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i.\({{1}\over \sqrt{2}}\)

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

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

Answer :

i.\({{1}\over \sqrt{2}}\)

Let us suppose that \({{1}\over \sqrt{2}}\) is rational

So there should be two co- prime integers (p and q, where q \(\ne\) 0 ) such that:

\({{1}\over \sqrt{2}} = {{p}\over{q}} \)

\(\sqrt{2} = {{q}\over{p}} \)

p and q are integers which means \({{q}\over{p}}\) is a rational number.

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

This is not possible since LHS \(\ne\) RHS.

This contradiction has risen because of our incorrect assumption that \({{1}\over \sqrt{2}}\) is rational.

Hence it is proved that \({{1}\over \sqrt{2}}\) is irrational.

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

Let us suppose that \(7 \sqrt{5}\) is rational.

So there should be two co- prime integers (p and q, where q \(\ne\) 0 ) such that:

\(7 \sqrt{5} = {{p}\over{q}} \)

\(\sqrt{5} = {{p}\over{7q}} \)

p and q are integers which means \({{p}\over{7q}}\) is a rational number.

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

This is not possible since LHS \(\ne\) RHS.

This contradiction has arisen because of our incorrect assumption that \(7 \sqrt{5}\) is rational.

Hence it is proved that \(7 \sqrt{5}\) is irrational.

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

Let us suppose that \( 6 + \sqrt{2}\) is rational

So there should be two co- prime integers (p and q, where q \(\ne\) 0 ) such that:

\( 6 + \sqrt{2} = {{p}\over{q}} \)

\(\sqrt{2} = {{p}\over{q}} - 6 \)

p and q are integers which means \({{p}\over{q}}-6 \) is a rational number.

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

This is not possible since LHS \(\ne\) RHS.

That proves that our assumption that \( 6 + \sqrt{2}\) is rational is wrong.

Hence it is proved that \( 6 + \sqrt{2}\) is irrational.

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