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Suppose we start with the opposite view. That is, the square root of 5 is rational. This means that it can be expressed as a ratio of two integers. Let the square root of 5 in its simplest terms be a/b so that a and b are integers with no common factors and b > 0.
sqrt(5) = a/b so 5 = a2/b2
that is, 5b2= a2
5 is a factor of the left hand side (LHS) so 5 must be a factor of the right hand side (RHS).
Since 5 is a prime and a is an integer, 5 must be a factor of a. That is, a = 5c for some integer c and since a and b are coprime b and c must also be coprime.
But now,
5 = (5c)2/b2so that 5 = 25c2/b2and so b2= 5c2. As before, this implies that 5 must be a factor of b. But that contradicts the supposition that a and b are coprime.
The contradiction implies that the original assumption was incorrect. That is, sqrt(5) cannot be rational.
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No; you can prove the square root of any positive number that's not a perfect square is irrational, using a similar method to showing the square root of 2 is irrational.
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I linked a good resource that explains what you asked below.
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yes
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Prove that square root of 2 is an irrational number?
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