To prove that [tex]\(3 + \sqrt{5}\)[/tex] is an irrational number, let's proceed with a proof by contradiction. We'll assume, for the sake of contradiction, that [tex]\(3 + \sqrt{5}\)[/tex] is rational, and show that this assumption leads to a contradiction.
1. Assume [tex]\(3 + \sqrt{5}\)[/tex] is rational:
Suppose [tex]\(3 + \sqrt{5}\)[/tex] is a rational number. A number is rational if it can be expressed as a fraction of two integers. Therefore, assume there exist integers [tex]\(a\)[/tex] and [tex]\(b\)[/tex] (with [tex]\(b \neq 0\)[/tex]) such that:
[tex]\[
3 + \sqrt{5} = \frac{a}{b}.
\][/tex]
2. Isolate [tex]\(\sqrt{5}\)[/tex]:
To isolate the term involving [tex]\(\sqrt{5}\)[/tex], subtract 3 from both sides of the equation:
[tex]\[
\sqrt{5} = \frac{a}{b} - 3.
\][/tex]
3. Rewrite [tex]\(\sqrt{5}\)[/tex] in terms of rational numbers:
Since [tex]\(\frac{a}{b}\)[/tex] is a rational number and 3 is also a rational number, their difference must be a rational number. Hence, let:
[tex]\[
\sqrt{5} = \frac{a - 3b}{b}.
\][/tex]
The right-hand side of the equation, [tex]\(\frac{a - 3b}{b}\)[/tex], is a rational number because it is the ratio of two integers.
4. Reach a contradiction:
We have now expressed [tex]\(\sqrt{5}\)[/tex] as a rational number. However, it is a well-known result from number theory that [tex]\(\sqrt{5}\)[/tex] is an irrational number. This means it cannot be expressed as a ratio of two integers.
5. Conclusion:
Our assumption that [tex]\(3 + \sqrt{5}\)[/tex] is rational leads to the conclusion that [tex]\(\sqrt{5}\)[/tex] would be rational, which contradicts the established fact that [tex]\(\sqrt{5}\)[/tex] is irrational. Therefore, the initial assumption must be incorrect.
Thus, we conclude that [tex]\(3 + \sqrt{5}\)[/tex] is indeed an irrational number.
