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I know that the proof by contradiction is useful when we know that there is only one answer and we can pick from two exclusive options. Therefore, if one doesn't work, we know the other is the correct answer. It seems to me that to use the proof I require, beforehand, the knowledge that the result can be only one option or the other.

When proving, that the sum of an irrational number and a rational number is irrational, we can use proof by contradiction to demonstrate that it can't be rational. Doesn't this assume that the sum can't be sometimes rational and other times irrational?

However, with the sum of two irrational numbers, the result can be either rational or irrational based on the chosen numbers. How can I deduce, beforehand, that the proof by contradiction won't be useful in this case?

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  • $\begingroup$ The exclusive alternative is usually that a statement is either true or false, so one will not necessarily have a fine-grained dichotomy like rational vs. irrational. Indeed a "direct" proof could always be recast as a proof by contradiction, so I doubt that one can give a rule for when "the proof by contradiction is useful". $\endgroup$ – hardmath Jun 15 '20 at 19:22
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The trick is to use closure rules we know for rationals: they're closed under addition, subtraction, multiplication, and division if the quotient is nonzero. Let's discuss your examples:

  • If $i$ is irrational and $r$ is rational, suppose $i+r$ is rational: then $i=(i+r)-r$ is rational, a contradiction.
  • If $i,\,j$ are irrational, let $k:=i+j$; we can't write any one variable as a function of rationals, so an argument similar to the above is unavailable.
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When we prove that the sum of a rational and an irrational number cannot be rational, we are certainly not assuming that the sum cannot be rational in some cases and irrational in others: we are proving that this is the case. This really has nothing to do specifically with proofs by contradiction; it just happens that in this case we prove that something is true by proving that its negation is false. And in fact it’s entirely possible to prove the result without using proof by contradiction.

Proposition: If $x,y\in\Bbb R$, and $x$ is rational, then $x+y$ is rational if and only if $y$ is rational.

Proof: If $y$ is rational, then $x+y$ is the sum of two rational numbers and is therefore rational. If $x+y$ is rational, then $y=(x+y)+(-x)$ is the sum of two rational numbers, so $y$ is rational. $\dashv$

Knowing when a proof by contradiction or a proof of the contrapositive is likely to be a good way to proceed is in large part a matter of experience, but there is at least one moderately useful guide: if the hypotheses available for a proof of the contrapositive or a proof by contradiction seem to give you more to work with — more specific details, for instance — than the hypotheses available for a direct proof, then it’s definitely worth your while to consider an indirect proof.

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  • $\begingroup$ But using the proof by contradiction to prove it takes the premise that the result can only be rational or irrational for all the cases, right? I mean, could be the case that a proof by contradiction wouldn't work for something because this premise of either one option or the other isn't right? $\endgroup$ – Jon Jun 15 '20 at 20:25
  • $\begingroup$ @Jon: No, it doesn’t. It simply proves that the result is irrational in all cases. If there were some cases in which the sum was rational and others in which it was irrational, the result would simply be false, and no attempt to prove it, by any method whatsoever, would work. This has nothing to do with the method of proof that you’re attempting to use. $\endgroup$ – Brian M. Scott Jun 15 '20 at 20:29

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