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
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.
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
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.
No; you can prove the square root of any positive number that's not a perfect square is irrational, using a similar method to showing the square root of 2 is irrational.