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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?

mookid
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Tomas
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    If $x > a$, then $x > a$??? Maybe I'm missing out on something here... – Yiyuan Lee Mar 17 '14 at 06:08
  • You could prove "if $x>3$, then $x>1$" by contrapositive pretty easily if you really want to fully flush it out with a proof technique. Or just prove it directly (even more easily). But I really don't think this is something that it's necessary to prove... – foobar1209 Mar 17 '14 at 06:08
  • why don't you take the intersection $(3,\infty)\cap(1,\infty)$? – Guy Mar 17 '14 at 06:09
  • You actually need $$x>a,x>b\implies x>\max(a,b)$$ For the general case. – chubakueno Mar 17 '14 at 06:11

2 Answers2

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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 $x \in (1, \infty) \cap (3, \infty ) = (3 , \infty)$

Ishfaaq
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  • well to be fair, it is obvious. it really "grinds my gears" when people try to prove obvious stuff. – Guy Mar 17 '14 at 06:10
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    Spivak's calculus is an 'introduction to analysis' text book. It teaches you to complete proofs throughout the textbook, building up from only one axiom. I am sure he is just following the exercises as listed. – Display Name Mar 17 '14 at 06:15
  • I know it's obvious, I just thought there might be a proof to it, because the whole point of the chapter of the book I'm using is to prove very basic things from a set of assumptions; where that was not listed – Tomas Mar 17 '14 at 06:16
  • Ah, the above commenter beat me to it – Tomas Mar 17 '14 at 06:18
  • Thanks a bunch, the edit was really valuable for me – Tomas Mar 17 '14 at 06:26
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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.

foobar1209
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