-0.000 000 027 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 027 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
-0.000 000 027 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. Start with the positive version of the number:

|-0.000 000 027 9| = 0.000 000 027 9


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: 0.000 000 027 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) 0.000 000 027 9 × 2 = 0 + 0.000 000 055 8;
  • 2) 0.000 000 055 8 × 2 = 0 + 0.000 000 111 6;
  • 3) 0.000 000 111 6 × 2 = 0 + 0.000 000 223 2;
  • 4) 0.000 000 223 2 × 2 = 0 + 0.000 000 446 4;
  • 5) 0.000 000 446 4 × 2 = 0 + 0.000 000 892 8;
  • 6) 0.000 000 892 8 × 2 = 0 + 0.000 001 785 6;
  • 7) 0.000 001 785 6 × 2 = 0 + 0.000 003 571 2;
  • 8) 0.000 003 571 2 × 2 = 0 + 0.000 007 142 4;
  • 9) 0.000 007 142 4 × 2 = 0 + 0.000 014 284 8;
  • 10) 0.000 014 284 8 × 2 = 0 + 0.000 028 569 6;
  • 11) 0.000 028 569 6 × 2 = 0 + 0.000 057 139 2;
  • 12) 0.000 057 139 2 × 2 = 0 + 0.000 114 278 4;
  • 13) 0.000 114 278 4 × 2 = 0 + 0.000 228 556 8;
  • 14) 0.000 228 556 8 × 2 = 0 + 0.000 457 113 6;
  • 15) 0.000 457 113 6 × 2 = 0 + 0.000 914 227 2;
  • 16) 0.000 914 227 2 × 2 = 0 + 0.001 828 454 4;
  • 17) 0.001 828 454 4 × 2 = 0 + 0.003 656 908 8;
  • 18) 0.003 656 908 8 × 2 = 0 + 0.007 313 817 6;
  • 19) 0.007 313 817 6 × 2 = 0 + 0.014 627 635 2;
  • 20) 0.014 627 635 2 × 2 = 0 + 0.029 255 270 4;
  • 21) 0.029 255 270 4 × 2 = 0 + 0.058 510 540 8;
  • 22) 0.058 510 540 8 × 2 = 0 + 0.117 021 081 6;
  • 23) 0.117 021 081 6 × 2 = 0 + 0.234 042 163 2;
  • 24) 0.234 042 163 2 × 2 = 0 + 0.468 084 326 4;
  • 25) 0.468 084 326 4 × 2 = 0 + 0.936 168 652 8;
  • 26) 0.936 168 652 8 × 2 = 1 + 0.872 337 305 6;
  • 27) 0.872 337 305 6 × 2 = 1 + 0.744 674 611 2;
  • 28) 0.744 674 611 2 × 2 = 1 + 0.489 349 222 4;
  • 29) 0.489 349 222 4 × 2 = 0 + 0.978 698 444 8;
  • 30) 0.978 698 444 8 × 2 = 1 + 0.957 396 889 6;
  • 31) 0.957 396 889 6 × 2 = 1 + 0.914 793 779 2;
  • 32) 0.914 793 779 2 × 2 = 1 + 0.829 587 558 4;
  • 33) 0.829 587 558 4 × 2 = 1 + 0.659 175 116 8;
  • 34) 0.659 175 116 8 × 2 = 1 + 0.318 350 233 6;
  • 35) 0.318 350 233 6 × 2 = 0 + 0.636 700 467 2;
  • 36) 0.636 700 467 2 × 2 = 1 + 0.273 400 934 4;
  • 37) 0.273 400 934 4 × 2 = 0 + 0.546 801 868 8;
  • 38) 0.546 801 868 8 × 2 = 1 + 0.093 603 737 6;
  • 39) 0.093 603 737 6 × 2 = 0 + 0.187 207 475 2;
  • 40) 0.187 207 475 2 × 2 = 0 + 0.374 414 950 4;
  • 41) 0.374 414 950 4 × 2 = 0 + 0.748 829 900 8;
  • 42) 0.748 829 900 8 × 2 = 1 + 0.497 659 801 6;
  • 43) 0.497 659 801 6 × 2 = 0 + 0.995 319 603 2;
  • 44) 0.995 319 603 2 × 2 = 1 + 0.990 639 206 4;
  • 45) 0.990 639 206 4 × 2 = 1 + 0.981 278 412 8;
  • 46) 0.981 278 412 8 × 2 = 1 + 0.962 556 825 6;
  • 47) 0.962 556 825 6 × 2 = 1 + 0.925 113 651 2;
  • 48) 0.925 113 651 2 × 2 = 1 + 0.850 227 302 4;
  • 49) 0.850 227 302 4 × 2 = 1 + 0.700 454 604 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).


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:


0.000 000 027 9(10) =

0.0000 0000 0000 0000 0000 0000 0111 0111 1101 0100 0101 1111 1(2)

6. Positive number before normalization:

0.000 000 027 9(10) =

0.0000 0000 0000 0000 0000 0000 0111 0111 1101 0100 0101 1111 1(2)

7. 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:


0.000 000 027 9(10) =

0.0000 0000 0000 0000 0000 0000 0111 0111 1101 0100 0101 1111 1(2) =

0.0000 0000 0000 0000 0000 0000 0111 0111 1101 0100 0101 1111 1(2) × 20 =

1.1101 1111 0101 0001 0111 111(2) × 2-26


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): -26

Mantissa (not normalized):
1.1101 1111 0101 0001 0111 111


9. 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)


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;
  • 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;

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) =

101(10) =

0110 0101(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. 110 1111 1010 1000 1011 1111 =

110 1111 1010 1000 1011 1111


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 0101

Mantissa (23 bits) =
110 1111 1010 1000 1011 1111


Decimal number -0.000 000 027 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0101 - 110 1111 1010 1000 1011 1111

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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