Can the Induction axiom in the Peano arithmetic be replaced by the irrationality of $\sqrt{2}$? This is inspired by the Alexander Shen's post here: https://www.facebook.com/groups/mathpuz/permalink/1058782384297603/ (the post is in Russian, but it is easy Russian, and google translate should work fine).
All proofs of irrationality of $\sqrt{n}$ where $n$ is not a perfect square that I know are using (explicitly of implicitly) the induction axiom of the Peano arithmetic. 
Question: Can the induction axiom be replaced by the axiom that $\sqrt{2}$ is irrational? 
 A: One can prove that $\sqrt{2}$ is irrational and indeed that any particular $\sqrt{n}$ is irrational (if $n$ is not a perfect squares) assuming only the principle of induction for $\Delta_0$ formulas, which is much weaker than the full strength of PA. In this sense, the answer is negative.
The classical proof of the irrationality of $\sqrt{2}$ relies on the fact that every rational number has a representation in lowest terms, and this can be proved with a very weak induction principle, using no unbounded quantifiers. Namely, every $p/q$ has a lowest-term representation, since if this is true for all pairs of numbers whose numerator is smaller than $p$, then it is true for $p/q$, since otherwise we would immediately reduce to a case with a smaller numerator. Thus, we never need to embark on unbounded searches to find the lowest-terms representation of a rational number. 
And once one has the lowest-term representations, one can easily prove the classical result that $\sqrt{2}$ is irrational, since otherwise one gets $2q^2=p^2$ and so on, as we all know, ultimately showing that both $p$ and $q$ are even, a contradiction. 
For the case of $\sqrt{n}$, for fixed $n$, then one can also prove this using just the existence of lowest-terms representations. Basically, $\sqrt{n}$ is irrational, unless $n$ is a perfect square, since otherwise $nq^2=p^2$ and the prime exponent parities don't work out, as you know. 
To prove the universal statement about all such $n$, however, that is, the statement "for all numbers $n$, the square root $\sqrt{n}$ is irrational unless $n$ is a perfect square," it seems that one uses the fundamental theorem of arithmetic, asserting that every number has unique prime factorization, which can be proved using $\Sigma_1$ induction. Namely, if every number smaller than $n$ has a unique factorization into primes, then $n$ also has this, since otherwise one can factor $n$ and reduce to the smaller case. (Here, we have to search a little bit, but not much, to find the code of the factorization.) But once you have the factorization into primes, then again the classical argument shows as above that $\sqrt{n}$ is irrational unless $n$ is a perfect square. 
I'm not sure if one can prove the fundamental theorem using only $\Delta_0$-induction. And because we moved up to $\Sigma_1$ induction to get the fundamental theorem, it is conceivable to me that one might be able to have a bizarre model of $\Delta_0$-induction in which the universal claim about all $\sqrt{n}$ was not true, even if it was true in every standard-finite instance, and this would be an interesting answer to your question. I would want the proof theorists to weigh in on this.
Since PA proves the consistency of $\Sigma_0$ induction, and indeed of $\Sigma_n$-induction for any fixed $n$, it follows by the incompleteness theorem (as mentioned in the comment of Dan Piponi) that we cannot axiomatize PA by adding the assertion "$\sqrt{2}$ is irrational" or "$\sqrt{n}$ is irrational for all non-perfect squares $n$" to any finite list of axioms.
The essence of PA is the induction axiom for increasingly difficult arithmetic properties. No bounded amount of complexity for induction, limited to $\Sigma_n$ induction for fixed $n$, is sufficient to capture the full power of PA.
