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
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.