This wholes section is about how we can use algebra to prove that expressions are equivalent or that they have certain characteristics (such as being odd, a multiple of 6 etc). Whenever we are proving a statement or an equation, we need to prove that it is true all of the time. The most common way to prove a statement or an equation is to use algebra, but sometimes we will be able to prove it using words. If we are asked to prove that an equation or statement is not true, we only have to give one example where the statement or rule does not work; this is enough evidence to say that the equation or statement is incorrect.

Before we go into the algebraic proof, it is a good idea to quickly go over all of the rule for even and odd numbers. Here are the rules:

Usually when we divide even or odd numbers, we will rarely get an integer. This is why I have not included division in the rule above. When we are adding, subtracting or multiplying integers (even or odd), we will always get an integer, and this is why we have the rules above.

Before we go into the algebraic proof, it is a good idea to quickly go over all of the rule for even and odd numbers. Here are the rules:

**Addition or subtraction**

Here are the rules for addition or subtraction. The sign ± means “plus or minus”.- even ± even = even
- odd ± odd = even
- even ± odd = odd
- odd ± even = odd

**Multiplication**

These are the rules for multiplication:- even x even = even
- odd x odd = odd
- even x odd = even
- odd x even = even

Usually when we divide even or odd numbers, we will rarely get an integer. This is why I have not included division in the rule above. When we are adding, subtracting or multiplying integers (even or odd), we will always get an integer, and this is why we have the rules above.

**Word Proof**

We are now going to have a look at a few examples that we will prove by using words or disprove by using one example that does not work.

**Example 1**

If

*y*is an odd number, show that the expression below is even:

In order to prove that this statement is correct, we need to find whether the two brackets in the equation above are even or odd. After we have found this out, we can then use the rules for multiplying numbers that are odd, even or a combination of both.

Let’s find out if the first bracket is odd or even. The first bracket is (

Let’s find out if the first bracket is odd or even. The first bracket is (

*y*+ 2). The question tells us that*y*is an odd number and 2 is even. This means that we are adding an even and an odd number, which is going to give us an odd number. Therefore, the first bracket is odd.We now do the same for the second bracket, which is (

*y*+ 3).*y*is odd and 3 is also odd, which means that the second bracket is even because odd + odd is even.Therefore, we are multiplying an odd number (first bracket) by an even number (second bracket), which will result in an even number.

Therefore, if

*y*is an odd number, (*y*+ 2) x (*y*+ 3) is even.**Example 2**

*a*is an even number. Is the expression below odd or even?

To find out whether the outcome is odd or even, we need to find whether the two terms are even or odd.

3 is an odd number.

3 is an odd number.

We now move onto 7

*a*, which means 7 multiplied by*a*(7 x*a*). 7 is an odd number and we are told in the question that*a*is an even number. This means that 7*a*is even because odd x even = even.Therefore, we are adding an odd number (3) to an even number (7

*a*), which is going to give us an odd number (odd + even = odd).Therefore, 3 + 7

*a*is an odd number.**Example 3**

The expression below is always odd.

Disprove the above statement.

The question is asking us to disprove the statement and this means that we only need to find one time where the above statement does not work. Also, the question does not give us the value of

In order to disprove the statement, we want to show that the expression is even. 7 is odd and this means that we want 5

The odd number that I am going to choose for

The question is asking us to disprove the statement and this means that we only need to find one time where the above statement does not work. Also, the question does not give us the value of

*t*, which means that we can choose any value of*t*that disproves the above statement.In order to disprove the statement, we want to show that the expression is even. 7 is odd and this means that we want 5

*t*to be odd because two odd numbers added together are even (odd + odd = even). In order for 5*t*to be odd,*t*will have to be an odd number; this is because odd x odd is an odd number (if*t*was an even number, 5*t*would be even). This means that we can choose any odd number for*t*and this will prove that the above statement is wrong.The odd number that I am going to choose for

*t*is 3; we sub*t*as 3 into the expression.This is an even number, which proves that the above statement is wrong because 5

We could have let

*t*+ 7 is**NOT**always odd.We could have let

*t*be any odd number to prove that the statement was wrong. I am now going to let*t*be 9.52 is even, which means that we have disproved the statement.