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## The Should I Become a Mathematician? Thread

 Quote by micromass ... Rearranging again and we get $$\frac{(a^2-b^2)+c(a-b)}{a- b}$$ Eliminating a-b yields $$a+b+c$$
I'm not sure that I understand this step (even though it looks very simple). I did it a different way, so the way I got to that answer was different. What I'm not seeing is how you have a-b in the denominator, yet three separate occasions of a-b in the numerator (a2 - b2; and a-b), yet when you essentially cancel them out, you are somehow left with a + b + c.

In my mind, when you cancel out the a-b on the bottom with any of the three pairs of a-b on top, you are either left with:

(a-b) +c(a-b), or

(a2 - b2) + c

What do I seem to be missing, or not understanding?

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 Quote by AnTiFreeze3 I'm not sure that I understand this step (even though it looks very simple). I did it a different way, so the way I got to that answer was different. What I'm not seeing is how you have a-b in the denominator, yet three separate occasions of a-b in the numerator (a2 - b2; and a-b), yet when you essentially cancel them out, you are somehow left with a + b + c. In my mind, when you cancel out the a-b on the bottom with any of the three pairs of a-b on top, you are either left with: (a-b) +c(a-b), or (a2 - b2) + c What do I seem to be missing, or not understanding?
You know that $a^2-b^2=(a-b)(a+b)$

So

$$\frac{(a^2-b^2)+c(a-b)}{(a-b)}=\frac{(a-b)((a+b)+c)}{a-b}=a+b+c$$
 Recognitions: Gold Member I'm not going to try to right this with brevity like Micro, but instead I want to explain my thought process, because I feel that I may have done something incorrect. My solution: {a2(1/b - 1/c) + b2(1/c - 1/a) + c2(1/a - 1/b)} ____________________________________________________________________ { a(1/b - 1/c) + b(1/c - 1/a) + c(1/a - 1/b) I noticed from the start that the binomials would cancel out, so long as I was able to manipulate the problem and get them next to eachother, so I didn't see a reason to get rid of the fractions, since I knew they would cancel out anyways with their respective opposites. I then simplified a, b, and c to get rid of any multiplication in the denominator, and I used the commutative property to rearrange the denominator: { a(1/b - 1/c) + b(1/c -1/a) + c(1/a - 1/b)} ___________________________________ {(1/a - 1/a + 1/b - 1/b + 1/c - 1/c)} The denominator then cancels out to equal 1, so I am left with: { a(1/b - 1/c) + b(1/c - 1/a) + c(1/a - 1/b)} This next step is where I have broken math. I recognized what the answer should be, but I think that I may have cheated in order to get to that final result. As a result, I did this: {(a + b + c)(1/a - 1/a + 1/b -1/b + 1/c - 1/c) Then, similarly as before, the fractions cancel eachother out, so I was ultimately left with this: a + b + c I didn't peak at Micro's answer, and actually came to the correct answer myself. Regardless of that, I still feel as if that last step isn't allowed. Is it even possible to solve it correctly using the process that I used? EDIT: I messed it up in the first step, which is why I ended up in a situation where I couldn't correctly solve it.

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 Quote by AnTiFreeze3 I'm not going to try to right this with brevity like Micro, but instead I want to explain my thought process, because I feel that I may have done something incorrect. My solution: {a2(1/b - 1/c) + b2(1/c - 1/a) + c2(1/a - 1/b)} ____________________________________________________________________ { a(1/b - 1/c) + b(1/c - 1/a) + c(1/a - 1/b) I noticed from the start that the binomials would cancel out, so long as I was able to manipulate the problem and get them next to eachother, so I didn't see a reason to get rid of the fractions, since I knew they would cancel out anyways with their respective opposites. I then simplified a, b, and c to get rid of any multiplication in the denominator, and I used the commutative property to rearrange the denominator: { a(1/b - 1/c) + b(1/c -1/a) + c(1/a - 1/b)} ___________________________________ {(1/a - 1/a + 1/b - 1/b + 1/c - 1/c)}
Can you explain what you did to get to this step? It is clearly not correct, because:

 The denominator then cancels out to equal 1
Actually, it cancels to 0, not 1.

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 Quote by jbunniii Can you explain what you did to get to this step? It is clearly not correct, because: Actually, it cancels to 0, not 1.
I already mentioned that I messed up the first step, and that that is what threw off my whole solution. Thanks though.

EDIT: Although, if you are curious as to what was going through my mind, I embarrassingly forgot that I needed to simplify it before I could just eliminate a, b, and c. The rest of my problems stemmed from that.

I think it was coincidental that my answer ended up being a + b + c, even after making two big mistakes. Or maybe it wasn't coincidental, and I have just inadvertently invented a new form of Algebra where you break rules until you get the answer.
 Recognitions: Homework Help Science Advisor Very impressive micromass and Antifreeze! nice solutions! micromass and Antifreeze are very strong, but we can also make progress using some basic principles to help us. Here is a hint for other possible solutions: Generalized factor theorem: if f is an irreducible polynomial, and if f = 0 implies g = 0, then f divides g. (This is a basic result in “algebraic geometry”, and generalizes the basic result that x-r is a factor if r is a root.).) For instance, suppose a-b = 0, then what about a^3(c-b) + b^3(a-c) + c^3(b-a), does it vanish too? Then what? Now how did we guess to try a-b=0? Recall the "rational root theorem"? It says you look for roots of form X-r by trying factors r of the "constant term. As miromass observed, we can rewrite the top of the fraction after simplifying, as a^3(c-b) - a(c^3-b^3) + bc(c^2-b^2). Think of this as a polynomial in a. thus the constant term has prime factors ±b,±c, ±(c-b),±(c+b). (also other products of these factors, possibly.) So we should try setting a equal to those factors. e.g. a=b iff a-b = 0.
 I don't want to start a new topic for this question, so i post it here: How important is (euclidean) geometry in the higher (that is at the university) mathematics education? I'm currently in high school and feel that I've barely touched the subject, only simple computations with area, proportions, and some volume problems, together with a few "angle games". I'm thinking of maybe getting the following book: http://www.amazon.com/Elementary-Geo.../dp/0201508672 But maybe it's all too much, and not so important? I've enjoyed the little euclidean geometry I've done, but if I don't have very much use of it in the basic calculus and linear algebra courses, I'll probably skip it (for now). Thoughts on that? (Sorry for possible language errors, english is not my native, hope it's all readable )
 Recognitions: Homework Help Science Advisor its important, that's a good book: here' a cheaper one: http://www.abebooks.com/servlet/Sear...metry&x=74&y=9 and here's the all time great original version from euclid: http://www.amazon.com/s/ref=nb_sb_no...lid+green+lion and an excellent companion volume: http://www.abebooks.com/servlet/Sear...ts=t&x=52&y=12
 Mathwonk, after finishing Elementary Geometry from an Advanced Standpoint and Principles of Mathematics, what would you suggest next? I'm about halfway through A&O and Chartrand's proof book, which I should have finished up relatively soon, since most of my time has been devoted to my summer calculus class.
 Recognitions: Homework Help Science Advisor the natural continuation would be a strong calculus book like spivak or apostol. since you are already taking calculus that makes sense only if your course is at a lower level. other basic topics are topology and abstract algebra.
 Great, that will be my plan then! I just wanted to make sure there wasn't some other basic text I should work through after these. My calculus course is taught from Stewart and is almost purely computational, which is at a significantly lower level. I have done some supplementary work/reading from Apostol, but it does not line up 100% with my course in a manner that I can concurrently work through Apostol, though. I may be taking an honors, proof-based intro to linear algebra course this fall, though.