# Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.5

5. Use the well-ordering property to show that $\sqrt{3}$ is irrational.

Solution: We prove the statement by contradiction.

Now, we assume that $\sqrt{3}$ is rational.

Then there exist positive integers a and b such that $\sqrt{3}=\frac{a}{b}$

Let us consider the set $S=\left\{k\sqrt{3}:k\ and\ k\sqrt{3}\ are\ positive\ integers\right\}$

The set S is non-empty because $a=b\sqrt{3}$

Therefore, by the well ordering property, has a smallest element, say,

We have,
$s\sqrt{3}-s=s\sqrt{3}-t\sqrt{3}$
$s\sqrt{3}-s=(s-t)\sqrt{3}$

Since, $s\sqrt{3}=3t$ and s are both integers,$(s-t)\sqrt{3}$ must also be an integer.

Also, it is positive, since,
$s\sqrt{3}-s=s(\sqrt{3}-1)$
And
$\sqrt{3}>1$
It is less than s because $s=t\sqrt{3},s\sqrt{3}=3t,\sqrt{3}<3$.
This contradicts our assumption of s being the smallest positive integer in S.
Hence,$\sqrt{3}$ is irrational.

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