  • 104 ÷ 2 = 52 + 0;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

104(10) =

0110 1000(2)


12. 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. 000 1011 1001 0110 0010 0000 =

000 1011 1001 0110 0010 0000


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 1000

Mantissa (23 bits) =
000 1011 1001 0110 0010 0000


Decimal number -0.000 000 13 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1000 - 000 1011 1001 0110 0010 0000

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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