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# Prove that the following are irrationals. i.$${{1}\over \sqrt{2}}$$ ii.$$7 \sqrt{5}$$ iii.$$6 + \sqrt{2}$$

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.