0.000 000 016 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 016 1(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
0.000 000 016 1(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: 0.000 000 016 1.

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) 0.000 000 016 1 × 2 = 0 + 0.000 000 032 2;
  • 2) 0.000 000 032 2 × 2 = 0 + 0.000 000 064 4;
  • 3) 0.000 000 064 4 × 2 = 0 + 0.000 000 128 8;
  • 4) 0.000 000 128 8 × 2 = 0 + 0.000 000 257 6;
  • 5) 0.000 000 257 6 × 2 = 0 + 0.000 000 515 2;
  • 6) 0.000 000 515 2 × 2 = 0 + 0.000 001 030 4;
  • 7) 0.000 001 030 4 × 2 = 0 + 0.000 002 060 8;
  • 8) 0.000 002 060 8 × 2 = 0 + 0.000 004 121 6;
  • 9) 0.000 004 121 6 × 2 = 0 + 0.000 008 243 2;
  • 10) 0.000 008 243 2 × 2 = 0 + 0.000 016 486 4;
  • 11) 0.000 016 486 4 × 2 = 0 + 0.000 032 972 8;
  • 12) 0.000 032 972 8 × 2 = 0 + 0.000 065 945 6;
  • 13) 0.000 065 945 6 × 2 = 0 + 0.000 131 891 2;
  • 14) 0.000 131 891 2 × 2 = 0 + 0.000 263 782 4;
  • 15) 0.000 263 782 4 × 2 = 0 + 0.000 527 564 8;
  • 16) 0.000 527 564 8 × 2 = 0 + 0.001 055 129 6;
  • 17) 0.001 055 129 6 × 2 = 0 + 0.002 110 259 2;
  • 18) 0.002 110 259 2 × 2 = 0 + 0.004 220 518 4;
  • 19) 0.004 220 518 4 × 2 = 0 + 0.008 441 036 8;
  • 20) 0.008 441 036 8 × 2 = 0 + 0.016 882 073 6;
  • 21) 0.016 882 073 6 × 2 = 0 + 0.033 764 147 2;
  • 22) 0.033 764 147 2 × 2 = 0 + 0.067 528 294 4;
  • 23) 0.067 528 294 4 × 2 = 0 + 0.135 056 588 8;
  • 24) 0.135 056 588 8 × 2 = 0 + 0.270 113 177 6;
  • 25) 0.270 113 177 6 × 2 = 0 + 0.540 226 355 2;
  • 26) 0.540 226 355 2 × 2 = 1 + 0.080 452 710 4;
  • 27) 0.080 452 710 4 × 2 = 0 + 0.160 905 420 8;
  • 28) 0.160 905 420 8 × 2 = 0 + 0.321 810 841 6;
  • 29) 0.321 810 841 6 × 2 = 0 + 0.643 621 683 2;
  • 30) 0.643 621 683 2 × 2 = 1 + 0.287 243 366 4;
  • 31) 0.287 243 366 4 × 2 = 0 + 0.574 486 732 8;
  • 32) 0.574 486 732 8 × 2 = 1 + 0.148 973 465 6;
  • 33) 0.148 973 465 6 × 2 = 0 + 0.297 946 931 2;
  • 34) 0.297 946 931 2 × 2 = 0 + 0.595 893 862 4;
  • 35) 0.595 893 862 4 × 2 = 1 + 0.191 787 724 8;
  • 36) 0.191 787 724 8 × 2 = 0 + 0.383 575 449 6;
  • 37) 0.383 575 449 6 × 2 = 0 + 0.767 150 899 2;
  • 38) 0.767 150 899 2 × 2 = 1 + 0.534 301 798 4;
  • 39) 0.534 301 798 4 × 2 = 1 + 0.068 603 596 8;
  • 40) 0.068 603 596 8 × 2 = 0 + 0.137 207 193 6;
  • 41) 0.137 207 193 6 × 2 = 0 + 0.274 414 387 2;
  • 42) 0.274 414 387 2 × 2 = 0 + 0.548 828 774 4;
  • 43) 0.548 828 774 4 × 2 = 1 + 0.097 657 548 8;
  • 44) 0.097 657 548 8 × 2 = 0 + 0.195 315 097 6;
  • 45) 0.195 315 097 6 × 2 = 0 + 0.390 630 195 2;
  • 46) 0.390 630 195 2 × 2 = 0 + 0.781 260 390 4;
  • 47) 0.781 260 390 4 × 2 = 1 + 0.562 520 780 8;
  • 48) 0.562 520 780 8 × 2 = 1 + 0.125 041 561 6;
  • 49) 0.125 041 561 6 × 2 = 0 + 0.250 083 123 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:


0.000 000 016 1(10) =

0.0000 0000 0000 0000 0000 0000 0100 0101 0010 0110 0010 0011 0(2)

5. Positive number before normalization:

0.000 000 016 1(10) =

0.0000 0000 0000 0000 0000 0000 0100 0101 0010 0110 0010 0011 0(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:


0.000 000 016 1(10) =

0.0000 0000 0000 0000 0000 0000 0100 0101 0010 0110 0010 0011 0(2) =

0.0000 0000 0000 0000 0000 0000 0100 0101 0010 0110 0010 0011 0(2) × 20 =

1.0001 0100 1001 1000 1000 110(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):
1.0001 0100 1001 1000 1000 110


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 1010 0100 1100 0100 0110 =

000 1010 0100 1100 0100 0110


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 1010 0100 1100 0100 0110


Decimal number 0.000 000 016 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 000 1010 0100 1100 0100 0110

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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