One of the earliest known proofs in number theory tackles this elegant question:

Problem: Show that \( \sqrt{2} \) cannot be written as a ratio of two integers.
In other words, prove that \( \sqrt{2} \) is irrational.

Step-by-Step Proof (By Contradiction)

Suppose, for contradiction, that \( \sqrt{2} \) is rational.

Then we can write:

\[ \sqrt{2} = \frac{a}{b} \]

where \( a \) and \( b \) are integers with no common factors (i.e., the fraction is in lowest terms).

\[ \left(\frac{a}{b}\right)^2 = 2 \Rightarrow \frac{a^2}{b^2} = 2 \Rightarrow a^2 = 2b^2 \]

So \( a^2 \) is even, which means \( a \) must also be even (since the square of an odd number is odd).

Let \( a = 2k \) for some integer \( k \).

\[ (2k)^2 = 2b^2 \Rightarrow 4k^2 = 2b^2 \Rightarrow b^2 = 2k^2 \]

So \( b^2 \) is also even → \( b \) is even.

Now we’ve concluded that both \( a \) and \( b \) are even, which means they share a common factor of 2.

But that contradicts our assumption that \( \frac{a}{b} \) was in lowest terms.

Final Answer

\( \sqrt{2} \) cannot be expressed as a ratio of two integers.
Therefore, \( \sqrt{2} \) is irrational.