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Answer :

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

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

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

\(=>\sqrt{5} × q = p \)

On squaring both sides we'll get:

\(=>5q^2 = p^2 \) ............................(i)

From equation (i) it can be clearly seen that 5 is a factor of \(p^2\)

Therefore, 5 is also factor of p.

Let p = 5r, where r is an integer.

On substituting the value of p in equation (i),

\(=>5q^2 = 25r^2 \)

\(=>q^2 = 5r^2 \).............................(ii)

From equation (ii) it can be clearly seen that 5 is a factor of \(q^2\)

Therefore, 5 is also factor of q.

Which means 5 is a common factor of p and q which contradicts our assumption,

that p and q are co-prime and also that \(\sqrt{5}\) is rational.

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

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