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I have tried this problem five times but I keep getting stuck. I keep following the proof for $\sqrt{2}$. I know that I have to prove that the set is nonempty. Which I do by induction.

$2^1 > 1$

assume $2^n > n$ and prove $2^{n+1} > n+1$

$2^n + 2^n > n+n > n$

Then I think I have to prove that it has an upper bound. Also, I think it has something to do with prime numbers?

Ben Grossmann
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We need to show that if $n^{1/2}$ is a rational number, then $n^{1/2}$ is an integer.

Suppose that $n^{1/2}=\frac{a}{b}$, where $a$ and $b$ are integers. We may without loss of generality assume that $a$ and $b$ are relatively prime, and that $b\ge 1$. We show that $b=1$.

Suppose to the contrary that $b\gt 1$. Then some prime $p$ divides $b$.

From $n^{1/2}=\frac{a}{b}$, we get that $b^2n=a^2$. Since $p$ divides $b$, we conclude that $p$ must divide $a^2$, so $p$ divides $a$. This contradicts the fact that $a$ and $b$ are relatively prime.

So $n^{1/2}=\frac{a}{1}$, and therefore $n^{1/2}$ is an integer.

André Nicolas
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  • It is worth emphasis that, although not explicitly mentioned, the proof depends crucially on the fundamental theorem of arithmetic, i.e. the existence and uniqueness of prime factorizations. The statement generally fails in domains lacking this property. – Bill Dubuque Feb 13 '14 at 04:47
  • The proof depends strongly on "Euclid's Lemma," in this instance the fact that if a prime divides $xy$, then $p$ divides $x$ or $p$ divides $y$. I agree that after the Lemma, the proof of the Fundamental Theorem of Arithmetic is straightforward, but the proof above only uses the Lemma. – André Nicolas Feb 13 '14 at 04:52
  • Euclid's Lemma is equivalent to the uniqueness of prime factorizations (as are many closely related properties, e.g. see here for over 15 more equivalents) – Bill Dubuque Feb 13 '14 at 05:07