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I came across this question in a book. I tried proving the condition as the following: Suppose that a and b are rational. Clearly the sum of $a$ and $b$ is rational, which contradicts the condition, which is that $a+b$ is irrational. Therefore at least one of a or b is irrational.

I have a feeling that something is amiss there. I feel like using contrapositive proof here could be better but I'm not sure because I'm new to the world of proofs.

Arnav Mahajan
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Maths explorer
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    Everything is correct. – zkutch Aug 17 '20 at 12:25
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    The contrapositive is exactly what you have proven, namely$$a,b\in\mathbb{Q}\implies a+b\in\mathbb{Q}$$ – Peter Foreman Aug 17 '20 at 12:26
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    You are already basically proving the contrapositive "if $a$ and $b$ are rational then $a+b$ is rational", and then just bookending it with the language of proof by contradiction. So it's fine. If you were to turn this in as homework, or do it on an exam, then probably you would be expected to justify why the sum of two rational numbers is rational. – halrankard Aug 17 '20 at 12:27
  • If you have this feeling, perhaps there was another statement meant? For example, this one? – Dietrich Burde Aug 17 '20 at 12:35
  • I dont see how the contrapositive works here because technically speaking the contrapositive means that 'not Y -> not X' and I don't see how making this statement (at least one of a or b is irrational) becomes 'a and b are rational' when making it into a 'not Y' statement. – Maths explorer Aug 17 '20 at 12:57
  • @JiangMichael $Y$ here is "$a \notin \mathbb{Q}$ or $b \notin \mathbb{Q}$". Use DeMorgan's Laws - $\neg(a \notin \mathbb{Q} \vee b \notin \mathbb{Q}) \iff \neg a \notin \mathbb{Q}\wedge \neg b \notin \mathbb{Q} \iff a \in \mathbb{Q}\wedge b \in \mathbb{Q}$. – Jair Taylor Aug 17 '20 at 13:55
  • Not Y is: neither $a$ nor $b$ are irrational $\iff$ $a,b\in \mathbb Q$.

    Not X is: $a+b$ is rational $\iff a,b\in \mathbb Q$.

    Not Y $\implies$ Not X is exactly that $\mathbb Q$ is closed under the operation ,,$+$''

     – crush3dice Aug 17 '20 at 13:59

2 Answers2

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The statement you're trying to prove is $\forall a,b\, (a+b\notin \Bbb{Q} \implies a\notin \Bbb{Q} \text{ or } b \notin \Bbb{Q})$. This is simply the symbolic translation of the statement "for every $a,b$, if $a+b$ is irrational then atleast one of $a$ or $b$ is irrational".

Here, the statement $X$ is "$a+b\notin \Bbb{Q}$", and the statement $Y$ is "$a\notin \Bbb{Q} \text{ or } b \notin \Bbb{Q}$". So, the contrapositive of "for every $a,b$ ($X \implies Y$)" is "for every $a,b$ $(\neg Y \implies \neg X)$", which in this case is:

For every $a,b$ we have ($a\in \Bbb{Q}$ and $b\in \Bbb{Q} \implies a+b \in \Bbb{Q}$)

and this is what you argued.

peek-a-boo
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  • I understand how (a+b∉Q⟹a∉Q or b∉Q), but could you please explain how that relates to the question title, which is 'at least one of a or b is irrational'? – Maths explorer Aug 18 '20 at 11:18
  • @Jiang Michael what are you unclear on? 'at least one of $a$ or $b$ is irrational' means "$a$ is irrational or $b$ is irrational" (note that in math the word "or" is used in the inclusive sense opposed to everyday language when we use it in the exclusive sense so 'P or Q' means either P or Q or possibly both) – peek-a-boo Aug 18 '20 at 12:08
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I want to address your "I don't see how the contrapositive works here" comment.

Let $\mathbb{I} = \mathbb{R} \setminus \mathbb{Q}$ (the set of irrational numbers).

You want to show that

$$ a+b \in \mathbb{I} \implies a \in \mathbb{I} \vee b \in \mathbb{I}$$

Before switching to the contrapositive, note that for $a \in \mathbb{R}$ $$ \lnot (a \in \mathbb{I}) \Leftrightarrow a \in (\mathbb{R} \setminus \mathbb{I}) \Leftrightarrow a \in \mathbb{Q}$$

Now, the contrapositive becomes

$$ \lnot (a \in \mathbb{I} \vee b \in \mathbb{I}) \implies \lnot (a+b \in \mathbb{I})$$ which, in light of the observation above, is $$ a \in \mathbb{Q} \land b \in \mathbb{Q} \implies a+b \in \mathbb{Q}$$

which is a defining property of $\mathbb{Q}$.

Remember also that $\lnot (P \vee Q) = (\lnot P) \land (\lnot Q)$.

AnalysisStudent0414
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