Can someone please tell me how to prove that there is no last prime number? I am familiar with Euclid's one, but I'm looking for a different way. I'm in 11th grade and I can do intermediate algebra and basic calculus. Only a basic idea/guidelines would suffice.

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    $\begingroup$ Since, you have tagged it sequences-and-series, i will give you a hint along the same line. Try to think of a sequence of integers in which infinitely many terms(maybe all) are pairwise co-prime(relatively prime). $\endgroup$ Dec 21, 2015 at 13:30
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    $\begingroup$ en.wikipedia.org/wiki/Euclid%27s_theorem#Some_recent_proofs ....see,this must suffice... $\endgroup$
    – Soham
    Dec 21, 2015 at 13:31
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    $\begingroup$ Being in the 11th grade means you haven't been taught yet a variety of areas in math that can be used to prove this. I'm hoping the idea suggested by @MosBlack will be elaborated into an Answer. $\endgroup$
    – hardmath
    Dec 21, 2015 at 13:42
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    $\begingroup$ Likely not all the proofs there will be accessible to you, but still it should answer your question. $\endgroup$
    – quid
    Dec 21, 2015 at 13:43
  • $\begingroup$ I saw a proof that involved defining the "height" of a number $n=p_1^{a_1}p_2^{a_2}\dotsb p_N^{a_N}$ by ${\rm height}(n)=a_1+a_2+\dotsb+a_N$, where the $p_i$ are prime. We then find contradictory lower and upper bounds for how many primes have height less than some number $h$. $\endgroup$ Dec 21, 2015 at 13:53


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