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$a^{1/2}$ is either an integer or an irrational number

I would like to know the better proof for the following one.

question: non perfect square of any integer is an irrational. I given my answer at below. But I need better method. If any one can provide, I am very grateful to them.

my answer is: x=sqrt(2) is irratioonal by looking at x+1 and x-1. Then (x+1)(x-1)=1 and x-1=a/b gives a/b+2=b/a, so that (a+2b)/b=b/a. Without loss of generality (a,b)=1 and then also (a+2b,b)=1 and the conclusion (Euclid VII.20) is that a+2b=b, b=a, which gives a=b=0, a contradiction.

We can use a similar idea, although we won't be able to use reciprocals, generally. Suppose sqrt(n) is rational. Then for any a/b ((a,b)=1), sqrt(n)-a/b=c/d is rational, where w.l.o.g. (c,d)=1, d>0. We get c/d+2a/b=n-a^2/b^2. So cb^2=d(nb^2-a^2-2ab). Now b^2|RHS, but b^2 is coprime to nb^2-2ab-a^2, since any prime factor of b^2 is a prime factor of b and hence also of nb^2-2ab. If it were also of nb^2-2ab-a^2, then it would divide a^2 and hence a, contrary to (a,b)=1. Thus we have b^2|d. Write d=b^2r. Then c=r(nb^2-a^2-2ab). Now any prime factor of r divides c. Since r|d and (c,d)=1, r is a unit. Since d>0 and b^2>0, r>0. So r=1. Then d=b^2. So sqrt(n)=a/b+c/b^2=(ab+c)/b^2. Choosing a/b=0/1, sqrt(n)=c, an integer, so that n=c^2 is the square of an integer. Thus, if n is not a square, then sqrt(n) is irrational.

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marked as duplicate by JavaMan, Zev Chonoles Oct 15 '11 at 19:15

This question was marked as an exact duplicate of an existing question.

If any rational number is squared; you have two cases, if it's an integer, then its square is a perfect square. Otherwise, you need to prove that its square is not an integer, and then you're done.

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