0.000 000 013 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 013 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
0.000 000 013 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: 0.000 000 013 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) 0.000 000 013 2 × 2 = 0 + 0.000 000 026 4;
  • 2) 0.000 000 026 4 × 2 = 0 + 0.000 000 052 8;
  • 3) 0.000 000 052 8 × 2 = 0 + 0.000 000 105 6;
  • 4) 0.000 000 105 6 × 2 = 0 + 0.000 000 211 2;
  • 5) 0.000 000 211 2 × 2 = 0 + 0.000 000 422 4;
  • 6) 0.000 000 422 4 × 2 = 0 + 0.000 000 844 8;
  • 7) 0.000 000 844 8 × 2 = 0 + 0.000 001 689 6;
  • 8) 0.000 001 689 6 × 2 = 0 + 0.000 003 379 2;
  • 9) 0.000 003 379 2 × 2 = 0 + 0.000 006 758 4;
  • 10) 0.000 006 758 4 × 2 = 0 + 0.000 013 516 8;
  • 11) 0.000 013 516 8 × 2 = 0 + 0.000 027 033 6;
  • 12) 0.000 027 033 6 × 2 = 0 + 0.000 054 067 2;
  • 13) 0.000 054 067 2 × 2 = 0 + 0.000 108 134 4;
  • 14) 0.000 108 134 4 × 2 = 0 + 0.000 216 268 8;
  • 15) 0.000 216 268 8 × 2 = 0 + 0.000 432 537 6;
  • 16) 0.000 432 537 6 × 2 = 0 + 0.000 865 075 2;
  • 17) 0.000 865 075 2 × 2 = 0 + 0.001 730 150 4;
  • 18) 0.001 730 150 4 × 2 = 0 + 0.003 460 300 8;
  • 19) 0.003 460 300 8 × 2 = 0 + 0.006 920 601 6;
  • 20) 0.006 920 601 6 × 2 = 0 + 0.013 841 203 2;
  • 21) 0.013 841 203 2 × 2 = 0 + 0.027 682 406 4;
  • 22) 0.027 682 406 4 × 2 = 0 + 0.055 364 812 8;
  • 23) 0.055 364 812 8 × 2 = 0 + 0.110 729 625 6;
  • 24) 0.110 729 625 6 × 2 = 0 + 0.221 459 251 2;
  • 25) 0.221 459 251 2 × 2 = 0 + 0.442 918 502 4;
  • 26) 0.442 918 502 4 × 2 = 0 + 0.885 837 004 8;
  • 27) 0.885 837 004 8 × 2 = 1 + 0.771 674 009 6;
  • 28) 0.771 674 009 6 × 2 = 1 + 0.543 348 019 2;
  • 29) 0.543 348 019 2 × 2 = 1 + 0.086 696 038 4;
  • 30) 0.086 696 038 4 × 2 = 0 + 0.173 392 076 8;
  • 31) 0.173 392 076 8 × 2 = 0 + 0.346 784 153 6;
  • 32) 0.346 784 153 6 × 2 = 0 + 0.693 568 307 2;
  • 33) 0.693 568 307 2 × 2 = 1 + 0.387 136 614 4;
  • 34) 0.387 136 614 4 × 2 = 0 + 0.774 273 228 8;
  • 35) 0.774 273 228 8 × 2 = 1 + 0.548 546 457 6;
  • 36) 0.548 546 457 6 × 2 = 1 + 0.097 092 915 2;
  • 37) 0.097 092 915 2 × 2 = 0 + 0.194 185 830 4;
  • 38) 0.194 185 830 4 × 2 = 0 + 0.388 371 660 8;
  • 39) 0.388 371 660 8 × 2 = 0 + 0.776 743 321 6;
  • 40) 0.776 743 321 6 × 2 = 1 + 0.553 486 643 2;
  • 41) 0.553 486 643 2 × 2 = 1 + 0.106 973 286 4;
  • 42) 0.106 973 286 4 × 2 = 0 + 0.213 946 572 8;
  • 43) 0.213 946 572 8 × 2 = 0 + 0.427 893 145 6;
  • 44) 0.427 893 145 6 × 2 = 0 + 0.855 786 291 2;
  • 45) 0.855 786 291 2 × 2 = 1 + 0.711 572 582 4;
  • 46) 0.711 572 582 4 × 2 = 1 + 0.423 145 164 8;
  • 47) 0.423 145 164 8 × 2 = 0 + 0.846 290 329 6;
  • 48) 0.846 290 329 6 × 2 = 1 + 0.692 580 659 2;
  • 49) 0.692 580 659 2 × 2 = 1 + 0.385 161 318 4;
  • 50) 0.385 161 318 4 × 2 = 0 + 0.770 322 636 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).


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 2(10) =

0.0000 0000 0000 0000 0000 0000 0011 1000 1011 0001 1000 1101 10(2)

5. Positive number before normalization:

0.000 000 013 2(10) =

0.0000 0000 0000 0000 0000 0000 0011 1000 1011 0001 1000 1101 10(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 2(10) =

0.0000 0000 0000 0000 0000 0000 0011 1000 1011 0001 1000 1101 10(2) =

0.0000 0000 0000 0000 0000 0000 0011 1000 1011 0001 1000 1101 10(2) × 20 =

1.1100 0101 1000 1100 0110 110(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 0101 1000 1100 0110 110


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 0010 1100 0110 0011 0110 =

110 0010 1100 0110 0011 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 0100

Mantissa (23 bits) =
110 0010 1100 0110 0011 0110


Decimal number 0.000 000 013 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 110 0010 1100 0110 0011 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