Struggling with a proof when $x > 0, x > a$, then $x > a$

Question

This is coming from a question in spivak's calculus, solving $(x-1)(x-3) > 0$.

There are two cases where this is true, when both brackets are positive, or when both are negative.

But when I look at the positive case, I get $x>1, x> 3$.

I know intuitively that the only valid $x$ values where both are true is when $x > 3$. But can't seem to be able to prove it.

I realize that I should try and get a general proof for when $x> 0, x > a$, then $x > a$.

But i've tried for the last couple days and I haven been able to get anything remotely on a good track.

Can anyone give me a solution, or maybe some hints in the right direction?

Answer 1

The only logical solution is $x \gt 3$. You are right. And that is it. There is no need to prove this identity.

Say you are given that $x \gt a$ and $x \gt 0$. By solving this what you are require to do is to find a set of values for which both inequalities are satisfied. Clearly, $x \gt \text{Max} \{ 0, a \}$. That is it. It is intuitive and it is the only solution. This is what you have arrived at for the exercise too. There is no need to prove it.

Obvious is a word I hate to use in Mathematics. But in your case it seems appropriate. Stop overthinking is another advice.

Hope I helped.

EDIT: I checked the book out. Can't believe I forgot all of this. Maybe this would help.
$x \gt 1 \implies (x - 1) \in P \implies x \in (1, \infty)$ and $x \gt 3 \implies (x- 3) \in P \implies x \in (3, \infty)$ where $P$ is the set of all positive numbers as per the definition in the text.

For both inequalities to be satisfied $x \in (1, \infty) \cap (3, \infty ) = (3 , \infty)$

Answer 2

If you really want a proof:

We need to prove that $x>$ Max$(0, a)$ given that $x > 0$ and $x>a$ for $\exists a \in \mathbb{R}^+$.

Since $a \in \mathbb{R}^+$, $a > 0$. So, Max$(0, a) = a$. Therefore, we only need to show that $x > a$. But, we already know that $x > a$, so we're done.