Answer :
Answer:
To prove that \(7\sqrt{2}\) is an irrational number, given that \(\sqrt{2}\) is an irrational number, we can use proof by contradiction.
### Proof by Contradiction
1. **Assume the Opposite**: Suppose \(7\sqrt{2}\) is a rational number. By definition, a rational number can be expressed as a fraction \(\frac{a}{b}\) where \(a\) and \(b\) are integers and \(b \neq 0\), with \(a\) and \(b\) having no common factors other than 1 (i.e., \(\frac{a}{b}\) is in simplest form).
2. **Express the Assumed Rational Number**: If \(7\sqrt{2}\) is rational, then there exist integers \(a\) and \(b\) such that:
\[
7\sqrt{2} = \frac{a}{b}
\]
3. **Isolate \(\sqrt{2}\)**: Solve for \(\sqrt{2}\) by dividing both sides of the equation by 7:
\[
\sqrt{2} = \frac{a}{7b}
\]
4. **Rationality of \(\sqrt{2}\)**: Since \(a\) and \(b\) are integers, \(7b\) is also an integer. Thus, \(\frac{a}{7b}\) is a fraction where both numerator and denominator are integers. This implies \(\frac{a}{7b}\) is a rational number.
5. **Contradiction**: However, it is given that \(\sqrt{2}\) is an irrational number. This contradicts the assumption that \(\sqrt{2} = \frac{a}{7b}\) is rational.
6. **Conclusion**: Because our assumption that \(7\sqrt{2}\) is rational leads to a contradiction, we must conclude that \(7\sqrt{2}\) is indeed an irrational number.
Thus, \(7\sqrt{2}\) is an irrational number.
Answer:
According to class 10th CBSE (NCERT)
let us assume that 7√2 is rational
so , it can be written in the form of p/q were p and q are co-prime ( does not have a common factor)
& q≠0
then, 7√2 = p/q
take seven to opposite side
√2 = p/7q
give that √2 is irration
we know that irrational ≠ rational
so this contradiction is due to our wrong assumption that 7√2 is rational so , 7√2 is irrational
hence proved.