# 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 [closed]

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 $10$

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

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

• (+1) There is often little difference between proofs using the contrapositive and proofs using contradiction.
– robjohn
Apr 11, 2014 at 19:23

Suppose, toward a contradiction, that $0<a\leq 10$ and $0<b\leq 10$. Then $$100 < ab \leq 10\cdot 10 = 100,$$ contradiction.

• The question appears to have been changed ... I will edit my answer Apr 11, 2014 at 19:16
• I don't think a downvote is necessary. What is written is correct and would be helpful as a hint to the answer.
– robjohn
Apr 11, 2014 at 19:17
• Indeed, two downvotes in fact. Could the downvoters comment? Apr 11, 2014 at 19:19
• There is only one downvoter. Perhaps you were noting that I had removed my upvote temporarily.
– robjohn
Apr 11, 2014 at 19:21
• Oh, sorry, I wasn't aware. But thanks for helping me to improve the post! Apr 11, 2014 at 19:22

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.