0.000 000 027 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 027 7(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 7(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: 0.000 000 027 7.

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 7 × 2 = 0 + 0.000 000 055 4;
  • 2) 0.000 000 055 4 × 2 = 0 + 0.000 000 110 8;
  • 3) 0.000 000 110 8 × 2 = 0 + 0.000 000 221 6;
  • 4) 0.000 000 221 6 × 2 = 0 + 0.000 000 443 2;
  • 5) 0.000 000 443 2 × 2 = 0 + 0.000 000 886 4;
  • 6) 0.000 000 886 4 × 2 = 0 + 0.000 001 772 8;
  • 7) 0.000 001 772 8 × 2 = 0 + 0.000 003 545 6;
  • 8) 0.000 003 545 6 × 2 = 0 + 0.000 007 091 2;
  • 9) 0.000 007 091 2 × 2 = 0 + 0.000 014 182 4;
  • 10) 0.000 014 182 4 × 2 = 0 + 0.000 028 364 8;
  • 11) 0.000 028 364 8 × 2 = 0 + 0.000 056 729 6;
  • 12) 0.000 056 729 6 × 2 = 0 + 0.000 113 459 2;
  • 13) 0.000 113 459 2 × 2 = 0 + 0.000 226 918 4;
  • 14) 0.000 226 918 4 × 2 = 0 + 0.000 453 836 8;
  • 15) 0.000 453 836 8 × 2 = 0 + 0.000 907 673 6;
  • 16) 0.000 907 673 6 × 2 = 0 + 0.001 815 347 2;
  • 17) 0.001 815 347 2 × 2 = 0 + 0.003 630 694 4;
  • 18) 0.003 630 694 4 × 2 = 0 + 0.007 261 388 8;
  • 19) 0.007 261 388 8 × 2 = 0 + 0.014 522 777 6;
  • 20) 0.014 522 777 6 × 2 = 0 + 0.029 045 555 2;
  • 21) 0.029 045 555 2 × 2 = 0 + 0.058 091 110 4;
  • 22) 0.058 091 110 4 × 2 = 0 + 0.116 182 220 8;
  • 23) 0.116 182 220 8 × 2 = 0 + 0.232 364 441 6;
  • 24) 0.232 364 441 6 × 2 = 0 + 0.464 728 883 2;
  • 25) 0.464 728 883 2 × 2 = 0 + 0.929 457 766 4;
  • 26) 0.929 457 766 4 × 2 = 1 + 0.858 915 532 8;
  • 27) 0.858 915 532 8 × 2 = 1 + 0.717 831 065 6;
  • 28) 0.717 831 065 6 × 2 = 1 + 0.435 662 131 2;
  • 29) 0.435 662 131 2 × 2 = 0 + 0.871 324 262 4;
  • 30) 0.871 324 262 4 × 2 = 1 + 0.742 648 524 8;
  • 31) 0.742 648 524 8 × 2 = 1 + 0.485 297 049 6;
  • 32) 0.485 297 049 6 × 2 = 0 + 0.970 594 099 2;
  • 33) 0.970 594 099 2 × 2 = 1 + 0.941 188 198 4;
  • 34) 0.941 188 198 4 × 2 = 1 + 0.882 376 396 8;
  • 35) 0.882 376 396 8 × 2 = 1 + 0.764 752 793 6;
  • 36) 0.764 752 793 6 × 2 = 1 + 0.529 505 587 2;
  • 37) 0.529 505 587 2 × 2 = 1 + 0.059 011 174 4;
  • 38) 0.059 011 174 4 × 2 = 0 + 0.118 022 348 8;
  • 39) 0.118 022 348 8 × 2 = 0 + 0.236 044 697 6;
  • 40) 0.236 044 697 6 × 2 = 0 + 0.472 089 395 2;
  • 41) 0.472 089 395 2 × 2 = 0 + 0.944 178 790 4;
  • 42) 0.944 178 790 4 × 2 = 1 + 0.888 357 580 8;
  • 43) 0.888 357 580 8 × 2 = 1 + 0.776 715 161 6;
  • 44) 0.776 715 161 6 × 2 = 1 + 0.553 430 323 2;
  • 45) 0.553 430 323 2 × 2 = 1 + 0.106 860 646 4;
  • 46) 0.106 860 646 4 × 2 = 0 + 0.213 721 292 8;
  • 47) 0.213 721 292 8 × 2 = 0 + 0.427 442 585 6;
  • 48) 0.427 442 585 6 × 2 = 0 + 0.854 885 171 2;
  • 49) 0.854 885 171 2 × 2 = 1 + 0.709 770 342 4;

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:


0.000 000 027 7(10) =

0.0000 0000 0000 0000 0000 0000 0111 0110 1111 1000 0111 1000 1(2)

5. Positive number before normalization:

0.000 000 027 7(10) =

0.0000 0000 0000 0000 0000 0000 0111 0110 1111 1000 0111 1000 1(2)

6. 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 7(10) =

0.0000 0000 0000 0000 0000 0000 0111 0110 1111 1000 0111 1000 1(2) =

0.0000 0000 0000 0000 0000 0000 0111 0110 1111 1000 0111 1000 1(2) × 20 =

1.1101 1011 1110 0001 1110 001(2) × 2-26


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): -26

Mantissa (not normalized):
1.1101 1011 1110 0001 1110 001


8. 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)


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;
  • 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;

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) =

101(10) =

0110 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. 110 1101 1111 0000 1111 0001 =

110 1101 1111 0000 1111 0001


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 0101

Mantissa (23 bits) =
110 1101 1111 0000 1111 0001


Decimal number 0.000 000 027 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 110 1101 1111 0000 1111 0001

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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