Prove that \(\sqrt{3} + \sqrt{5}\) is irrational.
Ans:
Let's prove \(\sqrt{3} + \sqrt{5}\) is an irrational number.
Now prove by contradiction method.
| 1. | Assume \(\sqrt{3} + \sqrt{5}\) is | |
| 2. | Therefore, it can be written as | |
| 3. | And \(p\) and \(q\) are | |
| 4 |
Squaring on both sides, we get
|
|
| 5. | Simplifying the term, | |
| 6. | This implies that, | |
| 7. | This | contradicts our assumption |
| 8. | Thus, \(\sqrt{3} + \sqrt{5}\) is |
Answer variants:
\((\sqrt{3} + \sqrt{5})^2 = \left(\frac{q}{p} \right)^2\)
an irrational number
\(\sqrt{15}\) is a rational number
\((\sqrt{3} - \sqrt{5})^2 = \left(\frac{p}{q} \right)^2\)
co-primes
\(\sqrt{3} + \sqrt{5}\)
\(\sqrt{15} =\)\(\frac{8p^2 - q^2}{2q^2}\)
\(\sqrt{3} + \sqrt{5}=\frac{p}{q}\) where \(p\), \(q\) are integers and \(q \neq 0\)
composites
contradicts
\(\sqrt{15}\) is an rational number
Satisfies
\(\sqrt{15} =\)\(\frac{p^2 - 8q^2}{2q^2}\)
\((\sqrt{3} + \sqrt{5})^2 = \left(\frac{p}{q} \right)^2\)
\(\sqrt{3} + \sqrt{5}=\frac{p}{q}\) where \(p\), \(q\) are integers and \(p \neq 0\)
a rational number
\(\sqrt{5} =\)\(\frac{p^2 - 8q^2}{2q^2}\)
\(\sqrt{3} + \sqrt{5}=\frac{p}{q}\) where \(p\), \(q\) are integers and \(q = 0\)