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Question

Prove that $$\sqrt{5}$$ is irrational.


Solution

Let root $$\sqrt{5}$$ be rational.
Then it must in the form of $$\dfrac{p}{q}$$ [$$q$$ is not equal to $$0$$ and $$p, q$$ are co-prime].
Hence, $$\sqrt {5}=\dfrac{p}{q}$$
$$\implies\sqrt {5} q = p$$.
Squaring on both sides,
$$\implies 5q^2 = p^2$$  ------ (1)
$$\implies q^2=\dfrac{p^2}{5}$$.
Therefore, $$p^2$$ is divisible by $$5$$
and $$p$$ is divisible by $$5$$. ------{If $$p$$ is a prime no. and $$p$$ divides $$a^2$$, then $$p$$ divides $$a$$ also, where $$a$$ is a positive integer}

Then, $$p = 5c$$  [$$c$$ is a positive integer].
Squaring on both sides,
$$\implies p^2 = 25c^2$$  --------- (2)

Now, substitute for $$p^2$$ in (1),
we get, $$5q^2 = 25c^2$$
$$\implies q^2 = 5c^2$$.
Therefore, $$q^2$$ is divisible by $$5$$
and $$q$$ is divisible by $$5$$.

Thus $$q$$ and $$p$$ have a common factor $$5$$.
There is a contradiction.
Therefore, $$p$$ and $$q$$ are not co-prime.

Hence, $$\sqrt{5}$$ is irrational.

Mathematics

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