Are all of the following options being irrational?

1) $\frac{1}{\sqrt{2}}$

2) $7 \sqrt{5}$

3) $6 + \sqrt{2}$

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Detailed Solution

In irrational number the denominator q is not equal to zero (q ≠ 0). (i) 1/√2
Let us assume that 1/√2 is a rational number.
Then, 1/√2 = a/b, where a and b have no common factors other than 1.
√2 × a = b
√2 = b/a
Since b and a are integers, b/a is a rational number and so, √2 is rational.
But we know that √2 is irrational. So, our assumption was wrong. Therefore, 1/√2 is an irrational number.
(ii) 7√5
Let us assume that 7√5 is a rational number.
Then, 7√5 = a/b, where a and b have no common factors other than 1.
(7√5) b = a
√5 = a/7b
Since, a, 7, and b are integers, so, a/7b is a rational number. This means √5 is rational. But this contradicts the fact that √5 is irrational.
So, our assumption was wrong. Therefore, 7√5 is an irrational number.
(iii) 6 + √2
Let us assume that 6 + √2 is rational.
Then, 6 + √2 = a/b, where a and b have no common factors other than 1.
√2 = (a/b) - 6
Since, a, b, and 6 are integers, so, a/b - 6 is a rational number. This means √2 is also a rational number.
But this contradicts the fact that √2 is irrational. So, our assumption was wrong.
Therefore,