Question

Prove that \(2 + 3\sqrt{5}\) is an irrational number given that \(\sqrt{5}\) is an irrational number.

Hint:

In such "prove that" questions, always isolate the root term on one side. If the other side consists entirely of known rational operations (addition, subtraction, multiplication, division) on integers, that side is rational, leading to the contradiction.

Answer 1

Step 1: Understanding the Concept:
The method of contradiction is used to prove irrationality.
We assume the number is rational and show that this leads to a logical inconsistency with the given facts.
Step 2: Key Formula or Approach:
A rational number is any number that can be expressed in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).
Rational numbers are closed under subtraction and division (by non-zero numbers).
Step 3: Detailed Explanation:
Let us assume, to the contrary, that \(2 + 3\sqrt{5}\) is a rational number.
Then, there exist co-prime integers \(a\) and \(b\) (\(b \neq 0\)) such that:
\[ 2 + 3\sqrt{5} = \frac{a}{b} \]
Rearranging the terms to isolate the irrational part:
\[ 3\sqrt{5} = \frac{a}{b} - 2 \]
\[ 3\sqrt{5} = \frac{a - 2b}{b} \]
\[ \sqrt{5} = \frac{a - 2b}{3b} \]
Here, since \(a\) and \(b\) are integers, \(a - 2b\) and \(3b\) are also integers.
This implies that \(\frac{a - 2b}{3b}\) is a rational number.
Consequently, \(\sqrt{5}\) must also be a rational number.
However, this contradicts the given fact that \(\sqrt{5}\) is an irrational number.
Our assumption was wrong.
Step 4: Final Answer:
Therefore, \(2 + 3\sqrt{5}\) is an irrational number.