Class 10

### Revisiting Irrational Numbers

● In this section, we will prove that sqrt2 , sqrt3 , sqrt5 and, in general sqrtp is "irrational," where p is a prime. One of the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic.
=> Recall, a number ‘s’ is called irrational if it cannot be written in the form p/q where p and q are integers and q ≠ 0.
=> Some examples of irrational numbers, with which you are already familiar, are :

sqert2 , sqrt3 , sqrt(15) , pi , - sqrt2/sqrt3 , 0.10110111011110........ etc.

● Before we prove that sqrt2 is irrational, we need the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic.

color{blue}{"Theorem 1.3:"} Let p be a prime number. If" p divides a^2, then p divides a, where a is a positive integer.

Proof : Let the prime factorisation of a be as follows :
a = p_1p_2 . . . p_n, where p_1,p_2, . . ., p_n are primes, not necessarily distinct. Therefore, a^2 = ( p_1p_2 . . . p_n)( p_1p_2 . . . p_n) = p_1^2 p_2^2 . . . p_n^2.
=> Now, we are given that p divides a^2. Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a^2. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a^2 are p_1, p_2, . . ., p_n. So p is one of p_1, p_2, . . ., p_n.

Now, since a = p_1 p_2 . . . p_n, p divides a.
We are now ready to give a proof that sqrt2 is irrational.

The proof is based on a technique called ‘proof by contradiction’. (This technique is discussed in some detail in Appendix 1).

color{blue}{"Theorem 1.4 :"} sqrt2 is irrational.

Proof : Let us assume, to the contrary, that sqrt2 is rational.
So, we can find integers r and s (≠ 0) such that sqrt2 = r/s. Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get sqrt2 = a/b where a and b are coprime.

● So, b sqrt2 = a.
Squaring on both sides and rearranging, we get 2b^2 = a^2. Therefore, 2 divides a2.
=> Now, by Theorem 1.3, it follows that 2 divides a.
So, we can write a = 2c for some integer c.
* Not from the examination point of view.

● Substituting for a, we get 2b^2 = 4c^2, that is, b^2 = 2c^2.
This means that 2 divides b^2, and so 2 divides b (again using Theorem 1.3 with p = 2).
Therefore, a and b have at least 2 as a common factor.

●But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that sqrt2 is rational.
So, we conclude that sqrt2 is irrational

Q 3119856719

Prove that sqrt3 is irrational.
Class 10 Chapter 1 Example 9
Solution:

Let us assume, to the contrary, that sqrt3 is rational.

That is, we can find integers a and b (≠ 0) such that sqrt3 = a/b

common factor, and assume that a and b are coprime.
So, b sqrt3 = a.
Squaring on both sides, and rearranging, we get 3b^2 = a^2.
Therefore, a^2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b^2 = 9c^2, that is, b^2 = 3c^2.
This means that b^2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3
with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that sqrt3 is rational.
So, we conclude that sqrt3 is irrational.
In Class IX, we mentioned that :
• the sum or difference of a rational and an irrational number is irrational and
• the product and quotient of a non-zero rational and irrational number is irrational.
We prove some particular cases here.
Q 3129056811

Show that 5 - sqrt3 is irrational.
Class 10 Chapter 1 Example 10
Solution:

Let us assume, to the contrary, that 5-sqrt3 is rational.

That is, we can find coprime a and b (b ≠ 0) such that 5-sqrt3 = a/b

Therefore, 5-a/b = sqrt3

Rearranging this equation, we get sqrt3 = 5-a/b = ( 5b -a)/b

Since a and b are integers, we get 5-a/b is rational, and so sqrt3 is rational.
But this contradicts the fact that sqrt3 is irrational.

This contradiction has arisen because of our incorrect assumption that 5-sqrt3 is rational.

So, we conclude that 5-sqrt3 is irrational.
Q 3179056816

Show that 3sqrt2 is irrational.
Class 10 Chapter 1 Example 11
Solution:

Let us assume, to the contrary, that 3 sqrt2 is rational.

That is, we can find coprime a and b (b ≠ 0) such that 3 sqrt2 = a/b .

Rearranging, we get sqrt2 = a/(3b)

Since 3, a and b are integers a/(3b) is rational, and so sqrt2 is rational.

But this contradicts the fact that sqrt2 is irrational.

So, we conclude that 3 sqrt2 is irrational.