To prove that [tex]\( 5 + 2 \sqrt{3} \)[/tex] is irrational, we will use a proof by contradiction.
Step-by-step solution:
1. Assume the opposite. Suppose [tex]\( 5 + 2 \sqrt{3} \)[/tex] is rational. This means it can be expressed as a ratio of two integers:
[tex]\[
5 + 2 \sqrt{3} = \frac{p}{q}
\][/tex]
where [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are integers and [tex]\( q \neq 0 \)[/tex]. Additionally, we can assume that [tex]\(\frac{p}{q}\)[/tex] is in its lowest terms (so that [tex]\( \gcd(p, q) = 1 \)[/tex]).
2. Isolate the irrational part. Subtract 5 from both sides to isolate the term involving [tex]\(\sqrt{3}\)[/tex]:
[tex]\[
2 \sqrt{3} = \frac{p}{q} - 5
\][/tex]
Simplify the right-hand side:
[tex]\[
2 \sqrt{3} = \frac{p - 5q}{q}
\][/tex]
3. Divide by 2 to further isolate [tex]\(\sqrt{3}\)[/tex].
[tex]\[
\sqrt{3} = \frac{p - 5q}{2q}
\][/tex]
4. Analyze the implication. Since [tex]\( \sqrt{3} \)[/tex] is irrational, the left side of the equation is irrational. For the right side to be equal to [tex]\(\sqrt{3}\)[/tex], it must also be irrational. However, the right side is a fraction where [tex]\( p \)[/tex], [tex]\( q \)[/tex], and [tex]\( 5 \)[/tex] are all integers. Hence [tex]\(\frac{p - 5q}{2q}\)[/tex] represents a ratio of two integers, which is a rational number.
5. Contradiction arises. Since we have shown that [tex]\(\sqrt{3}\)[/tex] is irrational but expressed [tex]\(\sqrt{3}\)[/tex] as a ratio of two integers (which is rational), we have reached a contradiction. This contradiction arises from the assumption that [tex]\( 5 + 2 \sqrt{3} \)[/tex] is rational.
6. Conclusion. Since our initial assumption that [tex]\( 5 + 2 \sqrt{3} \)[/tex] is rational leads to a logical contradiction, we must conclude that [tex]\( 5 + 2 \sqrt{3} \)[/tex] is irrational.
Thus, [tex]\( 5 + 2 \sqrt{3} \)[/tex] is irrational.
