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does anyone know how to proof by contradiction that if $a$ and $b$ are positive integars and $ab >100$ then at least one of the integars $a$ and $b$ is greater than $10$

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  • $\begingroup$ yes but you have to prove it for a and b $\endgroup$
    – user142405
    Apr 11, 2014 at 19:09
  • $\begingroup$ Sorry, I misread the last line. $\endgroup$ Apr 11, 2014 at 19:10
  • $\begingroup$ @AndréNicolas: no, you read correctly; it was edited. $\endgroup$
    – robjohn
    Apr 11, 2014 at 19:12

3 Answers 3

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Do you need to use contradiction? If not, the contrapositive is straightforward: $$ a\le 10, \ b\le 10 \implies ab\le 100 $$

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  • $\begingroup$ (+1) There is often little difference between proofs using the contrapositive and proofs using contradiction. $\endgroup$
    – robjohn
    Apr 11, 2014 at 19:23
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Suppose, toward a contradiction, that $0<a\leq 10$ and $0<b\leq 10$. Then $$ 100 < ab \leq 10\cdot 10 = 100, $$ contradiction.

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  • $\begingroup$ The question appears to have been changed ... I will edit my answer $\endgroup$
    – user139388
    Apr 11, 2014 at 19:16
  • $\begingroup$ I don't think a downvote is necessary. What is written is correct and would be helpful as a hint to the answer. $\endgroup$
    – robjohn
    Apr 11, 2014 at 19:17
  • $\begingroup$ Indeed, two downvotes in fact. Could the downvoters comment? $\endgroup$
    – user139388
    Apr 11, 2014 at 19:19
  • $\begingroup$ There is only one downvoter. Perhaps you were noting that I had removed my upvote temporarily. $\endgroup$
    – robjohn
    Apr 11, 2014 at 19:21
  • $\begingroup$ Oh, sorry, I wasn't aware. But thanks for helping me to improve the post! $\endgroup$
    – user139388
    Apr 11, 2014 at 19:22
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You need to assume the opposite. What is the negation of the statement "At least one of $a$ and $b$ are greater than 10"? After you form the negation, you need to simply use elementary algebra to get a contradiction.

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