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I'm practicing and Found this question.

If $ a $ and $b$ are irrational, either prove or disprove that $a + b$ is irrational.

So I tried contradiction (to a + b is irrational).

Let $a$ and $b$ be arbitrary irrational numbers. Assume that$ a + b $is rational.

Then $ a + b = x/y$ for some integers $x$ and $y$.

then $y(a + b) = x$

and $ay + by = x$

Because $x$ was an integer $ay$ is an integer and $by$ is an integer.

then $a$ divides $ay$ and $b$ divides $by$. But that's impossible because a is irrational and b is irrational and y is an integer.

So $a+b$ must be irrational as well.

Now I know this is wrong. Because I found a counterexample as the solution.

$sqrt(2)$ + $-sqrt(2)$ = 0.

Can someone point out my logic mistake? Thank you very much in advance!

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    $\begingroup$ Since you know a counterexample just examine your proof in this case till you reach a line that is false. You'll reach "Because $x$ was an integer $y\sqrt 2$ is an integer", which yields the contradiction that $\,\sqrt{2}\in\Bbb Q,\,$ since $\,y\neq 0.\,$ So that inference is false, which invalidates the proof. So your counterexample is also a counterexample to that claimed inference, i.e $\,x+y\in\Bbb Z\,\Rightarrow\, x,y\in\Bbb Z.\ \ $ $\endgroup$ – Bill Dubuque Dec 18 '19 at 18:54
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    $\begingroup$ The above method works generally to debug proofs when you know a counterexample, e.g. here and here and here for some worked examples and further discussion. $\endgroup$ – Bill Dubuque Dec 18 '19 at 19:00
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    $\begingroup$ You can use "\sqrt n" instead of "sqrt n" to get $\sqrt n$, in case you didn't know about it. $\endgroup$ – Yong Hao Ng Dec 19 '19 at 7:39
  • $\begingroup$ @Yao Hao Ng, thanks I didn't know about it! $\endgroup$ – oliver Dec 19 '19 at 16:50
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$ay$ and $by$ need not be integers in your proof.

$0=\sqrt 2 +(-\sqrt 2)$. If sum of two numbers is an integer you cannot say that both numbers are integers.

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  • $\begingroup$ Thank you @Kabo Murphy, I'm sorry for being slow. If I would let x not be zero, can ay + by be not an integer and equal an integer? Like if x/y is a non zero rational number. $\endgroup$ – oliver Dec 18 '19 at 12:31
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    $\begingroup$ $1 =(\sqrt 2) +(1-\sqrt 2)$; any integer can be written as the sum of non-integers. @oliver $\endgroup$ – Kavi Rama Murthy Dec 18 '19 at 12:33
  • $\begingroup$ I see it now. Somewhat... :-D thank you! $\endgroup$ – oliver Dec 18 '19 at 12:35
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The mistake is in the step when you say "Because $x$ was an integer $ay$ is an integer and $by$ is an integer."

As your counterexample shows, the sum of two non-integer real numbers may be an integer.

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  • $\begingroup$ I got it from Kabo's kind second comment, thanks! $\endgroup$ – oliver Dec 18 '19 at 12:32
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The mistake is that $\ ay\ $ and $\ by\ $ cannot be integers since $\ a\ $ and $\ b\ $ are irrational and $\ y\ $ a non-zero integer.

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  • $\begingroup$ Hi @Peter thank you. Yes, that was the idea of my proof that they cannot be integers. I just went on to go further. But if they can not be integers a + b can not be rational so it has to be irrational? I'm sorry for being slow $\endgroup$ – oliver Dec 18 '19 at 12:28
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    $\begingroup$ If $\ a\ $ and $\ b\ $ are irrational , $\ a+b\ $ can be anything : irrational, rational and even an integer. But in your proof, you apparently used that $\ ay\ $ and $\ by\ $ are integers which is not the case. $\endgroup$ – Peter Dec 18 '19 at 12:40

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