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

Question

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 $10$

Answer 1

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

Answer 2

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.

Answer 3

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