Show That the Following Numbers Are Irrational

Video Explanation

Question

Show that the following numbers are irrational:

(i) \( \dfrac{1}{\sqrt{2}} \)
(ii) \( 7\sqrt{5} \)
(iii) \( 6 + \sqrt{2} \)
(iv) \( 3 – \sqrt{5} \)

Solution

(i) \( \dfrac{1}{\sqrt{2}} \)

We know that \( \sqrt{2} \) is irrational.

If \( \dfrac{1}{\sqrt{2}} \) were rational, then its reciprocal \( \sqrt{2} \) would also be rational.

But this is a contradiction.

\[ \therefore \quad \dfrac{1}{\sqrt{2}} \text{ is irrational.} \]

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(ii) \( 7\sqrt{5} \)

Assume \( 7\sqrt{5} \) is rational.

Then, \[ \sqrt{5} = \frac{1}{7}(7\sqrt{5}) \]

Since a rational number multiplied by a rational number is rational, this would imply \( \sqrt{5} \) is rational.

But \( \sqrt{5} \) is irrational.

This is a contradiction.

\[ \therefore \quad 7\sqrt{5} \text{ is irrational.} \]

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(iii) \( 6 + \sqrt{2} \)

Assume \( 6 + \sqrt{2} \) is rational.

Then, \[ \sqrt{2} = (6 + \sqrt{2}) – 6 \]

Since the difference of two rational numbers is rational, this would imply \( \sqrt{2} \) is rational.

But \( \sqrt{2} \) is irrational.

This is a contradiction.

\[ \therefore \quad 6 + \sqrt{2} \text{ is irrational.} \]

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(iv) \( 3 – \sqrt{5} \)

Assume \( 3 – \sqrt{5} \) is rational.

Then, \[ \sqrt{5} = 3 – (3 – \sqrt{5}) \]

This implies \( \sqrt{5} \) is rational, which is a contradiction.

\[ \therefore \quad 3 – \sqrt{5} \text{ is irrational.} \]

Conclusion

All the given numbers are irrational:

\[ \dfrac{1}{\sqrt{2}},\quad 7\sqrt{5},\quad 6+\sqrt{2},\quad 3-\sqrt{5} \]

\[ \therefore \quad \text{The given numbers are irrational.} \]