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The proof is by the method of reductio ad absurdum.

We start by assuming that sqrt(3) is rational.

That means that it can be expressed in the form p/q where p and q are co-prime integers.

Thus sqrt(3) = p/q.

This can be simplified to 3*q^2 = p^2

Now 3 divides the left hand side (LHS) so it must divide the right hand side (RHS).

That is, 3 must divide p^2 and since 3 is a prime, 3 must divide p.

That is p = 3*r for some integer r.

Then substituting for p gives,

3*q^2 = (3*r)^2 = 49*r^2

Dividing both sides by 3 gives q^2 = 3*r^2.

But now 3 divides the RHS so it must divide the LHS.

That is, 3 must divide q^2 and since 3 is a prime, 3 must divide q.

But then we have 3 dividing p as well as q which contradict the requirement that p and q are co-prime.

The contradiction implies that sqrt(3) cannot be rational.

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