Certainly! Let's prove that [tex]\(\sqrt{5}\)[/tex] is irrational using a method called proof by contradiction. Here's a detailed, step-by-step explanation:
Step 1: Assume the opposite (i.e., [tex]\(\sqrt{5}\)[/tex] is rational).
Assume that [tex]\(\sqrt{5}\)[/tex] is rational. This means it can be expressed as a fraction [tex]\(\frac{a}{b}\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers with no common factors other than 1 (i.e., [tex]\(\frac{a}{b}\)[/tex] is in its simplest form).
[tex]\[
\sqrt{5} = \frac{a}{b}
\][/tex]
Step 2: Square both sides to eliminate the square root.
[tex]\[
5 = \left(\frac{a}{b}\right)^2
\][/tex]
[tex]\[
5 = \frac{a^2}{b^2}
\][/tex]
Step 3: Multiply both sides by [tex]\(b^2\)[/tex] to clear the fraction.
[tex]\[
5b^2 = a^2
\][/tex]
Step 4: Analyze the implications for [tex]\(a^2\)[/tex].
Since [tex]\(5b^2 = a^2\)[/tex], this means that [tex]\(a^2\)[/tex] is divisible by 5. Consequently, [tex]\(a\)[/tex] itself must be divisible by 5 (because if a square of a number is divisible by a prime, the number itself must also be divisible by that prime).
Let’s express [tex]\(a\)[/tex] as [tex]\(5k\)[/tex] for some integer [tex]\(k\)[/tex].
[tex]\[
a = 5k
\][/tex]
Step 5: Substitute [tex]\(a = 5k\)[/tex] back into the equation.
[tex]\[
5b^2 = (5k)^2
\][/tex]
[tex]\[
5b^2 = 25k^2
\][/tex]
Step 6: Simplify the equation.
[tex]\[
b^2 = 5k^2
\][/tex]
Step 7: Analyze the implications for [tex]\(b^2\)[/tex].
Since [tex]\(b^2 = 5k^2\)[/tex], this means that [tex]\(b^2\)[/tex] is also divisible by 5. Consequently, [tex]\(b\)[/tex] itself must be divisible by 5.
Step 8: Arrive at a contradiction.
Since both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are divisible by 5, this contradicts our original assumption that [tex]\(\frac{a}{b}\)[/tex] is in its simplest form (with [tex]\(a\)[/tex] and [tex]\(b\)[/tex] having no common factors other than 1).
Conclusion:
This contradiction implies that our initial assumption—that [tex]\(\sqrt{5}\)[/tex] is rational—is false. Therefore, [tex]\(\sqrt{5}\)[/tex] must be irrational.
Thus, we have proved that [tex]\(\sqrt{5}\)[/tex] is irrational.
