# Prove: If $x+y>22$ then $x>11$ and $y>11$

Prove: If $x+y>22$ then $x>11$ and $y>11$

I'm not sure but I believe you have to use proof by contradiction.

EDIT: Proof: By contradiction, let $x+y>22$ and $x \leq 11$ or $y \leq 11$. Adding both sides, we get $x+y \leq 22$ which is a contradiction since we said $x+y>22$.

Any corrections?

• What about $x=3$, $y=20$? Or do you want an"or" instead of "and?"
– Paul
Mar 2 '16 at 21:42
• You mean $x>11$ or $y>11$? Mar 2 '16 at 21:43
• Did you mean $x\gt11$ OR $y\gt11$ Mar 2 '16 at 21:43
• ok let me edit.
– kero
Mar 2 '16 at 21:44
• As per Paul's comment, your proof must be wrong, due to the fact that the conditional in the title and in the first line of the question is false. Mar 2 '16 at 21:46

• I mean that the statement "if $x+y>22$ then $x>11$ and $y>11$" is logically false, as has been amply illustrated by commenters. Mar 4 '16 at 12:31
Ok, if $x<11$ AND $y<11$ you can sum it into $x+y<22$, but if its OR you have a complex, not a system, you can't sum it(even it would be eqs). For example, x=13 or y=2 doesn't mean x+y=15.