Prove that the following are irrational. $$\displaystyle\frac{1}{\sqrt{2}}$$.
To prove $$\dfrac{1}{\sqrt{2}}$$ is irrational
Let us assume that $$\sqrt{2}$$ is irrational
$$\dfrac{1}{\sqrt{2}} = \dfrac{p}{q}$$ (where $$p$$ and $$q$$ are co prime)
$$\dfrac{q}{p} = \sqrt{2}$$
$$q = \sqrt{2}p$$
Squaring both sides
$$q^2 = 2p^2$$.....................(1)
By theorem - If $$p$$ is a prime no. and $$p$$ divides $$a^2$$, then $$p$$ divides $$a$$ also, where $$a$$ is a positive integer}
$$q$$ is divisible by $$2$$
$$\therefore q = 2c$$ (where $$c$$ is an integer)
Putting the value of q in equitation (1)
$$2p^2 = q^2 = 4c^2$$
$$p^2 =\dfrac{4c^2}{2} =2c^2$$
$$\dfrac{p^2}{2} = c^2$$
$$p$$ is also divisible by $$2$$
But $$p$$ and $$q$$ are coprime
This is a contradiction which has arisen due to our wrong assumption
$$\dfrac{1}{\sqrt{2}}$$ is irrational
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