Q: Proof of Nth root is irrational?

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Yes. The square root of a positive integer can ONLY be either:* An integer (in this case, it isn't), OR * An irrational number. The proof is basically the same as the proof used in high school algebra, to prove that the square root of 2 is irrational.

Square root of 10 is irrational.

The argument why the square root of 2 is irrational can be found in most high school algebra books. You can also find this proof, and several other proofs, that the square root of 2 is irrational, in the Wikipedia article "Square root of 2".The same argument can be applied to the square root of any natural number that is not a perfect square.

irrational

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.

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The square root of a positive integer can ONLY be:* Either an integer, * Or an irrational number. (The proof of this is basically the same as the proof, in high school algebra books, that the square root of 2 is irrational.) Since in this case 32 is not the square of an integer, it therefore follows that its square root is an irrational number.

Most high school algebra books show a proof (by contradiction) that the square root of 2 is irrational. The same proof can easily be adapted to the square root of any positive integer, that is not a perfect square. You can find the proof (for the square root of 2) on the Wikipedia article on "irrational number", near the beginning of the page (under "History").

Yes. The square root of a positive integer can ONLY be either:* An integer (in this case, it isn't), OR * An irrational number. The proof is basically the same as the proof used in high school algebra, to prove that the square root of 2 is irrational.

sqrt(2) is irrational. 3 is rational. The product of an irrational and a non-zero rational is irrational. A more fundamental proof would follow the lines of the proof that sqrt(2) is irrational.

See http://mathforum.org/library/drmath/view/52619.html for a proof of the irrationality of sqrt(3). The proof that sqrt(5) is irrational is identical (substituting 5 for 3 in the proof).

A direct proof of the infinity of primes would require what is essentially a formula to calculate the Nth prime number; such a formula isn't even guaranteed to exist. It's possible to formulate a proof of the infinity of primes that would be, in a sense, direct. A direct proof that the square root of 2 is irrational is impossible, because the irrational numbers aren't defined in any direct way - just as the real numbers which aren't rational. So to prove that the square root of 2 is irrational, we have to prove that it's not rational, which requires indirect techniques.

Root of '3' is NOT rational. It is an IRRATIONAL Number. To 9 d.p. it is sqrt(3) = 1.732050808.... NB THE square roots of prime numbers are irrational , just like 'pi = 3.141592....'. NNB A irrational number, put casually, is a number were the decimals go to inifinty and there is no regular order in the number2.

The square root of 94 is an irrational number

The square root of 200 is irrational.

It is a prime number that has only factors of itself and one therefore it is an irrational number like all prime numbers are.

Search for the proof for the irrationality of the square root of 2. The same reasoning applies to any positive integer that is not a perfect square. In summary, the square root of any positive integer is either a whole number, or - as in this case - it is irrational.

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.