# Working with binary numbers in converter

What would the binary number be in decimal notation? Click here to see the answer Try converting these numbers from binary to decimal: Since 11 is greater than 10, a one is put into the 10's column carried , and a 1 is recorded in the one's column of the sum.

Thus, the answer is Binary addition works on the same principle, but the numerals are different. Begin with one-bit binary addition:. In binary, any digit higher than 1 puts us a column to the left as would 10 in decimal notation. Record the 0 in the ones column, and carry the 1 to the twos column to get an answer of " The process is the same for multiple-bit binary numbers: Record the 0, carry the 1.

Add 1 from carry: Multiplication in the binary system works the same way as in the decimal system: Follow the same rules as in decimal division. For the sake of simplicity, throw away the remainder. Converting from decimal to binary notation is slightly more difficult conceptually, but can easily be done once you know how through the use of algorithms. Begin by thinking of a few examples.

Almost as intuitive is the number 5: Then we just put this into columns. This process continues until we have a remainder of 0. Let's take a look at how it works. To convert the decimal number 75 to binary, we would find the largest power of 2 less than 75, which is Subtract 8 from 11 to get 3. Thus, our number is Making this algorithm a bit more formal gives us: Find the largest power of two in D.

Let this equal P. Put a 1 in binary column P. Subtract P from D. Put zeros in all columns which don't have ones. This algorithm is a bit awkward. Particularly step 3, "filling in the zeros.

Now that we have an algorithm, we can use it to convert numbers from decimal to binary relatively painlessly. Our first step is to find P. Subtracting leaves us with Subtracting 1 from P gives us 4. Next, subtract 16 from 23, to get 7.

Subtract 1 from P gives us 3. Subtract 1 from P to get 1. Subtract 1 from P to get 0. Subtract 1 from P to get P is now less than zero, so we stop. Another algorithm for converting decimal to binary However, this is not the only approach possible.

For the sake of simplicity, throw away the remainder. Converting from decimal to binary notation is slightly more difficult conceptually, but can easily be done once you know how through the use of algorithms.

Begin by thinking of a few examples. Almost as intuitive is the number 5: Then we just put this into columns. This process continues until we have a remainder of 0. Let's take a look at how it works. To convert the decimal number 75 to binary, we would find the largest power of 2 less than 75, which is Subtract 8 from 11 to get 3.

Thus, our number is Making this algorithm a bit more formal gives us: Find the largest power of two in D. Let this equal P. Put a 1 in binary column P. Subtract P from D. Put zeros in all columns which don't have ones. This algorithm is a bit awkward. Particularly step 3, "filling in the zeros.

Now that we have an algorithm, we can use it to convert numbers from decimal to binary relatively painlessly. Our first step is to find P.

Subtracting leaves us with Subtracting 1 from P gives us 4. Next, subtract 16 from 23, to get 7. Subtract 1 from P gives us 3. Subtract 1 from P to get 1. Subtract 1 from P to get 0. Subtract 1 from P to get P is now less than zero, so we stop. Another algorithm for converting decimal to binary However, this is not the only approach possible. We can start at the right, rather than the left.

This gives us the rightmost digit as a starting point. Now we need to do the remaining digits. One idea is to "shift" them. It is also easy to see that multiplying and dividing by 2 shifts everything by one column: Similarly, multiplying by 2 shifts in the other direction: Take the number Dividing by 2 gives Since we divided the number by two, we "took out" one power of two.

Also note that a1 is essentially "remultiplied" by two just by putting it in front of a[0], so it is automatically fit into the correct column. Now we can subtract 1 from 81 to see what remainder we still must place Dividing 80 by 2 gives We can divide by two again to get