Answer
See explanation
Work Step by Step
**Proof by Unique Factorization (Counting 2’s):**
Suppose, for the sake of contradiction, that \(\sqrt{2}\) is rational. Then we can write
\[
\sqrt{2} = \frac{m}{n},
\]
where \(m\) and \(n\) are positive integers with no common factors (i.e. the fraction is in lowest terms). Squaring both sides gives:
\[
2 = \frac{m^2}{n^2} \quad \Longrightarrow \quad m^2 = 2n^2.
\]
Now, express \(m\) and \(n\) in their prime factorizations. In particular, write:
\[
m = 2^r \cdot s \quad \text{and} \quad n = 2^t \cdot u,
\]
where \(r\) and \(t\) are nonnegative integers, and \(s\) and \(u\) are odd integers.
Then,
\[
m^2 = 2^{2r} \cdot s^2 \quad \text{and} \quad n^2 = 2^{2t} \cdot u^2.
\]
Substitute these into the equation \(m^2 = 2n^2\):
\[
2^{2r} \cdot s^2 = 2 \cdot \left(2^{2t} \cdot u^2\right) = 2^{2t+1} \cdot u^2.
\]
By the Fundamental Theorem of Arithmetic (unique factorization), the exponent of the prime \(2\) must be the same on both sides of the equation. Therefore, we must have:
\[
2r = 2t + 1.
\]
However, the left-hand side \(2r\) is an even number, while the right-hand side \(2t + 1\) is odd. This is impossible.
Thus, our initial assumption that \(\sqrt{2}\) is rational leads to a contradiction. Therefore, \(\sqrt{2}\) is irrational.
