OCTAL SUBTRACTION

Since we can't subtract 0 by 6, we have to borrow from 2. Subtract 1 from 2 and add 8 to 0. We're adding 8 because we're dealing with OCTAL number systems. So if we were to do this in DECIMAL, we add 10; HEXADECIMAL we add 16. This is actually what happens no matter what number system you're dealing with.

So, subtract 6 from 8 to get 2.

We can't subtract 3 from 1 so we borrow 1 from 1 making the 3rd digit zero. Now we add 8 to 1 to have 9.

Subtract 3 from 9 to get 6. The final answer is 62.

OCTAL ADDITION

The octal number system consists of 8 digits ranging from 0 to 7.

So, we add these numbers as we usually would. But there's a slight difference in the process.

5 + 6 = 11. We write 11 below and not write 1 carry 1. Do the same for the other numbers. 7 + 3 = 10. 6 + 4 = 10.

Here's what we do with the 2 digit numbers. We find multiples of 8 ( 0, 8, 16, 24, 32, 40...) closest to the sum. For the first digit , we have 11. The closest would be 8 so we subtract 8 to 11 to get 3. We subtracted it by 8 once so we add 1 to the next digit. 10 + 1 will be 11. Find the multiple closest to 11 which is 8. Perform the subtraction to get 3 and carry 1 to the next digit once again. Repeat the process until all digits are below 8.

The final answer is 1 3 3 3 (Octal).

If we were to do this normally in decimal number system ( 0 - 9) , every time we reach a value of 10 or above, we always carry a 1 and input the remainder. So in octal number system ( 0 - 7) we carry a 1 every time we reach a value of 8 or above. This concept is applied to ANY number system that you can think of.

BINARY DIVISION

This is the basics of division. Note that any number divided by 0 is undefined.

Here is an example:



Step1. 1 0 is divisible by 1 0. Enter a quotient of 1. Multiply 1 by 1 0 . Subtract 1 0 by 1 0.

Step2. 1 0 - 1 0 is 0. We bring down 1. Since 0 1 is not divisible by 1 0. We enter a 0 into the quotient.
Then step3. We bring down the last digit 1. Now we'd have to divide 0 1 1 by 1 0, so we enter 1 to have 0 1 1 - 1 0.

Perform the subtraction operation of 0 1 1 and 1 0 to get 1. This will be our remainder.

The final answer is 1 0 1 remainder 1.

BINARY MULTIPLICATION

This is the basics of binary multiplication.

Step1. We multiplied 1 0 1 by 0. This will yield 0 0 0.


Step2. We leave a space on the first digit. We multiply 1 0 1 by 1 to get 1 0 1 _.

Step3. Multiply 1 0 1 by 1 to get 1 0 1.

Just like in any other multiplication process, we add all the products of each digit.
> Bring down 0.
> 0 + 1 = 1
> 0 + 0 + 1 = 1
> 1 + 0 = 1
> Bring down 1

We get a final answer of 1 1 1 0.

BINARY SUBTRACTION

This is the basics of binary subtraction.

We can't subtract 1 from 0 so we'd have to borrow from the 2nd digit.
Every time we borrow, we're actually getting 1 and 1. So we'd have:
1 0 <--- this can be written as--> 1
- 1 + 1
--------- - 1
1 --------
1

Then continue subtracting the next digits, 0 - 0 will be 0. 1 - 0 will be 1.
Final Answer is 1 0 1.

BINARY ADDITION

The binary number systems consists of only 2 digits : these are 0 and 1.

In this number system, we apply the same rules as adding numbers in the decimal system.

This is the basics of binary addition.

You're probably wondering why 1 + 1 is 1 0 and not 2. That's because there is no number 2 (or 3, 4, 5, 6...) in the system. Every time a 1 + 1 appears, we carry 1 to the following digit and leave a 0 in the first digit.

Now let's try it with a four digit binary number :

So, we add these numbers as we normally would, column by column.

Zero plus zero will be 0.
One plus one will be 1 0. So we write the 0 while the 1 is carried over to the third column.
One plus zero plus one is equal to 1 0. Bring down 0 and carry the 1 over to the 4th column.
1 plus 1 will be 1 0. Add another 1 and the result will be 1 1.
Bring down 1 while the other 1 is carried over.
Bring down the last 1 to get the answer.