-0.000 001 06 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 001 06(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 06(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:

|-0.000 001 06| = 0.000 001 06


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: 0.000 001 06.

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 06 × 2 = 0 + 0.000 002 12;
  • 2) 0.000 002 12 × 2 = 0 + 0.000 004 24;
  • 3) 0.000 004 24 × 2 = 0 + 0.000 008 48;
  • 4) 0.000 008 48 × 2 = 0 + 0.000 016 96;
  • 5) 0.000 016 96 × 2 = 0 + 0.000 033 92;
  • 6) 0.000 033 92 × 2 = 0 + 0.000 067 84;
  • 7) 0.000 067 84 × 2 = 0 + 0.000 135 68;
  • 8) 0.000 135 68 × 2 = 0 + 0.000 271 36;
  • 9) 0.000 271 36 × 2 = 0 + 0.000 542 72;
  • 10) 0.000 542 72 × 2 = 0 + 0.001 085 44;
  • 11) 0.001 085 44 × 2 = 0 + 0.002 170 88;
  • 12) 0.002 170 88 × 2 = 0 + 0.004 341 76;
  • 13) 0.004 341 76 × 2 = 0 + 0.008 683 52;
  • 14) 0.008 683 52 × 2 = 0 + 0.017 367 04;
  • 15) 0.017 367 04 × 2 = 0 + 0.034 734 08;
  • 16) 0.034 734 08 × 2 = 0 + 0.069 468 16;
  • 17) 0.069 468 16 × 2 = 0 + 0.138 936 32;
  • 18) 0.138 936 32 × 2 = 0 + 0.277 872 64;
  • 19) 0.277 872 64 × 2 = 0 + 0.555 745 28;
  • 20) 0.555 745 28 × 2 = 1 + 0.111 490 56;
  • 21) 0.111 490 56 × 2 = 0 + 0.222 981 12;
  • 22) 0.222 981 12 × 2 = 0 + 0.445 962 24;
  • 23) 0.445 962 24 × 2 = 0 + 0.891 924 48;
  • 24) 0.891 924 48 × 2 = 1 + 0.783 848 96;
  • 25) 0.783 848 96 × 2 = 1 + 0.567 697 92;
  • 26) 0.567 697 92 × 2 = 1 + 0.135 395 84;
  • 27) 0.135 395 84 × 2 = 0 + 0.270 791 68;
  • 28) 0.270 791 68 × 2 = 0 + 0.541 583 36;
  • 29) 0.541 583 36 × 2 = 1 + 0.083 166 72;
  • 30) 0.083 166 72 × 2 = 0 + 0.166 333 44;
  • 31) 0.166 333 44 × 2 = 0 + 0.332 666 88;
  • 32) 0.332 666 88 × 2 = 0 + 0.665 333 76;
  • 33) 0.665 333 76 × 2 = 1 + 0.330 667 52;
  • 34) 0.330 667 52 × 2 = 0 + 0.661 335 04;
  • 35) 0.661 335 04 × 2 = 1 + 0.322 670 08;
  • 36) 0.322 670 08 × 2 = 0 + 0.645 340 16;
  • 37) 0.645 340 16 × 2 = 1 + 0.290 680 32;
  • 38) 0.290 680 32 × 2 = 0 + 0.581 360 64;
  • 39) 0.581 360 64 × 2 = 1 + 0.162 721 28;
  • 40) 0.162 721 28 × 2 = 0 + 0.325 442 56;
  • 41) 0.325 442 56 × 2 = 0 + 0.650 885 12;
  • 42) 0.650 885 12 × 2 = 1 + 0.301 770 24;
  • 43) 0.301 770 24 × 2 = 0 + 0.603 540 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).


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:


0.000 001 06(10) =

0.0000 0000 0000 0000 0001 0001 1100 1000 1010 1010 010(2)

6. Positive number before normalization:

0.000 001 06(10) =

0.0000 0000 0000 0000 0001 0001 1100 1000 1010 1010 010(2)

7. 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 06(10) =

0.0000 0000 0000 0000 0001 0001 1100 1000 1010 1010 010(2) =

0.0000 0000 0000 0000 0001 0001 1100 1000 1010 1010 010(2) × 20 =

1.0001 1100 1000 1010 1010 010(2) × 2-20


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): -20

Mantissa (not normalized):
1.0001 1100 1000 1010 1010 010


9. 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)


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;
  • 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;

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) =

107(10) =

0110 1011(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 1110 0100 0101 0101 0010 =

000 1110 0100 0101 0101 0010


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 1011

Mantissa (23 bits) =
000 1110 0100 0101 0101 0010


Decimal number -0.000 001 06 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1011 - 000 1110 0100 0101 0101 0010

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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