0.000 000 015 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 015 2(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
0.000 000 015 2(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: 0.000 000 015 2.

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) 0.000 000 015 2 × 2 = 0 + 0.000 000 030 4;
  • 2) 0.000 000 030 4 × 2 = 0 + 0.000 000 060 8;
  • 3) 0.000 000 060 8 × 2 = 0 + 0.000 000 121 6;
  • 4) 0.000 000 121 6 × 2 = 0 + 0.000 000 243 2;
  • 5) 0.000 000 243 2 × 2 = 0 + 0.000 000 486 4;
  • 6) 0.000 000 486 4 × 2 = 0 + 0.000 000 972 8;
  • 7) 0.000 000 972 8 × 2 = 0 + 0.000 001 945 6;
  • 8) 0.000 001 945 6 × 2 = 0 + 0.000 003 891 2;
  • 9) 0.000 003 891 2 × 2 = 0 + 0.000 007 782 4;
  • 10) 0.000 007 782 4 × 2 = 0 + 0.000 015 564 8;
  • 11) 0.000 015 564 8 × 2 = 0 + 0.000 031 129 6;
  • 12) 0.000 031 129 6 × 2 = 0 + 0.000 062 259 2;
  • 13) 0.000 062 259 2 × 2 = 0 + 0.000 124 518 4;
  • 14) 0.000 124 518 4 × 2 = 0 + 0.000 249 036 8;
  • 15) 0.000 249 036 8 × 2 = 0 + 0.000 498 073 6;
  • 16) 0.000 498 073 6 × 2 = 0 + 0.000 996 147 2;
  • 17) 0.000 996 147 2 × 2 = 0 + 0.001 992 294 4;
  • 18) 0.001 992 294 4 × 2 = 0 + 0.003 984 588 8;
  • 19) 0.003 984 588 8 × 2 = 0 + 0.007 969 177 6;
  • 20) 0.007 969 177 6 × 2 = 0 + 0.015 938 355 2;
  • 21) 0.015 938 355 2 × 2 = 0 + 0.031 876 710 4;
  • 22) 0.031 876 710 4 × 2 = 0 + 0.063 753 420 8;
  • 23) 0.063 753 420 8 × 2 = 0 + 0.127 506 841 6;
  • 24) 0.127 506 841 6 × 2 = 0 + 0.255 013 683 2;
  • 25) 0.255 013 683 2 × 2 = 0 + 0.510 027 366 4;
  • 26) 0.510 027 366 4 × 2 = 1 + 0.020 054 732 8;
  • 27) 0.020 054 732 8 × 2 = 0 + 0.040 109 465 6;
  • 28) 0.040 109 465 6 × 2 = 0 + 0.080 218 931 2;
  • 29) 0.080 218 931 2 × 2 = 0 + 0.160 437 862 4;
  • 30) 0.160 437 862 4 × 2 = 0 + 0.320 875 724 8;
  • 31) 0.320 875 724 8 × 2 = 0 + 0.641 751 449 6;
  • 32) 0.641 751 449 6 × 2 = 1 + 0.283 502 899 2;
  • 33) 0.283 502 899 2 × 2 = 0 + 0.567 005 798 4;
  • 34) 0.567 005 798 4 × 2 = 1 + 0.134 011 596 8;
  • 35) 0.134 011 596 8 × 2 = 0 + 0.268 023 193 6;
  • 36) 0.268 023 193 6 × 2 = 0 + 0.536 046 387 2;
  • 37) 0.536 046 387 2 × 2 = 1 + 0.072 092 774 4;
  • 38) 0.072 092 774 4 × 2 = 0 + 0.144 185 548 8;
  • 39) 0.144 185 548 8 × 2 = 0 + 0.288 371 097 6;
  • 40) 0.288 371 097 6 × 2 = 0 + 0.576 742 195 2;
  • 41) 0.576 742 195 2 × 2 = 1 + 0.153 484 390 4;
  • 42) 0.153 484 390 4 × 2 = 0 + 0.306 968 780 8;
  • 43) 0.306 968 780 8 × 2 = 0 + 0.613 937 561 6;
  • 44) 0.613 937 561 6 × 2 = 1 + 0.227 875 123 2;
  • 45) 0.227 875 123 2 × 2 = 0 + 0.455 750 246 4;
  • 46) 0.455 750 246 4 × 2 = 0 + 0.911 500 492 8;
  • 47) 0.911 500 492 8 × 2 = 1 + 0.823 000 985 6;
  • 48) 0.823 000 985 6 × 2 = 1 + 0.646 001 971 2;
  • 49) 0.646 001 971 2 × 2 = 1 + 0.292 003 942 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:


0.000 000 015 2(10) =

0.0000 0000 0000 0000 0000 0000 0100 0001 0100 1000 1001 0011 1(2)

5. Positive number before normalization:

0.000 000 015 2(10) =

0.0000 0000 0000 0000 0000 0000 0100 0001 0100 1000 1001 0011 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:


0.000 000 015 2(10) =

0.0000 0000 0000 0000 0000 0000 0100 0001 0100 1000 1001 0011 1(2) =

0.0000 0000 0000 0000 0000 0000 0100 0001 0100 1000 1001 0011 1(2) × 20 =

1.0000 0101 0010 0010 0100 111(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):
1.0000 0101 0010 0010 0100 111


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. 000 0010 1001 0001 0010 0111 =

000 0010 1001 0001 0010 0111


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) =
000 0010 1001 0001 0010 0111


Decimal number 0.000 000 015 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 000 0010 1001 0001 0010 0111

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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