We want to proove that:

If $p_1,...,p_n$ are positive, distinct, prime numbers then $\sqrt{p_1\cdots p_n} \notin \Bbb{Q}$.

Let's assume that $\sqrt{p_1\cdots p_n} \in \Bbb{Q}$. Then, $\exists (a,b)\in \mathbb{Z^*\times Z^*}:\sqrt{p_1\cdots p_n}=\frac{a}{b}$ with $\gcd (a,b)=1.$ So, $a^2=p_1\cdots p_n \cdot b^2. $ But how do we continue? Is this technique right or should we follow something different?

PS: This is a part this proof, and I would like to discuss it.

Thank you.


marked as duplicate by JonMark Perry, Siong Thye Goh, Lord Shark the Unknown, user91500, mlc Aug 25 '17 at 7:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ $p_1$ appears with odd exponent in the right-hand side, but with even exponent in the left-hand side. $\endgroup$ – egreg Aug 24 '17 at 20:08
  • $\begingroup$ @Chris See each of the following: math.stackexchange.com/questions/189130/… or math.stackexchange.com/questions/4467/… $\endgroup$ – Davood Khajehpour Aug 24 '17 at 20:21
  • $\begingroup$ I don't think this question is a duplicate of the one indicated. The one indicated states "an integer or an irrational number", this one "$\notin \mathbb{Q}$". So, any answer should also show $\sqrt{p_1...p_n}$ is not an integer. I also was confused by this subtlety ... $\endgroup$ – rtybase Aug 25 '17 at 13:00

The classic proof that $\sqrt 2 \notin \Bbb Q$ readily extends to this case, to wit:


$\sqrt{p_1p_2 . . . p_n} = \dfrac{a}{b} \tag 1$

with $a, b \in \Bbb N$, $\gcd(a, b) = 1$, then

$p_1p_2 \ldots p_n b^2 = a^2, \tag 2$

whence $p_1 \mid a^2$; thus $p_1 \mid a$ and so $a = p_1c$; thus

$a^2 = p_1^2c^2 = p_1p_2 \ldots p_n b^2, \tag 3$


$p_1c^2 = p_2p_3 \ldots p_n b^2; \tag 4$

since the $p_i$ are distinct, we must have $p_1 \mid b^2$, whence $p_1 \mid b$, contradicting our assumption that $\gcd(a, b) = 1$.

I think the ancient roots or this proof are worthy or respect, and I also note the method readily extends to show many similar propositions hold.

I presented this answer because it seems to me that the linked proof is pretty complex for this particular problem, although it is certainly engaging in and of itself, and leads in entaging directions.

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    $\begingroup$ Thank you for this clear proof. I 'm sure this will help and the other proof (which I linked) above. $\endgroup$ – Chris Aug 24 '17 at 20:48
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    $\begingroup$ You should justify the fundamental claim that prime $\,p\mid a^2\Rightarrow\,p\mid a.\ $ $\endgroup$ – Bill Dubuque Aug 24 '17 at 23:07

Hint: think about the prime factorizations of $a$ and $b$, and note that $\gcd(a,b)=1$ implies that they do not share any prime factors. Use this to show that $a^2 = p_1 \cdots p_n b^2$ cannot happen.

  • $\begingroup$ Thank you for your comment. Let's assume that $a={a_1}^{n_1}\cdots {a_k}^{n_k}$ and $b={b_1}^{m_1}\cdots {b_l}^{m_l}$ are the prime factorazations of $a,b$ respectively. So, ${a_1}^{2n_1} \cdots {a_k}^{2n_k}=p_1 \cdots p_n \cdot {b_1}^{2m_1} \cdots {b_l}^{2m_l}$ where $a_i\neq b_j \forall i,j$. Now? Does this help? $\endgroup$ – Chris Aug 24 '17 at 20:23
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    $\begingroup$ @Chris Yes you are almost there. Now use the hint given by the other answerers/commenters: since $a_i \ne b_j$ for all $i,j$, it must be that one of the $a_i$ is $p_1$. But the exponent of $p_1$ on the right-hand side is odd while the exponents on the left-hand side are all even. $\endgroup$ – angryavian Aug 24 '17 at 20:35
  • $\begingroup$ So, you say that $\exists i\in \{1,...,k \}: a_i=p_1$. But, from this $2n_i=1$, contradiction. Right? $\endgroup$ – Chris Aug 24 '17 at 20:45

If you look at the factorization of $q^2$ for some $q \in \mathbb Q$, you will see that all the prime powers are indeed even. In your case, they're not, hence there is no rational square root.

  • $\begingroup$ Worth emphasis: the above hint implicitly uses the Fundamental Theorem of Arithmetic, i.e. existence and uniqueness of prime factorizations of integers $\neq 0.\,$ In any complete rigorous proof, we should explicitly mention any use of such properties (or closely related properties such as Euclid's or Bezout's Lemma), esp. when one is first learning these matters. $\endgroup$ – Bill Dubuque Aug 26 '17 at 2:52

A ready consequence of Fundamental theorem of arithematic is that A perfect square must have even powers of primes. Since all the primes are coprime with each other so there is no chance for thing in you square root to be a perfect square since every prime will have an odd power $(1)$


A somewhat different result is that if $n$ is not a perfect square then $\sqrt{n}$ is irrational.

Here is one of the many proofs: Follow-up Question: Proof of Irrationality of $\sqrt{3}$

Of course you then have to show that $\prod p_i$ is not a perfect square.


It is worth mentioning Rational root theorem (RRT). From $$\sqrt{\prod\limits_{i=1}^{n} p_i}=x \Rightarrow x^2 - \prod\limits_{i=1}^{n} p_i=0 \tag{1}$$ which is a polynomial with integer coefficients, leading coefficient is $1$, thus $x$, the solution, can only be integer or irrational. Let's assume it's an integer, then (from the same RRT) it should divide the constant term $\prod\limits_{i=1}^{n} p_i$ or $\gcd\left(x,\prod\limits_{i=1}^{n} p_i \right)=x$. Then either $x$ is a prime $p_k$ or divisible by such a prime, in both cases $p_k \mid x$ and $p_k \mid \prod\limits_{i=1}^{n} p_i$. Using Euclid's lemma we will conclude that $p_k \in \{p_1,p_2,...,p_n\}$ and from $(1)$ $$p_k \mid \prod\limits_{i=1,i\ne k}^{n} p_i$$ which, from the same Euclid's lemma, is not possible.

  • $\begingroup$ There is a big gap: RRT implies only that any rational root is an integer. To complete the proof you need to prove it has no integer roots (which requires using strong properties of integer arithmetic). $\endgroup$ – Bill Dubuque Aug 24 '17 at 23:00
  • $\begingroup$ @BillDubuque Yes, I missed that gap. Fixed now with just $\gcd$ and Euclid Lemma. $\endgroup$ – rtybase Aug 24 '17 at 23:55

Correct me if wrong:


Assume :

$ a^2= p_1 p_2p_3 .....p_n b^2$.

$\rightarrow$ $p_1$ divides $a^2$;

$\rightarrow$ $p_1$ divides $a,$ by the Theorem below

$ \rightarrow$ $ (p_1)^2$ divides $a^2$.


$(p_2)^2,(p_3)^2.,.....,(p_n)^2$ divide $a^2$.


$a^2 =$

$(p_1p_2p_3......p_n)^2 q^2 =$

$p_1p_2p_3.....p_n b^2$;

$\rightarrow$ $ b^2 = p_1p_2p_3...p_n q^2$

$\rightarrow$ $p_1,p_2,p_3,....,p_ n$ divide $b^2$;

$\rightarrow$ $p_1,p_2,p_3,....,p_n$ divide $ b$ by the Theorem below.

Hence $p_1,p_2,p_3,.....,p_n$ are common divisors of $a$ and $b$,

in contradiction to the assumption

gcd$(a,b) =1$.

$ $

Used: Theorem of Number Theory:

If $p$ is a prime and $p$ | $ab$, then

$p$ | $a$ or $p$ | $b$.

It follows: If $p$ | $a^2$ then $p$ | $a$.

  • $\begingroup$ You should justify the fundamental claim that prime $\,p\mid a^2\Rightarrow\,p\mid a.\ $ $\endgroup$ – Bill Dubuque Aug 24 '17 at 23:13
  • $\begingroup$ Bill. Thanks for your comment. $\endgroup$ – Peter Szilas Aug 25 '17 at 13:11
  • $\begingroup$ Reading further there are a couple more significant gaps. You need to justify the claims that $\,p_i^3\mid a^2 \,\Rightarrow\, m := p_1^2\cdots p_n^2\mid a^2.\,$ Also needing justifcation is the claim that $\,a^2/m = q^2$ is the square of an integer. Both proofs depend crucially on special properties of the ring of integers that are not generally true for other number systems, so such properties should be explicitly invoked when that are employed. $\endgroup$ – Bill Dubuque Aug 25 '17 at 14:13
  • $\begingroup$ Bill. Thank you for pointing out weak points.Attempt to remedy: p_1 | a, then a = r× p_1. Now p_2 | a then p_2 divides r or p_1. Since p_2 does not divide p_1, it divides r, I.e. r = s× p_2, or a = s× p_1 p_2 . p_3 | a , p_3 | s, or p_3 | (p_1p_2) .p_3 does not divide (p_1p_2) , hence divides s .Continuing we get a= q× p_1p_2...p_n. Then a^2 = q^2× (p_1)^2 (p_2 )^2...(p_n)^2, [hence (p_1)^2(p_2)^2...(p_n)^2 , the product, divides a^2]. Kindly let me know, thanks again, Peter. $\endgroup$ – Peter Szilas Aug 25 '17 at 19:20

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