Prove that √3 is an irrational number.

1.	Prove that √3 is an irrational number.
Answer» Let us assume, to the contrary, that √3 is rational. That is, we can find integers a and b (≠0) such that √3 =a/b Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime. So, b√3 =a Squaring on both sides, and rearranging, we get 3b² = a². Therefore, a² 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 3b² = 9c², that is, b² = 3c². This means that b² 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 3 is rational. So, we conclude that √3 is irrational.

Answer»

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

That is, we can find integers a and b (≠0) such that √3 =a/b

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So, b√3 =a

Squaring on both sides, and rearranging, we get 3b² = a².

Therefore, a² 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 3b² = 9c², that is, b² = 3c².

This means that b² 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 3 is rational.

So, we conclude that √3 is irrational.

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