Prove that $$\sqrt{3}$$ is irrational.
Let us assume, to the contrary, that $$\sqrt{3}$$ is rational.
That is, we can find integers a and b $$(\neq 0)$$ such that
$$\sqrt{3}=\displaystyle\frac{a}{b}$$.
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, $$b\sqrt{3}=a$$.
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 $$\sqrt{3}$$ is rational. So, we conclude that $$\sqrt{3}$$ is irrational.
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