Prove that is `sqrt3` irrational number.
Let us assume, to contrary, that is rational. That is, we can find integers a and b (≠0) such that `sqrt2=a/b`
Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b`sqrt3`= a
Squaring on both sides, and rearranging, we get 3b2 = a2.
Therefore, a2 is divisible by 3, and by Theorem, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2.
This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 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 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.
Let `sqrt3` is rational
∴`sqrt3 = "p"/"q"` where p and q are co-prime integers and q ≠ 0.
⇒ `sqrt3"q" = "p"`
⇒ 3q2 = p2..........(1)
⇒ 3 divides p2
⇒ 3 divides p .....(A)
Let p = 3c where c is an integer
⇒ p2 = 9c2
⇒ 3q2 = 9c2 [from (1)]
⇒ q2 = 3c2
⇒ 3 divides q2
⇒ 3 divides q .....(B)
From statements (A) and (B), 3 divides p and q both that means p and q are not co-prime which contradicts our assumption.
So, our assumption is wrong.
Hence `sqrt3` is irrational.
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