I can’t add anything to the excellent analyses of the problem at hand, but I would like to say that whomever wrote that calendar should be ashamed.
“Cannot be determined” is a damn stupid thing to print as the solution. It should, of course, have said “x may equal any value except -10.”
Well, magellan, I hope you’re happy, because I just spent the past hour proving to myself that you can’t disprove algebra by excluding x from possibly being any real number, and did the “mathematician’s headsmack” at the end for not having realized why in the first place. 
I won’t post everything I did, both for reasons of length and because I’m kind of ashamed to admit that I bothered, but:
I started by conceptualizing 1/(x + 1) = y/(x + 2). I assumed that it must be impossible to solve that for y in terms of x, because then I could construct an inductive proof that for any real number n, you could construct an expression y/(x+n) equal to 1/(x + 1) such that the denominator was equal to zero. Of course, as soon as I wrote this down, it became obvious that y = (x + 2) / (x + 1).
I then proceeded to spend far, far too long algebraically manipulating 1/(x + 1) to obtain denominators of (x - 3) through (x + 3) in equivalent expressions, and adapting that procedure into an inductive proof using n.
Of course, as soon as I wrote out the equation for the inductive proof, I realized what this board’s less-out-of-practice-than-I math whizzes probably noticed from the outset: I was right the first time. You can’t solve my original equation for y in terms of x and substitute the answer back in, because if you do, the resulting expression is always equal to 0/0…in other words, undefined. You can no more set up an equation this way than you can answer “-10” for the original problem. :smack:
Blargh. Well, it was a good question, and at least I have an answer for the next guy who asks me this. (I occasionally tutor algebra students, so it’s more likely than you’d imagine.) Thanks for the midday mental workout!
This particular problem isn’t likely to be an issue in engineering, since it’s only a single exact point out of the entire number line of possible solutions, and nothing in engineering is ever perfectly precise. That is to say, you wouldn’t end up with some bridge falling down because one beam is exactly 10 times the length of another beam, because that beam would probably actually be 10.000367289 times the length of the other, or some such.
I recall seeing a neat proof that 1 = 2. Unfortunately, it had to go thru a relatively hidden step where 0 divided by 0 lurked, which invalidated everything after that step. I’ll see if I can dredge that one up and post it.
0/0 generally screws everything up.
Wait, what are you saying, Roland Orzabal? If 1/(x+1) = y/(x+2), then it most certainly is the case that y = (x+2)/(x+1). Or, to put it more meticulously, so long as x is neither -1 nor -2 (so that all the divisions are well-defined), it is the case that 1/(x+1) = y/(x+2) if and only if y = (x+2)/(x+1).
Found it.
step 1: Let a = b
step 2: Multiply both sides by a, which gives
a^2=a*b
step 3: Now lets subtract b^2 from both sides giving
a^2 - b^2 = a*b-b^2
This is equal to
(a+b)(a-b)= b(a-b)
step 4: dividing both sides by a-b gives
a+b = b
step 5: since a=b, we have
a+a = a
or
2a = a
QED
2=1
Of course, this blew up at step 4 where we not only divided both sides by a-b (which is 0), but also had a-b on the numerator of the left side of the equation which resulted in 0/0. So everything after that step was bogus.
Why does he evaluation have to be pointwise? Why can’t we claim that (x+10)/(x+10) = 1 even at x=-10, by evaluating the function symbolically (is there another term which would be better?) over the range of all x, then performing the evaluation for its value at x=-10, rather than evaluating (x+10) at x = -10 , then attempting to find 0/0 (which of course isn’t defined)?
I realize this isn’t how things are defined, but could this approach be taken and used rigorously? Is there an example showing this would lead to ill-defined behavior?
Also, Chronos’s example of why 0/0 isn’t a problem shouldn’t make you feel any safer, because it’s just as likely that in the real world, what happens is you end up with (x+10.00001)/(x+9.9997), which is hardly any better near x=-10 than 0/0.
What you speak about is perfectly reasonable, ZenBeam; for example, we can identify partial functions with each other if they are equal wherever they are both defined and this common domain is a dense subset of each function’s full domain. Then, we can pick canonical representatives from each equivalence class as unique continuous extensions, thus turning (x+10)/(x+10) into 1. Indeed, people do do this sort of thing not infrequently, with perfect rigor; you just have to remember that this is what you’re doing, and not something else. But so it is with everything.
Unfortunately, I have nothing more to add then to merely give the confirmation that “Yes, this sort of thing works, perfectly well”.