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So, I thought of something nerdy...I was making an algebra problem to disguise it so that to see the joke, you'd have to a bit of work, but I think that I might be doing something wrong, because neither Wolfram|Alpha nor MyAlgebra can solve it...I fixed a few things that I had wrong initially, but I don't see anything wrong're supposed to solve for i. $$\begin{array}{rcl} (5u)^2-(i+2u)^2 &\gt& 20u^2+4u(i+2u)\\ (5u+(i+2u))(5u-(i+2u))&\gt& 4u(5u+(i+2u))\\ 5u-(i+2u)&\gt& 4u\\ 3u-i&\gt& 0\\ i&\lt& 3u \end{array} $$

I figured that you could factor the left side and then divide (5u+(i+2u)) out of both sides, but maybe not?

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I fixed your missing parenthesis on the third line; I will LaTeX up the equations as well. – Arturo Magidin Oct 20 '10 at 20:37
There is also an error in the fourth line of your comptuations. – Arturo Magidin Oct 20 '10 at 20:54
+1 for embedding algebra problems into jokes. – anon Oct 21 '10 at 19:30
up vote 8 down vote accepted

Your mistake lies in trying to cancel $5u +(i+2u)$ without being careful about whether it is positive, negative, or $0$. Added: There is also a miscalculation in the third line.

Remember that "cancelling factors" is really multiplying by the inverse. But when you multiply an inequality by a number, you have to be careful about whether the number is positive or negative: if you have $a\gt b$ and $c\gt 0$, then $ac\gt bc$; but if $c\lt 0$, then you have $ac\lt bc$ (multiplying by negative numbers reverses the inequality).

When you try cancelling $5u+(i+2u) = 7u+i$, by not reversing the inequality you are implicitly assuming that $7u+i\gt 0$; that is, that $-7u\lt i$. So in fact, your conclusion is that if $-7u\lt i$ and the inequality holds, then $i\lt 3u$. But what if $7u+i\lt 0$? Then when you divide both sides by $7u+i$ in order to cancel the factor, you must reverse the inequality, so you actually get $5u -(i+2u) \lt 4u$.

Added: At this point you made a slight mistake with the signs. The left hand side is $5u - (i+2u) = 3u-i$; you had $5u - i + 2u$ (missing the parenthesis) so you incorrectly got $7u - i$.

So, if $7u+i \gt 0$, then you get $3u-i\gt 4u$, or $i\lt -u$. Thus, if $i\gt -7u$, then you must also have $i\lt -u$, or $-7u\lt i \lt -u$. Note that this can only occur if $u\gt 0$.

If $7u+i \lt 0$, then when you cancel that factor you get $3u-i \lt 4u$, or $i\gt -u$; so you have that both $i\lt -7u$ and $i\gt -u$ must hold. That is, $-u\lt i\lt -7u$. Note that this can only occur if $u\lt 0$.

So in fact you have two parts of the solution: you either have $u\gt 0$ and $-7u \lt i\lt -u$, or else you have $u\lt 0$ and $-u\lt i\lt-7u$.

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Oh, yeah...I'm starting to remember that from algebra class now...any suggestions on how to fix it or just do something simpler? – Ullallulloo Oct 20 '10 at 20:38
Nothing wrong with what you did as long as you keep track of the sign. Or you could expand and cancel summands (sign does not matter with that). But that will leave a quadratic in i, which you would have to factor and then consider the signs at issue; I don't think it will be easier than what you did here. – Arturo Magidin Oct 20 '10 at 21:03

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