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If I have two algebraic quantities, what is the correct way to determine if they are equal, or if not which is greater than the other? For example, if I have $\frac{2d}{\sqrt{s^2-c^2}}$ and $\frac{2ds}{s^2-c^2}$, and I want to prove $\frac{2d}{\sqrt{s^2-c^2}} < \frac{2ds}{s^2-c^2}$ for all positive $d$, $s$, and $c$ I know that I can set them "equal" and treat them as an equation like so: $$\frac{2d}{\sqrt{s^2-c^2}}=\frac{2ds}{s^2-c^2}$$ $$\frac{2d}{\sqrt{s^2-c^2}}\cdot\left(s^2-c^2\right) = \frac{2ds}{s^2-c^2}\cdot\left(s^2-c^2\right)$$ $$2d\sqrt{s^2-c^2}=2ds$$ $$\sqrt{s^2-c^2}=s$$ $$s^2-c^2=s^2$$ $$0<c^2$$ because we are given $c>0$. Then you can replace all the equals signs all the way up with less thans since I didn't multiply by $-1$. This doesn't strike me as particularly formal, so what is the formal way to accomplish this, and in addition, is this even a valid line of reasoning?

JJW5432
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    It is not too often that the word "the" belongs in front of "correct way." Anyway, no, you can't turn an inequality into an equality and back again willy-nilly, this is a special case of the fact that you can't change whatever you want in a proposition. What you have to do is manipulate the inequality in reversible ways (ones that could have accompanying $\Leftrightarrow$ symbols), or potentially splitting into cases, until you wind up at some more basic inequalities, like "$0<s$ and $0<d$." – anon Sep 04 '14 at 22:06
  • Can you provide the specific steps of applying that method to the example I gave, as well as the correct notation? – JJW5432 Sep 04 '14 at 22:21

2 Answers2

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$$\begin{align}\frac{2ds}{s^2-c^2}-\frac{2d}{\sqrt{s^2-c^2}}&=\frac{2d}{\sqrt{s^2-c^2}}\left(\frac{s}{\sqrt{s^2-c^2}}-1\right)\\&=\frac{2d}{\sqrt{s^2-c^2}}\cdot \frac{s-\sqrt{s^2-c^2}}{\sqrt{s^2-c^2}}\\&=\frac{2d}{\sqrt{s^2-c^2}}\cdot \frac{s^2-(s^2-c^2)}{\sqrt{s^2-c^2}(s+\sqrt{s^2-c^2})}\\&=\frac{2dc^2}{(s^2-c^2)(s+\sqrt{s^2-c^2})}\\&\gt 0.\end{align}$$

Or easily, $$\frac{2ds}{s^2-c^2}\gt\frac{2d}{\sqrt{s^2-c^2}}\iff \frac{2ds}{s^2-c^2}\cdot \frac{s^2-c^2}{2d}\gt\frac{2d}{\sqrt{s^2-c^2}}\cdot \frac{s^2-c^2}{2d}\iff s\gt\sqrt{s^2-c^2}$$ $$\iff s^2\gt s^2-c^2\iff c^2\gt 0.$$ Since $c^2\gt 0$ is true, the first inequality is also true.

mathlove
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After messing around by doing scratch work like what you've done, I like to prove what I originally wanted by using a single chain of operations: \begin{align*} \frac{2d}{\sqrt{s^2-c^2}} &= \frac{2d}{\sqrt{s^2-c^2}} \cdot \frac{\sqrt{s^2-c^2}}{\sqrt{s^2-c^2}} \\ &= \frac{2d\sqrt{s^2-c^2}}{s^2-c^2} \\ &< \frac{2d\sqrt{s^2 + 0}}{s^2-c^2} &\text{since } c > 0 \iff c^2 > 0 \iff -c^2 < 0\\ &= \frac{2ds}{s^2-c^2} &\text{since } s > 0\\ \end{align*}

Adriano
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  • Can you explain in a little more detail exactly how to go from my scratch work to this chain of operations? – JJW5432 Sep 04 '14 at 22:24