Checking work on Proof by Contradiction?

Question

I am looking for any tips on how to go about proof by contradiction. I understand that we start by assuming the opposite of our proof, then proving the original by ending up with a double negative. Although when attempting my first proof by contradiction I felt helpless trying anything besides trial and error, then going from there. Here is how I tried to solve the following proof, I would appreciate any advice on how to continue. Proof: If a·b = n, then a or b ≤ √n where a,b and n are nonnegative real numbers.

We use proof by contradiction. Suppose if a·b = n is true, then a and b > √n is also true.

I came up with a and b > √n, because if both are greater than √n is false, we end up with the original two statements being true. Next I arrived at a > √n and if this is true, then a^2 > n.

I'm at a loss how to continue or even how to check if i'm going in the right direction. I know that if a=2, b=3, n=6 then a^2 > n is false, but im not sure if plugging random numbers in is the best way to solve this and even then im still not sure how to apply it.

Answer 1

Here is a proof by contradiction that if $a\cdot b=n$ then $a\le \sqrt n$ or $b \le\sqrt n$.

Assume to the contrary $a\cdot b=n$ and neither $a$ nor $b$ is $\le\sqrt n$; i.e., $a\gt\sqrt n$ and $b\gt \sqrt n$.

Then $n=ab\gt \sqrt n \sqrt n = n$, which is a contradiction.

Therefore the original assumption must be wrong, so $a$ or $b$ is $\le\sqrt n.$

Answer 2

Let $$ab=n\tag{1},$$

and we wish to prove that $a,b\leq\sqrt{n}$. For a contradiction assume the opposite, i.e. $$a,b>\sqrt{n}.\tag{2}$$

We can write $a=\sqrt{n}+x$ and $b=\sqrt{n}+y$ with $x,y>0$ so that $$ab=(\sqrt{n}+x)(\sqrt{n}+y)=n+\sqrt{n}(x+y)+xy.\tag{3}$$ But $x,y>0$ implies $x+y>0$ and $xy>0$, so (3) implies $$ab>n.$$ This contradicts (1), so our original assumption (2) must be wrong, i.e. it must be that $a,b\leq\sqrt{n}$.