Lesson 2. Binary Arithmetic

Lesson Objective

Be able to explain highest and lowest assignable Number.
Be able to add together up to three binary numbers.
Apply binary shifts to multiply and divide numbers.
Explain underflow and overflow and describe the circumstances in which they occur.

Lesson Notes

Binary Addition - THE RULES!!

Work right to left and apply these simple rules:

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0 Carry 1
1 + 1 + 1 = 1 Carry 1

Here are examples of two 8 bit numbers being added together:

				1	1	1
	0	0	0	0	1	1	1	0	14
+	1	0	1	0	0	0	1	0	162
	1	0	1	1	0	0	0	0	176

...

	1				1	1	1
	0	1	0	0	0	1	1	1	71
+	0	1	1	0	0	0	0	1	79
	1	0	1	0	1	0	0	0	150

Overflow Error

When and extra bit is created to represent a number.

Here is an example of an overflow error:

1			1	1	1	1
	1	1	0	0	1	1	0	0	204
+	1	0	0	1	1	1	0	1	157
1	0	1	1	0	1	0	0	1	361

Adding 3 Binary Numbers

Method 1.

Add the first two, then add the third to the result.

We will carry out the addition 1011 + 0111 + 101

We can can see the 1011 (11) + 0111 (7) + 101 (5) = 23

1	1	1	1
	1	0	1	1	11
+	0	1	1	1	7
1	0	0	1	0
+		1	0	1	5
1	0	1	1	1	23

Method 2.

Logical Binary Shifts

Left Shift = Multiply. Each shift is the number multiplied by a power of 2

0 Shift	0	0	0	1	Original
1 Shift	0	0	1	0	*2
2 Shift	0	1	0	0	*4
3 Shift	1	0	0	0	*8

Right Shift = Divide. Each shift is the number division by a power of 2

0 Shift	1	0	0	0	Original
1 Shift	0	1	0	0	/2
2 Shift	0	0	1	0	/4
3 Shift	0	0	0	1	/8

Binary Multiplication

How do I multiply with number that aren't 2, 4, 8, 16, 32, 64, 128, 256...?

	128	64	32	16	8	4	2	1
	0	0	0	1	1	0	0	1	25
x	0	0	0	0	1	0	1	0	10
	1	1	0	0	1	0	0	0	200
+	0	0	1	1	0	0	1	0	50
	1	1	1	1	1	0	1	0	250

Take the first number as the multipler. Multiply the multiplier by each digit of the multiplicand to achieve intermediate products, whose last digit is in the position of the corresponding multiplicand digit. Then add the intermediate values.

Example:

First x8, Left Shift 25 by 3 (11001000)

Then x2, Left Shift 25 by 1 (00110010)

Final add them together (11001000 + 00110010 = 11111010)

Underflow and Overflow

Overflow occurs when the result of a calculation is too large to be held in the number of bits allocated.

For example, adding two integers in an 8-bit byte (ignore the sign bit).

1						1	1
	1	0	0	0	0	0	0	1	129
+	1	0	0	0	0	0	1	1	131
1	0	0	0	0	0	1	0	0	260

Underflow occurs when a number is too small to be represented in the number of bits allocated.

It may occur if a very small number is divided by a number greater than 1.

Example: Shows a 8 bit binary number shifting one space to the right.

	1	0	0	0	0	0	0	1
÷2	0	1	0	0	0	0	0	0	1

	128	64	32	16	8	4	2	1
	0	0	0	1	1	0	0	1	25
x	0	0	0	0	1	0	1	0	10
	1	1	0	0	1	0	0	0	200
+	0	0	1	1	0	0	1	0	50
	1	1	1	1	1	0	1	0	250

	128	64	32	16	8	4	2	1
	0	0	0	1	1	0	0	1	25
x	0	0	0	0	1	0	1	0	10
	1	1	0	0	1	0	0	0	200
+	0	0	1	1	0	0	1	0	50
	1	1	1	1	1	0	1	0	250

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