-1
$\begingroup$

Prove that the equation $\big(\frac{p}{q}\big)^2 = 3$ has no solution for $p,q$ that belong to $\mathbb{N}$. Can anyone please provide a solution to this problem?

  • 0
    $\frac{p}{q} = \sqrt{3}$. – 2017-02-15
  • 1
    @O.VonSeckendorff That's a circular argument if ever I saw one. – 2017-02-15
  • 0
    @Arthur I didn't read the question correctly, for "no solution for p,q belongs to N". – 2017-02-15

3 Answers 3

2

WLOG, assume that $p$ and $q$ are coprime. $$\left(\frac{p}{q} \right)^2=3 \iff p^2=3q^2 \equiv 0 \pmod {3}$$ Thus $$p \equiv 0 \pmod {3} \iff p=3k, k \in \mathbb{N}$$ So $$3q^2=p^2=9k^2 \iff q^2=3k^2 \implies q \equiv 0 \pmod {3}$$ So $p \equiv q \equiv 0 \pmod {3}$. Contradiction to the assumption that $p$ and $q$ are coprime.

So there are no solutions.

  • 0
    Simple adaptation of the standard, classical proof that $\left(\frac{p}{q}\right)^2 = 2$ has no integer solutions. – 2017-02-15
  • 0
    @Arthur Yeah. And more generally, the proof that $\sqrt{n}$ is irrational for non square $n$. – 2017-02-15
  • 0
    Well, yes. But one usually does it only for $2$. – 2017-02-15
0

Suppose there are $p_0, q_0 \in \mathbb{N}$ so that $(p_0 / q_0 )^2 = 3 $ and assume $\gcd( p_0, q_0) = 1$. Then,

$$ p_0^2 = 3 q_0^2 \implies 3 | p_0 \implies p_0 = 3 n, \; \; \; \text{some} \; \; n \in \mathbb{Z}$$

Now, show that $3$ also divides $q_0$.

0

If $p^2= 3q^2$, then the exponent of $3$ in the factorization of the LHS is even but is odd on the RHS.

  • 1
    Should say "in the *unique prime* factorization...." to emphasize the use of the Fundamental Theorem of Arithmetic. In fact the proof needs only a much weaker case: every integer $\neq 0$ has a *unique* factorization of the form $\,3^k m\,$ where $\,3\nmid m.\ \ $ – 2017-02-15