0.000 000 013 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 013 3(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
0.000 000 013 3(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: 0.000 000 013 3.

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) 0.000 000 013 3 × 2 = 0 + 0.000 000 026 6;
  • 2) 0.000 000 026 6 × 2 = 0 + 0.000 000 053 2;
  • 3) 0.000 000 053 2 × 2 = 0 + 0.000 000 106 4;
  • 4) 0.000 000 106 4 × 2 = 0 + 0.000 000 212 8;
  • 5) 0.000 000 212 8 × 2 = 0 + 0.000 000 425 6;
  • 6) 0.000 000 425 6 × 2 = 0 + 0.000 000 851 2;
  • 7) 0.000 000 851 2 × 2 = 0 + 0.000 001 702 4;
  • 8) 0.000 001 702 4 × 2 = 0 + 0.000 003 404 8;
  • 9) 0.000 003 404 8 × 2 = 0 + 0.000 006 809 6;
  • 10) 0.000 006 809 6 × 2 = 0 + 0.000 013 619 2;
  • 11) 0.000 013 619 2 × 2 = 0 + 0.000 027 238 4;
  • 12) 0.000 027 238 4 × 2 = 0 + 0.000 054 476 8;
  • 13) 0.000 054 476 8 × 2 = 0 + 0.000 108 953 6;
  • 14) 0.000 108 953 6 × 2 = 0 + 0.000 217 907 2;
  • 15) 0.000 217 907 2 × 2 = 0 + 0.000 435 814 4;
  • 16) 0.000 435 814 4 × 2 = 0 + 0.000 871 628 8;
  • 17) 0.000 871 628 8 × 2 = 0 + 0.001 743 257 6;
  • 18) 0.001 743 257 6 × 2 = 0 + 0.003 486 515 2;
  • 19) 0.003 486 515 2 × 2 = 0 + 0.006 973 030 4;
  • 20) 0.006 973 030 4 × 2 = 0 + 0.013 946 060 8;
  • 21) 0.013 946 060 8 × 2 = 0 + 0.027 892 121 6;
  • 22) 0.027 892 121 6 × 2 = 0 + 0.055 784 243 2;
  • 23) 0.055 784 243 2 × 2 = 0 + 0.111 568 486 4;
  • 24) 0.111 568 486 4 × 2 = 0 + 0.223 136 972 8;
  • 25) 0.223 136 972 8 × 2 = 0 + 0.446 273 945 6;
  • 26) 0.446 273 945 6 × 2 = 0 + 0.892 547 891 2;
  • 27) 0.892 547 891 2 × 2 = 1 + 0.785 095 782 4;
  • 28) 0.785 095 782 4 × 2 = 1 + 0.570 191 564 8;
  • 29) 0.570 191 564 8 × 2 = 1 + 0.140 383 129 6;
  • 30) 0.140 383 129 6 × 2 = 0 + 0.280 766 259 2;
  • 31) 0.280 766 259 2 × 2 = 0 + 0.561 532 518 4;
  • 32) 0.561 532 518 4 × 2 = 1 + 0.123 065 036 8;
  • 33) 0.123 065 036 8 × 2 = 0 + 0.246 130 073 6;
  • 34) 0.246 130 073 6 × 2 = 0 + 0.492 260 147 2;
  • 35) 0.492 260 147 2 × 2 = 0 + 0.984 520 294 4;
  • 36) 0.984 520 294 4 × 2 = 1 + 0.969 040 588 8;
  • 37) 0.969 040 588 8 × 2 = 1 + 0.938 081 177 6;
  • 38) 0.938 081 177 6 × 2 = 1 + 0.876 162 355 2;
  • 39) 0.876 162 355 2 × 2 = 1 + 0.752 324 710 4;
  • 40) 0.752 324 710 4 × 2 = 1 + 0.504 649 420 8;
  • 41) 0.504 649 420 8 × 2 = 1 + 0.009 298 841 6;
  • 42) 0.009 298 841 6 × 2 = 0 + 0.018 597 683 2;
  • 43) 0.018 597 683 2 × 2 = 0 + 0.037 195 366 4;
  • 44) 0.037 195 366 4 × 2 = 0 + 0.074 390 732 8;
  • 45) 0.074 390 732 8 × 2 = 0 + 0.148 781 465 6;
  • 46) 0.148 781 465 6 × 2 = 0 + 0.297 562 931 2;
  • 47) 0.297 562 931 2 × 2 = 0 + 0.595 125 862 4;
  • 48) 0.595 125 862 4 × 2 = 1 + 0.190 251 724 8;
  • 49) 0.190 251 724 8 × 2 = 0 + 0.380 503 449 6;
  • 50) 0.380 503 449 6 × 2 = 0 + 0.761 006 899 2;

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:


0.000 000 013 3(10) =

0.0000 0000 0000 0000 0000 0000 0011 1001 0001 1111 1000 0001 00(2)

5. Positive number before normalization:

0.000 000 013 3(10) =

0.0000 0000 0000 0000 0000 0000 0011 1001 0001 1111 1000 0001 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 013 3(10) =

0.0000 0000 0000 0000 0000 0000 0011 1001 0001 1111 1000 0001 00(2) =

0.0000 0000 0000 0000 0000 0000 0011 1001 0001 1111 1000 0001 00(2) × 20 =

1.1100 1000 1111 1100 0000 100(2) × 2-27


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): -27

Mantissa (not normalized):
1.1100 1000 1111 1100 0000 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-27 + 2(8-1) - 1 =

(-27 + 127)(10) =

100(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;
  • 100 ÷ 2 = 50 + 0;
  • 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) =

100(10) =

0110 0100(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 0100 0111 1110 0000 0100 =

110 0100 0111 1110 0000 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) =
0110 0100

Mantissa (23 bits) =
110 0100 0111 1110 0000 0100


Decimal number 0.000 000 013 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 110 0100 0111 1110 0000 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