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There are a lot of good proofs already available for this question here, but I am trying to write the proof on my own. It'll be a great help if I can get an idea whether my proof is right or not.

$Proof.$ The given statement is equivalent to its contrapositive. $$If\;\sqrt{pq}\;is\;rational\;then\;p\;or\;q\;is\;composite\; or\; p = q$$ $\sqrt{pq}$ is rational only when $pq$ is a perfect square.
Every natural number can be written as a product of its prime factors.
So, $p = p_1^{r_1} p_2^{r_2} p_3^{r_3}...$, where $p_i$ represent a prime and $r_i$ represent its power.
Similarly, $q = p_1^{s_1} p_2^{s_2} p_3^{s_3}...$, where again $p_i$ represent a prime and $s_i$ represent its power.
Each $r_i \geq 0$ and $s_i \geq 0$. Also, the above series will be finite as $p$ and $q$ are finite.
So, $pq = p_1^{r_1 + s_1} p_2^{r_2 + s_2} p_3^{r_3 + s_3}...$
Now, $pq$ can be a perfect square only when for each $p_i$, $r_i + s_i$ is even.

Now, let us take $p\; or\; q\; is\; composite\; or \; p = q$ to be false, i.e $p\; and\; q\; both\; must\; be\; distinct\;prime$ is the true statement.
Now, if $p$ and $q$ are prime then $$p = p_k^{r_j}$$where $p_k = p$ and $r_j = 1$.
Similarly, $$q = p_l^{s_t}$$where $p_l = q$ and $s_t = 1$.
Hence, $pq = p_k^{r_j}p_l^{s_t}$, where both $r_j$ and $s_t$ are odd and $gcd(p_k, p_l) = 1$, which means $pq$ is not a perfect square and hence irrational.(a contradiction).
So, $p\; or\; q\; is\; composite$ is true, which implies
$$If\; \sqrt{pq}\;is\;rational \;then \;p \;or\; q \;is\; composite.$$
Hence, proved.

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  • $\begingroup$ The contrapositive is "If $\sqrt{pq}$ is rational, then $p$ or $q$ is composite or $p = q$". $\endgroup$ – eyeballfrog Dec 20 '20 at 15:24
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    $\begingroup$ What you are trying to prove is false, since $5$ and $7$ are prime while $\sqrt{35}$ isn't prime. It isn't even an integer or rational. $\endgroup$ – Bernard Massé Dec 20 '20 at 15:26
  • $\begingroup$ Thanks, corrected the title and contrapositive statement now. $\endgroup$ – die_wolf Dec 20 '20 at 15:29
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    $\begingroup$ Does this answer your question? If $p_1,...,p_n$ are positive prime numbers then $\sqrt{p_1\cdots p_n} \notin \Bbb{Q}$ $\endgroup$ – rtybase Dec 20 '20 at 16:44
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    $\begingroup$ @rtybase Thank-you...I know that there is simpler proof given in the link you provided, and some members have already answered it below. But I'm more interested in knowing whether the idea I'm using is valid or not. $\endgroup$ – die_wolf Dec 20 '20 at 16:54
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If $\sqrt{pq}=a/b$ for naturals $a, \, b$ then $a^2=b^2pq$. But $p$ divides the left (right) side an even (odd) number of times.

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