I haven't found a way, except using advanced methods, to prove the irrationality of $\pi$.
It is possible to prove that $\pi$ is irrational, using only elementary algebra?
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$\begingroup$ There isn't a truly easy proof (that i am aware of), but Hermite's argument isn't too bad. You can find it here $\endgroup$ – lulu Aug 29 '16 at 23:00
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$\begingroup$ An algebraic definition of $\pi$ (as opposed to a geometric definition like "the ratio of any circle's circumference to its diameter) is a bit involved and so any proof of its irrationality would likely also be involved. $\endgroup$ – D_S Aug 29 '16 at 23:07
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2$\begingroup$ I would typically expect a proof to use the fact that $\pi$ is a positive root (in fact, the least positive root) of $\sin,$ but then defining $\sin$ in a satisfactory way is already involved (and then there's the additional point of proving that there actually exists a least positive root of $\sin$). Typically, satisfactory definitions already involved a fair amount of analysis, so that elementary algebra is already insufficient. The fact is that, since $\pi$ is (in fact) transcendental, it isn't an altogether algebraic object. Some analysis would seem to be needed. $\endgroup$ – Will R Aug 29 '16 at 23:18
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$\begingroup$ Elementary algebra only? No. There isn't. $\endgroup$ – fleablood Aug 30 '16 at 0:33
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$\begingroup$ I think neither math.stackexchange.com/questions/1364970/… nor math.stackexchange.com/questions/1076966/… have an answer that answers this question. Maybe the task was to give a short simple proof that doesn't rely on a previously proven result that's difficult to prove without proving it. $\endgroup$ – Timothy Jul 28 '18 at 17:51
