To prove that [tex]\(2 - 5\sqrt{2}\)[/tex] is irrational, we'll assume the opposite and derive a contradiction. Here's the detailed step-by-step proof:
1. Assume [tex]\(2 - 5\sqrt{2}\)[/tex] is rational.
- If [tex]\(2 - 5\sqrt{2}\)[/tex] is rational, it can be expressed as the ratio of two integers, say [tex]\( k = \frac{a}{b} \)[/tex] where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers, and [tex]\(b \neq 0\)[/tex].
2. Express the assumption mathematically:
- Assume [tex]\(2 - 5\sqrt{2} = \frac{a}{b}\)[/tex].
3. Rewrite the equation to isolate [tex]\(\sqrt{2}\)[/tex]:
[tex]\[
2 - \frac{a}{b} = 5\sqrt{2}
\][/tex]
[tex]\[
2b - a = 5b\sqrt{2}
\][/tex]
4. Isolate the irrational part, [tex]\(\sqrt{2}\)[/tex]:
[tex]\[
\sqrt{2} = \frac{2b - a}{5b}
\][/tex]
5. Analyze both sides of the equation:
- The right side of the equation, [tex]\(\frac{2b - a}{5b}\)[/tex], is a ratio of integers and hence rational.
- The left side of the equation, [tex]\(\sqrt{2}\)[/tex], is known to be irrational.
6. Reach a contradiction:
- We have an irrational number ([tex]\(\sqrt{2}\)[/tex]) equated to a rational number ([tex]\(\frac{2b - a}{5b}\)[/tex]).
- This is a contradiction because a number cannot be both rational and irrational simultaneously.
7. Conclusion:
- Our initial assumption that [tex]\(2 - 5\sqrt{2}\)[/tex] is rational leads to a contradiction.
- Therefore, [tex]\(2 - 5\sqrt{2}\)[/tex] must be irrational.
Hence, [tex]\(2 - 5\sqrt{2}\)[/tex] is proven to be an irrational number.
