Verify that \(\sqrt{2} + \sqrt{83}\) is irrational.
Ans:
Let's prove \(\sqrt{2} + \sqrt{83}\) is an irrational number.
Now prove by contradiction method.
| 1. | Assume \(\sqrt{2} + \sqrt{83}\) 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 | our assumption |
| 8. | Thus, \(\sqrt{2} + \sqrt{83}\) is |
Answer variants:
contradicts
\(\sqrt{2}+\sqrt{83}\)
\((\sqrt{2}+\sqrt{83})^2=(\frac{p}{q})^2\)
\(\frac{p^{2}-2q^{2}}{q^{2}}\) is a rational number
co-prime numbers
\(\frac{a^{2}-2b^{2}}{b^{2}}\) is an irrational number
an irrational number
\((\sqrt{2}+\sqrt{83})^2=(\frac{q}{p})^2\)
\(\sqrt{2}+\sqrt{83}=\frac{q}{p}\) where \(p\), \(q\) are integers and \(q=0\)
\(83=\frac{p^{2}}{q^{2}}-2\frac{p}{q}\sqrt{2}+2\)
a rational number
\(\sqrt{2}+\sqrt{83}=\frac{p}{q}\) where \(p\), \(q\) are integers and \(q\neq0\)