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If $n$ is a positive integer and is not a perfect square, how do you prove that $n^{1/2}$ is irrational?

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  • $\begingroup$ do you know how to prove $\sqrt{2}$ is irrational? $\endgroup$ – user87543 Jan 15 '14 at 17:04
  • $\begingroup$ Yes, I do. But I was told to use the Fundamental Theorem of Arithmetic. $\endgroup$ – Lin Jan 15 '14 at 17:08
  • $\begingroup$ fine... once you prove that $\sqrt{p}$ is irrational for each prime $p$ then you can write $n$ as product of primes using fundamental theorem of arithmetic and then it is done! $\endgroup$ – user87543 Jan 15 '14 at 17:11
  • $\begingroup$ @PraphullaKoushik, you need an additional argument, since in general the product of two irrational numbers can be rational.. $\endgroup$ – Joachim Jan 15 '14 at 17:30
  • $\begingroup$ @Joachim : Of course additional argument is needed.. That is just a kind of hint! $\endgroup$ – user87543 Jan 15 '14 at 17:36
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Since $n$ is not a perfect number, there exists at least one prime number $p$ such that $$n=p^\alpha q$$ where $q\in\mathbb N$ is coprime to $p$ and $\alpha\ge 1$ is odd.

Now, suppose that $n$ is a rational number, namely, $$n^{1/2}=\frac{b}{a}$$ where $a,b$ are natural numbers and they are coprime to each other.

Then, we have $$n=\frac{b^2}{a^2}\Rightarrow a^2n=b^2.$$

By the fundamental theorem of arithmetic, this implies that the number of $p$ in the left hand side is odd, and that the number of $p$ in the right hand side is even. This is a contradiction.

Hence, $n^{1/2}$ is an irrational number.

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  • $\begingroup$ And we know that the number of primes in the left hand side is odd because of what? I'm having serious trouble understanding the Fundamental Theorem of Arithmetic. $\endgroup$ – Lin Jan 15 '14 at 17:28
  • $\begingroup$ @Courtney: Note that the number of the prime number of $p$ is odd in the left hand side, not all prime numbers. By the fundamental theorem of arithmetic, we have only one representation as $a={p_1}^{a_1}{p_2}^{a_2}\cdots {p_k}^{a_k}$ where $p_1\lt p_2\lt\cdots\lt p_k$ are prime numbers. So, $a^2={p_1}^{2a_1}\cdots{p_k}^{2a_k}.$ So, the number of each prime number of $a$ is even. $\endgroup$ – mathlove Jan 15 '14 at 17:39
  • $\begingroup$ You mean "perfect square", not "perfect number". $\endgroup$ – Bill Dubuque Jan 17 '14 at 3:05
  • $\begingroup$ @BillDubuque: Thanks for pointing it out. A typo. I know very little about "perfect number":) $\endgroup$ – mathlove Jan 17 '14 at 4:49
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$\,n = (a/b)^2\Rightarrow\, \color{#0a0}{a^2} = \color{#c00}nb^2.\, $ All primes occur to $\rm\color{#0a0}{even}$ power in the prime factorization of $\,\color{#0a0}{a^2},\,$ but, since $\,n\,$ is not a square, some prime occurs to $\color{#c00}{\rm odd}$ power in $\,\color{#c00}n,\,$ so odd power in $\,nb^2\,\Rightarrow\Leftarrow$

Remark $\ $ This parity argument depends crucially on the uniqueness of prime factorizations, hence it is essential to mention the use of this strong property when presenting the proof. One can also give proofs using closely related properties, e.g. the Rational Root Test (e.g. here) or Euclid's Lemma (e.g. here), or Bezout's gcd identity. Below is a simple proof employing Bezout that I discovered as a teenager (motivated by Dedekind).

Theorem $\quad \rm r = \sqrt{n}\;\;$ is integral if rational, $\:$ for $\:\rm n\in\mathbb{N}$

Proof $\ \ $ Note that $\rm\,\ \color{#0a0}{r = a/b},\ \ \gcd(a,b) = 1\ \Rightarrow\ \color{#C00}{ad\!-\!bc \,=\, \bf 1}\;$ for some $\:\rm c,d \in \mathbb{Z}\,\ $ by Bezout.

$\rm\color{#C00}{That\,}$ and $\rm\: r^2\! = \color{orange}{\bf n}\:\Rightarrow\ \color{#0a0}{0\, =\, (a\!-\!br)}\, (c\!+\!dr) \ =\ ac\!-\!bd\color{orange}{\bf n} \:+\: \color{#c00}{\bf 1}\cdot r \ \Rightarrow\ r \in \mathbb{Z}\ \ \ $ QED

The proof easily generalizes to roots of monic quadratic polynomials (and to higher degrees).

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Assume that $n=(a/b)^2$ for some positive integers $a$ and $b$. Let $p_1\lt p_2\lt\dots\lt p_k$ be all the prime divisors of $nab$. The Fundamental Theorem of Arithmetic tells you that, if $x_1,\dots,x_k$ and $y_1,\dots,y_k$ are nonnegative integers, then $$p_1^{x_1}p_2^{x_2}\cdots p_k^{x_k}=p_1^{y_1}p_2^{y_2}\cdots p_k^{y_k}\ \Rightarrow x_1=y_1,x_2=y_2,\dots,x_k=y_k.$$ Write $$n=p_1^{x_1}p_2^{x_2}\cdots p_k^{x_k},\ a=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k},\ b=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}$$ and use $n=(a/b)^2$ and the Fundamental Theorem to show that the $x_i$ are even, thus showing that $n$ is a perfect square.

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HINT:

Assuming the opposite, you have $\big(\frac ab\big)^2=\frac{a^2}{b^2}=n$ integer, so $b^2|a^2$ and $\frac ab$ not integer, so $b \nmid a$.

If $b \nmid a$, there is a prime factor $p$ of $a$ such that $p|a$ but $p \nmid b$. So $p|a^2$ and $p \nmid b^2$, a contradiction.

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  • $\begingroup$ That's what I was thinking, but I have no idea where to go from there. $\endgroup$ – Lin Jan 15 '14 at 17:07
  • $\begingroup$ @Courtney now is it OK? $\endgroup$ – LeeNeverGup Jan 15 '14 at 17:16
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If $n$ is not a perfect square, then there is a prime $p$ dividing $n$ which occurs in the expansion of $n$ an odd number of times. From the equation $\sqrt n=a/b$ you get $b^2n=a^2$, in which that prime occurs evenly many times on the right, oddly many times on the left. Contradiction.

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As shown in this answer, any algebraic integer that is rational is an integer. Since all roots of $x^2-n=0$ are algebraic integers (since the lead coefficient of $x^2-n$ is $1$), if they are not integers, they are not rational. Thus, since $n$ is not a perfect square, $n$ is irrational.

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