● 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
● 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
Prove that `sqrt3` is irrational.
