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for the inequality$$x+4 > 2$$ if you take the inverse you get$$\frac{1}{x+4} < \frac 12$$ The sign switches when you multiply or divide by a negative. My question is, what are you multiplying or dividing the expressions by? All I can see is e.g. $$2 \times \frac 14 = \frac 12$$ but 1/4 is positive, so the sign shouldn't switch yet it does, leading me to think something else is going on. Thanks

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  • $\begingroup$ Does this answer your question? If you take the reciprocal in an inequality, would it change the $&gt;/&lt; $ signs? $\endgroup$
    – Max0815
    Sep 12, 2022 at 21:51
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    $\begingroup$ Hint: It takes two steps to get from $a>b$ to $\frac{1}{a} < \frac{1}{b}$. $\endgroup$
    – Joe
    Sep 12, 2022 at 21:51
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    $\begingroup$ Divide both sides by $2(x+4)$ which we know to be positive. $\endgroup$
    – lulu
    Sep 12, 2022 at 21:52
  • $\begingroup$ No, I read that thread before asking this question. $\endgroup$
    – Rolomoto
    Sep 13, 2022 at 2:35
  • $\begingroup$ @Joe That's a wonderful way of viewing it. $\endgroup$ Sep 13, 2022 at 4:25

2 Answers 2

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Alternative approach:

Let $y = x+4.$

Then $~y > 2 \iff ~$ both of the following are true:

  • $\displaystyle y > 0 \implies \frac{1}{y} > 0$
  • $\displaystyle\frac{1}{y} < \frac{1}{2}.$
    This follows, since $\displaystyle ~y \times \frac{1}{y} = 2 \times \frac{1}{2}.$
    Therefore, if $~\dfrac{1}{y}~$ was $~\geq \dfrac{1}{2},~$ then (since $y > 2$), you would have that
    $~y \times \dfrac{1}{y}~$ would be greater than $~2 \times \dfrac{1}{2},~$ thus generating a contradiction.

Therefore,

$$x + 4 > 2 \implies 0 < \frac{1}{x+4} < \frac{1}{2}.$$

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  • $\begingroup$ I don't get the part: y(1/y)= 2(1/2). I mean if y<2 you could still use the same formula and get the same result, 1=1. $\endgroup$
    – Rolomoto
    Sep 14, 2022 at 4:12
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    $\begingroup$ @Rolomoto That portion of my answer was designed to prove that if $y > 2$ then $0 < (1/y) < (1/2).$ I used the concept that if $p > s$ and $q \geq t$, with $p,q,s,t$ all positive then $pq > st$. This was a proof by contradiction. In other words, with $p,q,s,t$ all positive, if $p > s$ and $pq$ equals $st$, then you must have that $q < t.$ $\endgroup$ Sep 14, 2022 at 4:16
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As mentioned in the comments, it effectively takes two steps (or the nice combo-deal shown by @lulu):

$$x+4 > 2 \implies 1 > \frac{2}{x+4} \;\text{ (this is OK because we know $x+4 > 0)$}$$

$$\implies \frac12 > \frac{1}{x+4}\;\square$$

So the sign didn't move at all and no negatives were involved :)

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  • $\begingroup$ where does the denominator 1 + x come from? $\endgroup$
    – Rolomoto
    Sep 13, 2022 at 3:46
  • $\begingroup$ @Rolomoto -- nowhere -- it was a typo -- fixed :) $\endgroup$
    – Annika
    Sep 13, 2022 at 3:47
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    $\begingroup$ I'd like to point out that the implication above is really only an implication. For $x <-4$, the left hand side has no solutions, while all $x < -4$ are solutions of the right hand side. So if you want to solve an inequality, you will very often need to make case distinctions, as user2661923 did below. The core of the problem is that multiplying by something that is not $0$ is equivalent only for equations, but needs to be $> 0$ for inequalities. $\endgroup$
    – Ingix
    Sep 13, 2022 at 6:34
  • $\begingroup$ @Ingix -- correct -- I kept this very straightforward here and took advantage of the positivity of the LHS, hence my use of $\implies$ and not $\iff$ $\endgroup$
    – Annika
    Sep 13, 2022 at 16:22
  • $\begingroup$ @Ingix What are you referring to when you say for 𝑥<−4, the left hand side has no solutions? $x + 4 > 2$? $\endgroup$
    – Rolomoto
    Sep 15, 2022 at 20:20

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