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I probably phrased it very bad.

This is what I mean: $$\sqrt{x} + \sqrt{y} \neq R$$ x and y being non-square coprime natural numbers.
And: $$\sqrt{xy} \neq R$$ x, y, AND R being coprime.
Let's try to prove the second one, the product of irrational coprime square roots, by contradiction. $$\sqrt{xy}=R$$ Then $$xy=R^2$$ Since they are coprime this has been proven, I note that if x and y are coprime, but each one is not coprime with R, this would be true if and only if x AND y were perfect squares; example: $$\sqrt{4*9}=6$$ Now I know that, but how do I apply it to try to prove the first hypothesis? I'm thinking of somehow using the fact that $$\sqrt{x} + \sqrt{y} \neq R$$ can be transformed into $$x+y+2\sqrt{xy} \neq R^2$$
Help please?

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You have the right idea for the second part, but you need to be much more precise to obtain a rigorous proof. Suppose $\,\sqrt{XY} = A/B\in \Bbb Q.\,$ Wlog we may assume that $\,A/B\,$ is reduced, i.e. $\,A,B\,$ are coprime. Squaring $\,XY = A^2/B^2\,$ so $\,A^2 = XYB^2.\,$ Comparing the powers of a prime $\,p\,$ on both sides of the equation we obtain $\, 2a = x + y + 2b,\,$ where $\,a\,$ is the power of $\,p\,$ in $\,A,\,$ etc. $ $ By $\,X,Y\,$ coprime, one of $\,x,y\,$ is $\,0,\,$ say $\,x=0,\,$ so $\,y = 2a-2b-0\,$ is even. Therefore every prime factor of $\,X\,$ and $\,Y\,$ occurs to even power, so $\,X\,$ and $\,Y\,$ are squares.

Remark $\ $ The proof depends crucially on the Fundamental Theorem of Arithmetic, i.e. the existence and uniqueness of prime factorizations. Be sure to understand the places where this is (implicitly) used above (e.g. it implies power additivity: the power of $\,p\,$ in $XY$ is $\,x+y).\,$ See here for some other proofs, using gcd laws, or using induction and the prime divisor property $\,p\mid ab\,\Rightarrow\,p\mid a\,$ or $\,p\mid b.$

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As you've already proven the first part, we need only show that
$\sqrt{x} + \sqrt{y} \neq R$
Now as you suggested we can prove that $x + y + 2\sqrt{xy} \neq R^2$
Where R is any rational number, x and y are integers. As $\sqrt{xy}$ is not a rational number $x + y + 2\sqrt{xy} = V$ is not rational (as all terms are rational except $\sqrt{xy}$). So V is not rational, thus $\sqrt{V}$ is not rational. Thus there can exist no such rational R where $\sqrt{x} + \sqrt{y} = R$

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  • $\begingroup$ The OP certainly did not prove the first part. $\endgroup$ – Bill Dubuque Jun 11 '15 at 0:46

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