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Show that $\sqrt n$ is irrational if $n$ is not a perfect square, using the method of infinite descent.

I know how to prove this by doing a contradiction proof and using The Fundamental Theorem of Arithmetic, but now I'm asked to use infinite descent to prove it. Then the very next problem says "Why does the method of the text fail to show that $\sqrt n$ is irrational if $n$ is a perfect square?" I'm confused by this. Any hints or solutions are greatly appreciated.

I was thinking of the standard argument, let $\sqrt n = {a\over b}$ where $gcd(a,b)=1$ and then through some algebra arrive at a common factor for both $a$ and $b$ which contradicts the fact that $gcd(a,b)=1$ and so we can apply this over and over again, but then I don't understand how the next problem says to explain why this method fails.

ddswsd
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    As you provide no information about the method used (nor even the definition of vii), there's really no way to know what they had in mind. Why not provide the argument you are thinking of? The theorem is not difficult...the argument should be brief. – lulu Mar 10 '21 at 12:34
  • What do you mean by “vii is irrational”? – José Carlos Santos Mar 10 '21 at 12:35
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    take $\sqrt n = a/b$ where $n$ is not a perfect square and $(a,b)=1$, then $b^2n = a^2$. If $p$ is a prime factor of $n$, then $n=pm$ and $b^2pm=a^2$, so $a=pa'$, but $a,b$ are coprime so.... – Exodd Mar 10 '21 at 12:39
  • @JoséCarlosSantos I fixed that, I don't know how that got there. – ddswsd Mar 10 '21 at 12:52
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    It seems to me that the method that you called " proof by The Fundamental Theorem of Arithmetic" is also called "the method of infinite descent" https://en.wikipedia.org/wiki/Proof_by_infinite_descent#Irrationality_of_%E2%88%9Ak_if_it_is_not_an_integer – NN2 Mar 10 '21 at 12:58
  • As explained in the linked dupes, the descent is on possible denominators of a rational root of $,x^2 = n,,$ viz. if $,c/d,$ is a root then $,c,$ too is a denom, and denoms are closed under mod, so $,c\bmod d\neq 0,$ is a smaller denom than $d,,$ yielding descent. – Bill Dubuque Mar 10 '21 at 14:31
  • Note that any inductive proof can be transformed into an infinite descent, e.g. the common proof that cancels $p^2$ from both sides of $,b^2,n = a^2$ also yields a descent on denoms $b$ (the smaller $b$ from descent can never be $1$ by $n$ isn't a square). But this is essentially the same as the proof you wish to avoid (by unique factorization), i.e. by noting the parity contradiction of the powers of $p$ in the unique prime factorization of both sides of $,b^2,n = a^2,,$ LHS is odd vs. RHS even. – Bill Dubuque Mar 10 '21 at 15:11

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