Answer variants:
satisfies
co-primes
contradicts
rational Number
\(\sqrt{31} = \frac{p}{q}\)
composites
\(q^2\) is divisible by \(31\) and \(q\) is also divisible by \(31\)
\(31q^2 = p^2\)
cannot be expressed as p/q form
irrational Number
can be expressed as p/q form
Let's prove is an irrational number.
Now prove by contradiction method.
| 1. | Assume is a | |
| 2. | By the definition, | |
| 3. | And \(p\) and \(q\) are | |
| 4. | So we can write it as | |
| 5. | Simplifying the term, | |
| 6. | This implies that, | |
| 7. | This | our assumption. |
| 8. | Thus, is |