Answer: The square root of 6 is irrational. Reason: Just try to convince it otherwise, you will see their is no way to deal with it since it becomes angry and irrational! But seriously, you can't write it as a fraction of the from p/q with p and q being integers so yes it is irrational. The proof would be easy by contradiction.
No. The square root of 6 is irrational.
6 times the square root of 6
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)
The square root of 6 is an irrational number; it can only be approximated as a decimal. √6 ≈ 2.449
The square root of six is irrational. Do you mean a rational approximation?
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.
No. The square root of 6 is irrational.
6 times the square root of 6
It is an irrational number.
irrational
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)
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).
The square root of 6 is an irrational number; it can only be approximated as a decimal. √6 ≈ 2.449
Irrational: it amounts to roughly 2.449489743
yes.
No, it is not irrational because it is a square root of a negative number - which falls into the set of Complex numbers. Irrational numbers can not have an imaginary component.
It is irrational. * The square root of any positive integer, except of a perfect square, is irrational. * The product of an irrational number and a rational number (except zero) is irrational.