Please Wait... While Loading Full Video#### Class 10 chapter -1

● 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

`=>` 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

Class 10 Chapter 1 Example 9

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

Class 10 Chapter 1 Example 10

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

Class 10 Chapter 1 Example 11

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.