It is irrational.
The proof depends on the proof that sqrt(5) is irrational. However, judging by this question, I suggest that you are not yet ready for that proof.
So, assume that sqrt(5) is irrational. Any multiple of an irrrational number by a non-zero rational isirrational.
For suppose 2*sqrt(5) were rational
that is 2*sqrt(5) = p/q for some integers p and q, where q is nonzero.
then dividing both isdes by 2 gives sqrt(5) = p/(2q) where p and 2q are both integers and 2q in non-zero.
But that implies that sqrt(5) is rational!
That is a contradiction so 2*sqrt(5) cannot be rational.
No, the square root of an irrational number is not always rational. In fact, the square root of an irrational number is typically also irrational. For example, the square root of 2, which is an irrational number, is itself irrational. However, there are exceptions, such as the square root of a perfect square of an irrational number, which can be rational.
The square root of 4 is 2 which is a rational number
irrational
The square roots of 2 and 3 are irrational but not transcendent.
No; √2 is irrational.
The square root of 2 is an irrational number
It is rational. The root of a perfect square, such as 4, is rational; the root of any positive integer that is not a perfect square is an irrational number.
irrational
The square root of (any number that isn't a perfect square) is irrational.
the square root of 2 fifths of irrational is 0.565685425
The square root of 4 is 2 which is a rational number
irrational
sqrt(32) = 4sqrt(2) The square root of '2' is irrational, so the square root of '32' is irrational.
The square roots of 2 and 3 are irrational but not transcendent.
No; √2 is irrational.
1, 2 are rational and square root of 2 and pi are irrational.
The square root of 2 is irrational. In general, the square root of a positive integer is either an integer (if you take the square root of a perfect square), or it is irrational.