Answer

Verified

420.6k+ views

**Hint:**In this question, we need to show that $\left( {{a}^{2}}-{{b}^{2}} \right)$ is composite if a and b are odd prime. For this we just need to show that $\left( {{a}^{2}}-{{b}^{2}} \right)$ has factors which are not equal to 1 or itself. We will first write a and b in the form of odd numbers which are prime. Then we will use them in $\left( {{a}^{2}}-{{b}^{2}} \right)$ to prove that it has factors other than 1. We will use the property that ${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$.

**Complete step-by-step solution:**

Here a and b are two odd prime and we need to prove $\left( {{a}^{2}}-{{b}^{2}} \right)$ as composite. Thus we need to prove that $\left( {{a}^{2}}-{{b}^{2}} \right)$ has factors other than 1 or itself. As we know from the algebraic property that $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$. So (a+b) and (a-b) are the factors of $\left( {{a}^{2}}-{{b}^{2}} \right)$. But we need to prove that these are not equal to 1. For this let us use general notation to write any odd prime.

a is an odd prime, so it can be written as a = 2k+1. Similarly, b is also an odd prime. So it can also be written as b = 2k'+1 where k, k' are integers.

Now (a-b) will become,

$\left( a-b \right)=2k+1-\left( 2k'+1 \right)\Rightarrow 2k+1-2k'-1\Rightarrow 2k-2k'$.

Taking 2 common from both the terms we get, $\left( a-b \right)=2k-2k'$ which is an even number, so it cannot be equal to 1. (1 is not an even number).

Now let us find the value of (a+b) we get,

$\left( a+b \right)=2k+1+2k'+1$.

Simplifying it we get, $\left( a+b \right)=2k+2k'+2$.

Taking 2 common from all the terms on the right side we get $\left( a+b \right)=2\left( k+k'+1 \right)$ which is an even number. Hence it cannot be equal to 1.

Therefore, none of the factors of $\left( {{a}^{2}}-{{b}^{2}} \right)$ are equal to 1 and as we can say they are neither equal to $\left( {{a}^{2}}-{{b}^{2}} \right)$. So $\left( {{a}^{2}}-{{b}^{2}} \right)$ has two factors. Therefore it cannot be prime.

**Hence $\left( {{a}^{2}}-{{b}^{2}} \right)$ is composite even if a and b are both odd primes.**

**Note:**Students should know the mathematical way to represent an odd integer i.e. 2k+1 where k is an integer. If a number can be divided by 2 then it is an even number. Here both (a+b) and (a-b) have 2 as their factor, so they can be divided by 2 easily and hence they are even numbers.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths