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#5
+1750
0

What you are seeing is Rom’s resignation –retroactive to the late hours of the prior week.

Rom has left before, without ceremony, for months at a time, once (except for a single post) for over 16 months, and once (except for a single post) for over a year. In this current session of 13+ months, Rom has made over 1800 posts.

On Jan 29, 2014, six weeks after DavidQD joined, Rom became the second engineering-classed mathematician to join the forum, with Alan joining 26 days later as the third. Seventeen days later CPhill joined, taking some of the load off of Melody for the answers to geometry and other questions.  In a little over five months, after Melody drained the primordial swamp and cracked the professional teacher’s hickory stick over the heads of a few brats, and one particular troll, this forum was well on its way to becoming a university classed tutoring site while still catering to the needs of the high school and elementary students.

It was and still is amazing to me to see competent answers to advanced physics and engineering questions that would have previously gone unanswered. The answers are every bit as skilled as answers found on Stackexchange, (without the me-too pedantic comments). As the questions received skilled answers, more were asked, and at least one member completed his degree in Civil Engineering under the tutelage of these three (five), with many others contributing who would not have ever been here, if not for the first five.

Because Rom has left before and returned after a sabbatical, he may very well do so again. This time it is telling that his exit is followed by a candlelight farewell by several members and guests. This is a first on this forum, and this very much reminds me of the farewells (some were actually celebrations) given by the students of my parochial school for the teachers who were retiring after the end of the school year.

So Rom, it seems that some of us will be keeping a lit candle in the window, awaiting your return. I’m sure a troll or two may come along and blow it out on occasion, but that’s not a problem: we have lots of matches. It’s not exactly the eternal flame, but it’s about as good as it gets on here, which is actually pretty damn good!

GA

Now, what did I do with my bloody anti-nausea meds

4 hours ago
#4
+23170
+1

Piecewise function

If  $$z = \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }$$  find $$\lfloor z \rfloor$$.

The fractional part denoted by  $$\{x\}$$ for real x and defined by the formula $$\{x\}=x-\lfloor x\rfloor$$.

$$\begin{array}{|rcll|} \hline z &=& \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} } \quad | \quad \{ \sqrt{3} \} = \sqrt{3}- \lfloor \sqrt{3}\rfloor,\ \{ \sqrt{2} \} = \sqrt{2}- \lfloor \sqrt{2}\rfloor \\\\ &=& \dfrac{ \left(\sqrt{3}- \lfloor \sqrt{3}\rfloor\right)^2 - 2 \left(\sqrt{2}- \lfloor \sqrt{2}\rfloor\right)^2 }{ \left(\sqrt{3}- \lfloor \sqrt{3}\rfloor\right) - 2 \left(\sqrt{2}- \lfloor \sqrt{2}\rfloor\right) } \quad | \quad \lfloor \sqrt{3}\rfloor = \lfloor \sqrt{2}\rfloor = 1 \\\\ &=& \dfrac{ \left(\sqrt{3}- 1\right)^2 - 2 \left(\sqrt{2}- 1\right)^2 }{ \left(\sqrt{3}- 1\right) - 2 \left(\sqrt{2}- 1\right) } \\\\ &=& \dfrac{ 3-2\sqrt{3} + 1 -2(2-2\sqrt{2}+1) } { \sqrt{3}-1-2\sqrt{2} +2 } \\\\ &=& \dfrac{ 3-2\sqrt{3} + 1 -4 +4\sqrt{2}-2 } { \sqrt{3}-2\sqrt{2} +1 } \\\\ &=& \dfrac{ -2\sqrt{3} +4\sqrt{2}-2 } { \sqrt{3}-2\sqrt{2} +1 } \\\\ &=& \dfrac{ -2\left( \sqrt{3} -2\sqrt{2}+1\right) } { \left(\sqrt{3}-2\sqrt{2} +1\right) } \\\\ &=& -2 \\ \hline \mathbf{ \lfloor z\rfloor } &=& \mathbf{ -2 } \\ \hline \end{array}$$

9 hours ago