$$a,b \in R/Q$$ Assume $$a+b = \frac{p}{q} $$ $$r=\frac{k}{j}$$ Now $$ra+rb =\frac{pk}{qj}$$ Clearly $\frac{pk}{qj}$ is rational, and both $ra$ and $rb$ are not which completes the proof.
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I can't follow this at all. You seem to be arguing that if you had two irrationals that added to a rational then you could scale each of them up by rationals and find more examples. But so what? Your job was to show that there was an example in the first place. – lulu Sep 22 '19 at 16:03
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@lulu it's just that if my original statement was false, then multiplying by rational would produce irrationals on both sides, contradicting it.Hence we could say that there are no two irrationals that make up a rational – Misha.P Sep 22 '19 at 16:10
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Huh? You assumed what you wanted to prove. So in the and you have proven that if there are two irrations that add to an irrational you can find another pair by multiplying both by the same rational. (By the way... what if $r=0$....). Knowing that $a,b \not \in \mathbb Q$ and $a+b\in \mathbb Q \implies ar,br \not \in Q$ and $ra+rb\in \mathbb Q$ does not in anyway prove that $a,b$ exist in the first place. – fleablood Sep 22 '19 at 16:12
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But that's not a contradiction. If you want to prove $P$ is false, so you assume $P$ is true and you try to get a contradiction, getting another example where $P$ is true is not a contradiction..... Hint: (did it occur that maybe the statement is true. If $x$ is irrational. ANd $r$ is rational the is $s = r-x$ rational or irrational. And what of $s+x$? Is that rational or irrational. – fleablood Sep 22 '19 at 16:16
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Sorry, all you've done is to show (correctly) that if there is a single example then there are infinitely many. You still have to show that there is even one example though. – lulu Sep 22 '19 at 16:18
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You have shown that if $a + b = M$ then $M$ is rational if and only if $Mr = ar+br;r\in \mathbb Q$ is rational. That doesn't help us in any way in determining if either $M$ is rational is possible or impossible in the first place. – fleablood Sep 22 '19 at 16:20
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@lulu how can I do that with what I have? – Misha.P Sep 22 '19 at 16:20
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Just think about it. It's easy to produce example of two irrationals that add to $0$, say. $\sqrt 2+(-\sqrt 2)$ for instance. If you wanted two positive examples, just modify this example accordingly. – lulu Sep 22 '19 at 16:22
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You can not. What you have doesn't go anywhere. But if you want to prove $a+b$ can be rational you just have to give a SINGLE example. (Hint: $a = \sqrt{2}$ is irrational? Is $b = -\sqrt{2}$. What is $a+b$? Is it irrational?) Of course, if it turns out that $a+b$ must always be irrational then you can't give an example and you must give a prove. So first of all. Do you think this is true? or not? – fleablood Sep 22 '19 at 16:23
2 Answers
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Consider irrationals $a=1+\sqrt 2$ and $b=1-\sqrt 2$. Then you have $a+b=2,$ which is rational.
John Bentin
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