Yes.
It is irrational.
YES.... squre root of 3 is in irrational number..
No, because the square root of 3 is an irrational number
Because 3 is a prime number and as such its square root is irrational
The square roots of 2 and 3 are irrational but not transcendent.
The square root of 3 is an irrational number
The square root of -3 is an imaginary number and the square root of 3 is an irrational number that can't be expressed as a fraction
It is irrational.
sqrt(27) = sqrt(3 x 9) => sqrt(3) x sqrt(9) => sqrt(3) X 3 => 3*sqrt(3) 3* 1.732050808..... => 5.196152423..... NB 'Sqrt' of prime numbers are IRRATIONAL. Since '3' is a prime number, then its sqrt(3) is irrational at 1.73206080..... NNB The product of a rational number and an irrational number is irrational.
YES.... squre root of 3 is in irrational number..
No, because the square root of 3 is an irrational number
It is known that the square root of an integer is either an integer or irrational. If we square root2 root3 we get 6. The square root of 6 is irrational. Therefore, root2 root3 is irrational.
Because 3 is a prime number and as such its square root is irrational
The square roots of 2 and 3 are irrational but not transcendent.
Yes. For example, the square root of 3 (an irrational number) times the square root of 2(an irrational number) gets you the square root of 6(an irrational number)
square root of (2 ) square root of (3 ) square root of (5 ) square root of (6 ) square root of (7 ) square root of (8 ) square root of (9 ) square root of (10 ) " e " " pi "
Sometimes it is and sometimes it isn't. The square root of 2 and the square root of 3 are both irrational, as is their product, the square root of 6. The square root of 2 and the square root of 8 are both irrational, but their product, the square root of 16, is rational (in fact, it equals 4).