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.
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.
1, 2 are rational and square root of 2 and pi are irrational.
The square root of 2 is an irrational number
If the positive square root (for example, square root of 2) is irrational, then the corresponding negative square root (for example, minus square root of 2) is also irrational.
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
sqrt(32) = 4sqrt(2) The square root of '2' is irrational, so the square root of '32' is irrational.
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.