0.001 010 086 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.001 010 086 9(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
0.001 010 086 9(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: 0.001 010 086 9.

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) 0.001 010 086 9 × 2 = 0 + 0.002 020 173 8;
  • 2) 0.002 020 173 8 × 2 = 0 + 0.004 040 347 6;
  • 3) 0.004 040 347 6 × 2 = 0 + 0.008 080 695 2;
  • 4) 0.008 080 695 2 × 2 = 0 + 0.016 161 390 4;
  • 5) 0.016 161 390 4 × 2 = 0 + 0.032 322 780 8;
  • 6) 0.032 322 780 8 × 2 = 0 + 0.064 645 561 6;
  • 7) 0.064 645 561 6 × 2 = 0 + 0.129 291 123 2;
  • 8) 0.129 291 123 2 × 2 = 0 + 0.258 582 246 4;
  • 9) 0.258 582 246 4 × 2 = 0 + 0.517 164 492 8;
  • 10) 0.517 164 492 8 × 2 = 1 + 0.034 328 985 6;
  • 11) 0.034 328 985 6 × 2 = 0 + 0.068 657 971 2;
  • 12) 0.068 657 971 2 × 2 = 0 + 0.137 315 942 4;
  • 13) 0.137 315 942 4 × 2 = 0 + 0.274 631 884 8;
  • 14) 0.274 631 884 8 × 2 = 0 + 0.549 263 769 6;
  • 15) 0.549 263 769 6 × 2 = 1 + 0.098 527 539 2;
  • 16) 0.098 527 539 2 × 2 = 0 + 0.197 055 078 4;
  • 17) 0.197 055 078 4 × 2 = 0 + 0.394 110 156 8;
  • 18) 0.394 110 156 8 × 2 = 0 + 0.788 220 313 6;
  • 19) 0.788 220 313 6 × 2 = 1 + 0.576 440 627 2;
  • 20) 0.576 440 627 2 × 2 = 1 + 0.152 881 254 4;
  • 21) 0.152 881 254 4 × 2 = 0 + 0.305 762 508 8;
  • 22) 0.305 762 508 8 × 2 = 0 + 0.611 525 017 6;
  • 23) 0.611 525 017 6 × 2 = 1 + 0.223 050 035 2;
  • 24) 0.223 050 035 2 × 2 = 0 + 0.446 100 070 4;
  • 25) 0.446 100 070 4 × 2 = 0 + 0.892 200 140 8;
  • 26) 0.892 200 140 8 × 2 = 1 + 0.784 400 281 6;
  • 27) 0.784 400 281 6 × 2 = 1 + 0.568 800 563 2;
  • 28) 0.568 800 563 2 × 2 = 1 + 0.137 601 126 4;
  • 29) 0.137 601 126 4 × 2 = 0 + 0.275 202 252 8;
  • 30) 0.275 202 252 8 × 2 = 0 + 0.550 404 505 6;
  • 31) 0.550 404 505 6 × 2 = 1 + 0.100 809 011 2;
  • 32) 0.100 809 011 2 × 2 = 0 + 0.201 618 022 4;
  • 33) 0.201 618 022 4 × 2 = 0 + 0.403 236 044 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:


0.001 010 086 9(10) =

0.0000 0000 0100 0010 0011 0010 0111 0010 0(2)

5. Positive number before normalization:

0.001 010 086 9(10) =

0.0000 0000 0100 0010 0011 0010 0111 0010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.001 010 086 9(10) =

0.0000 0000 0100 0010 0011 0010 0111 0010 0(2) =

0.0000 0000 0100 0010 0011 0010 0111 0010 0(2) × 20 =

1.0000 1000 1100 1001 1100 100(2) × 2-10


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): -10

Mantissa (not normalized):
1.0000 1000 1100 1001 1100 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-10 + 2(8-1) - 1 =

(-10 + 127)(10) =

117(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;
  • 117 ÷ 2 = 58 + 1;
  • 58 ÷ 2 = 29 + 0;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

117(10) =

0111 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. 000 0100 0110 0100 1110 0100 =

000 0100 0110 0100 1110 0100


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) =
0111 0101

Mantissa (23 bits) =
000 0100 0110 0100 1110 0100


Decimal number 0.001 010 086 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 0101 - 000 0100 0110 0100 1110 0100

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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