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I'm currently working through D. Velleman's How to Prove it. I have a question regarding an absolute value proof by cases (#10; section 3.5).

The question asked is to prove that: $$ \forall x\in\mathbb{R} \left(\lvert x-3 \rvert \gt 3 \rightarrow x^2 \gt 6x \right) $$

The book recommends a proof by cases, using the following as cases:

  1. if $ x-3 \geq 0 \rightarrow \lvert x-3 \rvert = x-3 $
  2. if $ x-3 \lt 0 \rightarrow \lvert x-3 \rvert = 3 - x$

My attempt:

let $x \in \Bbb{R}$ be arbitrary real number
For case 1 where $\lvert x-3 \rvert = x-3 $ (because $x-3 \geq 0)$

    $x-3 \gt 3$
    $x \cdot \left(x-3 \gt 3\right)$
    $x^2 - 3x \gt 3x$
    $x^2 > 6x$

For case 2, $ \left(x-3\right) < 0 $ so $ \lvert x-3 \rvert = 3 - x$

    $3-x \gt 3$

but then I'm not quite sure. I tried multiplying both sides by $3-x$, giving

    $ \left( 3-x \right) \cdot \left( 3-x \right) \gt 3 \cdot \left( 3-x \right) $
    $9 -6x + x^2 \gt 9 - 3x$
    $-6x + x^2 \gt -3x$
    $x^2 \gt 3x$

Which looks similar to $x^2 \gt 6x$ but simply isn't.

If anyone could advise regarding where to proceed, I'd be very grateful.

Thanks!

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  • $\begingroup$ You can't multiply both sides unless you know what you are multiplying is positive. And $x (a >b) $ is meaningless. $a>b $ isn't a number so you can't multiply it by something. That's like saying "2 times 'I like ice cream'". Or "'Fred is happy' +2". $\endgroup$ – fleablood Nov 2 '18 at 0:35
  • $\begingroup$ If 3-x >3 the x <0 so $x^2>3x>6x $. $\endgroup$ – fleablood Nov 2 '18 at 0:43
  • $\begingroup$ in case 1, if $x-3 \gt 3$ then $x \gt 6$. I had felt like that was sufficient to comfortably multiply the inequality by a positive real number. Have I missed a step? $\endgroup$ – tom Nov 2 '18 at 1:02
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HINT:

$$|x-3|>3\to x-3>3\to x>6$$ OR $$|x-3|>3\to-(x-3)>3\to x<0$$

You just have to show that $x^2>6x$ holds for $x>6$ and $x<0$. If you get stuck with this, I'm happy to give a further hint, but give it a go first.

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  • $\begingroup$ thanks for the quick reply! I think i'm having a hard time seeing why $x \lt 0$ implies that $x^2 \gt 6x$. <br> I obviously understand that it's true because (negative$\cdot$negative) $\geq$ (negative$\cdot$positive). I'm not sure what I would do with $x \lt 0$ to make it look like $x^2 \gt 6x$. The first case felt very succinct in that I began with the inequality given, manipulated it, and the result was that which I was trying to prove. Do you think you could give me a clue regarding how to highlight this implication? $\endgroup$ – tom Nov 2 '18 at 1:00
  • $\begingroup$ The first case is straightforward by taking $x=6+t, t>0$ (this is the same as saying $x>6$). Then: $$x^2=(6+t)^2=t^2+12t+36$$ while $$6x=6(6+t)=36+6t$$. It's clear the first expression is bigger than the second as $t$ and $t^2$ are both positive. With regards to the second case, let $x=-u, u>0$. Then you are showing $$(-u)^2>6(-u)\to u^2>-6u\to u^2+6u>0$$ Again, we know $u$ and $u^2$ are positive, so this holds too. $\endgroup$ – Rhys Hughes Nov 2 '18 at 1:18
  • $\begingroup$ Ah, that’s brilliant! The x = -u for x < 0 sub is really elucidating. Thanks again $\endgroup$ – tom Nov 2 '18 at 1:41
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If $3-x <3$ then $x <0$ so $3 <6$ and $3x>6x $.

So $x^2>3x>6x$

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  • $\begingroup$ Where did $x^3$ come from? $\endgroup$ – Rhys Hughes Nov 2 '18 at 1:20
  • $\begingroup$ What $x^3$?........ $\endgroup$ – fleablood Nov 2 '18 at 4:05

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