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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