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studying for a final right now, and one of the my study questions is,

If $p$ is prime then $\sqrt{p}$ is not rational (i.e., irrational).

I understand the standard proof by contradiction, where the contradiction is that $a, b$ are not coprimes when you represent $$\sqrt{p} = \frac{a}{b},$$

but I attempted a different way and want to see if it is acceptable or not (I'm assuming the latter case).

So,

  1. Assume $\sqrt{p}$ as a rational number then,
  2. $\sqrt{p} = \frac{a}{b}$, where $a, b$ coprimes
  3. $p = {\left(\frac{a}{b}\right)}^{2}$

Now, since $p$ is a prime number, then the only way to represent $p$ as a rational number is by $p = \frac{p}{1}$. So, $a = p$ and $b = 1$.

  1. So, we have $p = \frac{p^2}{1^2}$

  2. $p = p^2$, Contradiction.

It follows that $\sqrt{p}$ cannot be rational (i.e., $\sqrt{p}$ is irrational). QED.

Is this acceptable? Or are steps (3) $\rightarrow$ (4) too much of a leap?

Any comments appreciated.

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  • $\begingroup$ Fixed comment I think there is a mistake, you have $p = ( \dfrac ab)^2$, so $p = a^2$ and $1 = b^2$, not $p = a$ $\endgroup$ – Ovi Dec 18 '15 at 5:24
  • $\begingroup$ Step 3. But you aren't representing p. You are representing $\sqrt p $. Also p being prime doesn't mean p/1 is the only rep. 2p/p is also. If you meant the only reduced rep, you'd be right but that would be true of all integers. As 4 can only be represented as 4/1 it doesn't follow that 4 =(a/b)^2 means a =4 b =1. And discovering that 4 $\ne 4^2$ doesn't mean $\sqrt 2$ is irrational. $\endgroup$ – fleablood Dec 18 '15 at 5:57
  • $\begingroup$ If n and m are coprime does it follow that n^2 and m^2 must also be coprime? Then if p = a^2/b^2 it follows the b =1 and a^2 = p. (Not p^2). but p is not a perfect square. $\endgroup$ – fleablood Dec 18 '15 at 7:22
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$p = \frac{p}{1}$ is not the only way to represent $p$; $\frac{2p}{2}$, etc. would also work.

I think you are getting to the idea that $p = \frac{a^2}{b^2}$, for integers $a,b \in \mathbb{Z}$, would imply that $a^2 = p b^2$; and if you assume unique factorization into primes, this is a contradiction because $a^2$ has an even power of $p$ while $p b^2$ has an odd power of $p$. You can make a valid proof with this argument.

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  • $\begingroup$ The author of the original post did assume that $a$ and $b$ were coprime - so, though saying that $\frac{p}1$ is the only representation of $p$ is not correct (as you note), it is not unsalvageable in this context. $\endgroup$ – Milo Brandt Dec 18 '15 at 5:26
  • $\begingroup$ But n = a/b a b coprime implies a = n b =1 is true for all integers. Not just primes. $\endgroup$ – fleablood Dec 18 '15 at 6:01
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It strikes me that your proof is generally fine until you set $a=p$ and $b=1$. In particular, assuming $a$ and $b$ are coprime, one can also prove that $a^2$ and $b^2$ are coprime. Thus, when you reach $$p=\frac{a^2}{b^2}$$ you would be allowed to say $a^2=p$ and $b^2=1$ because this is the only representation of $p$ as the quotient of two coprime natural numbers. Your method forgot to square $a$ and $b$ for using this equality. In fact, to this point in the proof, we have not used that $p$ is prime - this part of your argument establishes that the square root of an integer is either irrational or an integer.

From here, one merely needs to conclude that $p$ is not the square of an integer - which is easy since if $a^2=p$ then $a$ is a factor of $p$, meaning $a=1$ or $a=p$ since $p$ is prime, which of course are not solutions to $a^2=p$, finishing the proof.

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There is another way to prove that $\sqrt p$ is not rational.

$1)$ Show that if $n$ is not a perfect square, and if you assume$\sqrt n = \dfrac ab$, then $b \not = 1$

$2)$ If $a$ and $b$ are co-prime, $n = \dfrac ab \cdot \dfrac ab$, so $\dfrac {a^2}{b^2}$ is an unsimplifiable fraction

$3)$ Conclude that if the square root of a number $n$ can be represented by co-prime $\dfrac ab$, then the number $n$ cannot be an integer

Let me know if you want me to fill in the details!

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  • $\begingroup$ Nice proof Ovi, +1 for a beautiful proof. $\endgroup$ – user268307 Dec 18 '15 at 5:25
  • $\begingroup$ @AnuragJain Thank you! $\endgroup$ – Ovi Dec 18 '15 at 5:25
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Let $p \in \mathbb{Z}$ be a prime number.

Suppose that

$$p=\frac{a^2}{b^2}$$ with $a,b\in \mathbb Z$ coprimes. The only way of representing $p$ as a ratio of two coprime numbers number is $\frac p1$, so

$$a^2=p\implies a \mid p$$ But then either $a=p$ or $a=1$; both cases are absurd.

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The step from $3$ to $4$ seems to be false.

The following are all valid representations of $p$:

$\frac{p}{1},\frac{2p}{2},\frac{3p}{3}$ and in general $\frac{kp}{k}$ where $k$ is a non-zero integer.

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But you can represent primes in other ways. Remember, the only assumed that $a$ and $b$ are coprime. $3 = \displaystyle\frac{6}{2}$

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