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I’ve consistently been made to solve a rather absurd number of problems that involve proving the irrationality of some radical (Nth Root of X) by Contradiction, and since these problems often use a very similar method of proving irrationality (assuming the radical is equal to a fraction in simplest form, then proving that the Numerator and Denominator have some common multiple.

However, I was wondering if there exists some absolute definitive template to solve such a question, solved exclusively in variables to which numbers can be substituted. I do understand the naïveté of my question but I would greatly appreciate it if someone may explain this exists.

As a side note: I have attempted to do this by stating:

$$\mathrm{N^{th}-Root}(x) = \frac{a}{b}$$

$$x = \frac{a^n}{b^n}$$

$(b^n)x = (a^n)$, thereby proving that $a^n$ is a multiple of $b^n$, but I’ve gotten stuck. Again, I do not know what I am doing and would appreciate any help or explanation!

user79161
  • 499
Ahmed
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  • You could do this: https://proofwiki.org/wiki/Nth_Root_of_Integer_is_Integer_or_Irrational – morrowmh Mar 05 '21 at 18:53
  • See the linked dupes (and their linkls) for most of the common methods. – Bill Dubuque Mar 05 '21 at 19:04
  • "that a^n is a multiple of b^n, but I’ve gotten stuck" .... Hint: Assume $\frac ab$ is in lowest terms. SO if $a^n$ is a multiple $b^n$ but $a$ and $b$ have no factors in common...... [But or you assuming $x$ is an integer?] – fleablood Mar 05 '21 at 19:35
  • Also note $\sqrt{36}$ is not irrational. Kinda important that you take the counter examples into accont. If $\sqrt{36} =\frac ab$ then $b^2\cdot 36 =a^2$ and ... what becomes of that. – fleablood Mar 05 '21 at 19:39

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