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I am working out of Calculus by Spivak (4th ed.) and have come across a question which I would like some further insight on.

If $a$ and $b$ are both irrational, is $a+b$ necessarily irrational?

I have already proved that if $a$ is rational and $b$ is irrational then $a+b$ is irrational.

Thanks in advance!

user326159
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Jose
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    $(1+\sqrt{2})+(1-\sqrt{2})=2$. – diracula Mar 05 '17 at 17:00
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    i think no take $$a=\sqrt{2},b=1-\sqrt{2}$$ and $$a+b=1$$ rational – Dr. Sonnhard Graubner Mar 05 '17 at 17:00
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    $\sqrt{2} + (-\sqrt{2}) = 0$ – Augustin Mar 05 '17 at 17:00
  • Lots of counterexamples here, so maybe you could write your own answer! We love that! – The Count Mar 05 '17 at 17:01
  • Also: http://math.stackexchange.com/questions/499784/sum-of-irrational-numbers-are-there-nontrivial-examples. – Martin R Mar 05 '17 at 17:05
  • This is a very frustrating question. The obvious counter example is $r$ rational and $a$ irrational then $b =r-a$ is irrational and $a+b =r$. To which one very much wants to say "okay, but what about two irrationals that aren't related to each other; two independent irrationals" to which the answer is "what do you mean by independent" so you say "$b\ne r-a$ for any rational $r$" to which the response is "$b\ne r-a \iff a+b\ne r$. What you said is redundant". You're left feeling somehow you've been duped. But you haven't. Yes, I've been there too. – fleablood Mar 05 '17 at 17:23

3 Answers3

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No. For example $\sqrt{2}$ and $-\sqrt{2}$

Andrei
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$a=1+\sqrt{2}$ and $b=1-\sqrt{2}$ are both irrational and $a+b=2$

marwalix
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It might be easier to answer the contrapositive, if you don't just see the answer. The contrapositive is: "if $a+b$ is rational, is it the case that at least one of $a$ and $b$ is rational?". Since you've already proved that irrational plus rational is irrational, this is just "if $a+b$ is rational, is it the case that both $a$ and $b$ are rational?". This slightly different perspective might make it easier for you to find an example.

Patrick Stevens
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