To prove that [tex]\(3 \sqrt{5}\)[/tex] is irrational, we'll follow a proof by contradiction. We will assume that [tex]\(3 \sqrt{5}\)[/tex] is rational and then show that this leads to a contradiction with known properties of rational and irrational numbers.
### Step-by-Step Proof:
1. Assume the Contrary:
Suppose [tex]\(3 \sqrt{5}\)[/tex] is rational. By definition, a rational number can be expressed as the quotient of two integers [tex]\(a\)[/tex] and [tex]\(b\)[/tex] (with [tex]\(b \neq 0\)[/tex]), such that:
[tex]\[
3 \sqrt{5} = \frac{a}{b}
\][/tex]
where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers with no common factors other than 1 (i.e., in their simplest form).
2. Isolate [tex]\(\sqrt{5}\)[/tex]:
To isolate [tex]\(\sqrt{5}\)[/tex], divide both sides of the equation by 3:
[tex]\[
\sqrt{5} = \frac{a}{3b}
\][/tex]
3. Rationality of [tex]\(\sqrt{5}\)[/tex]:
Since [tex]\(a\)[/tex] and [tex]\(3b\)[/tex] are both integers, [tex]\(\frac{a}{3b}\)[/tex] is a rational number. This implies that [tex]\(\sqrt{5}\)[/tex] is rational.
4. Contradiction:
However, we know from the properties of numbers that [tex]\(\sqrt{5}\)[/tex] is irrational. The square root of a non-square positive integer (like 5) is always irrational. This is a well-established mathematical fact.
5. Conclusion:
Since assuming that [tex]\(3 \sqrt{5}\)[/tex] is rational led us to the false conclusion that [tex]\(\sqrt{5}\)[/tex] is rational, our initial assumption must be wrong. Therefore, [tex]\(3 \sqrt{5}\)[/tex] cannot be rational.
Thus, we have shown by contradiction that [tex]\(3 \sqrt{5}\)[/tex] is irrational.
