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Why I don't think I'll ever enjoy math

  1. Sep 24, 2013 #1
    For me it's a necessary evil, I need it to do what I want to do, and part if why math usually makes me want to break things is so often when I reach a higher level of math, it often feels as though I'm being taught how to do one thing, like factoring polynomials, and in the next level, I've got to relearn how to do them because I get a wrong answer when I try to do it the way it was done before. Now I'm sure this is just my perception, but it's part of the reason that I've struggled with math in the past and still struggle at some things. Is this a wrong perception? Is it about how it's taught rather than something else?
  2. jcsd
  3. Sep 24, 2013 #2


    Staff: Mentor

    Do you have an example of this? This is not the way things usually are. More often, what you do in a subsequent class builds on what you've learned in a preceding class. The only thing I can think of that might explain this, is that a way you were taught before is applicable only to certain types of problems, but the new way is applicable to a different set of problems. As you progress in mathematics, theorems play a more pronounced role, and it's important to make sure all of the specified conditions are met before you attempt to use the theorem.

    An example would be helpful.
  4. Sep 24, 2013 #3
    Right now I am working on complex fractions, and have been trying all day to factor them, the way it worked in the previous lessons, and, it's not working. I'm working on fraction a + fraction b divided by fraction c - fraction d, and nothing seems to give me a right answer. It's driving me nuts, because, as I said, I've been doing it the same way all along and now it's not working, and I can't figure out why. I tried giving an example earlier, but it got axed for being a homework type question.
  5. Sep 24, 2013 #4


    Staff: Mentor

    I saw what you posted - why don't you try posting it in the Precalc section under Homework and Coursework?

    One problem is that you weren't factoring the complex fractions - you were actually supposed to simplify them, which could mean that you need to factor the numerator or denominator or both. However, whatever you learned before should still be valid, whether you're working with fractions made up of numbers or fractions made up of algebraic expressions. The same ideas apply to both.

    A mistake that beginning students make often is cancelling when they shouldn't do so. What cancelling really is, is finding factors that are the same in the numerator and denominator, and getting rid of them. An important word here is factor, which implies two expressions that are multiplied.

    For example, this is incorrect:
    $$ \frac{10 + 8 }{5 + 4} = 2 + 2 = 4$$
    This is actually 18/9, which is nowhere near 4.

    And this is correct.
    $$ \frac{10 + 8}{5 + 4} = \frac{18}{9} = \frac{2 * 9}{9} = \frac{2}{1} * \frac{9}{9}= 2$$
    9 is a factor in the numerator and denominator, which means the whole fraction can be thought of as having a factor of 1. Multiplication by 1 always leaves you with exactly the same value.

    Things work the same way with variables, except you need to be careful of factors in the denominator that could be zero.
  6. Sep 24, 2013 #5


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    You are in maybe, Introductory Algebra, or Pre-Algebra?

    The semester just began recently, since this is September. You become accustomed to the process. You are learning properties of numbers, so as these become more familiar, you will improve at figuring what to do. Most of the time, you look for a way to divide a fraction by 1 or multiply a fraction by 1. Can you give more examples of a problem that gives you trouble?
  7. Sep 24, 2013 #6
    So it turns out I was making it more complicated, I was trying to factor things that couldn't be.
  8. Sep 24, 2013 #7


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    Not sure.
    Could you provide a couple of examples?
  9. Sep 25, 2013 #8


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    My recommendation is to focus on what manipulations are allowed and why. So if we have ##{a \over b} = {c \over d}##, one learns the technique of cross multiplication, but why does it work? It works because what is on the left is equal to what is on the right, they represent the same number, so I can do the same thing to both sides. Multiplying each side by ##bd## will give the answer. Or think of it in two steps, I first multiply by b, then d.

    So any simplification can be split up into tiny steps that are usually justified by both sides being the same number. Here is another example, we have ##ax = b##, ##cx = d##. We learn that we can subtract one equation from the other, to get ##(a-c) x = b-d##, but why does this work? Well, there is a rule that equals can always be substituted for equals. If two numbers or expressions are equal, one can ALWAYS take the place of the other. So if I start with ##ax - cx = ax - cx##, then on the right, I substitute equals for equals, I get ##(a-c)x = b-d##.

    This is my recommendation, find out why each technique works and what steps are involved.
  10. Sep 25, 2013 #9


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    One problem many people have with mathematics is that they see "formulas" and think they can learn by memorizing formulas rather than learning basic definitions and concepts and thinking. "Memorizing" is easier than "thinking" but won't get you very far.
  11. Sep 25, 2013 #10


    Staff: Mentor

    To follow up with what verty said, if you have an equation, you can do all sorts of things, as long as you do the same thing to each side (e.g., add the same quantity to both sides, multiply both sides by the same quantity, etc.).

    If you are simplifying an expression, though, you are much more limited in what you can do. You can multiply by 1 (in a variety of forms), or add 0 (also in a variety of forms), or factor the expression, or expand it. In no case can you change the value of the expression.

    Looking at the equation that verty showed, ##\frac a b = \frac c d##, one approach is to multiply each side by bd, the common denominator of the two fractions. Another way of looking at this is that you can multiply the fraction on the left by d/d (= 1) and the fraction on the right by b/b (=1). This gives you ## \frac{ad}{bd} = \frac{cb}{bd}##. Since the denominators are equal, and the fractions are equal, it must be that ad = bc.

    It sort of goes without saying that neither b nor d can be zero. If either variable were zero, the original equation would be meaningless, and multiplying by b/b or d/d would be meaningless as well.
  12. Sep 25, 2013 #11
    I have autism, which isn't a developmental issue as much as it is a social one, but I've noticed that I tend to think in straight lines, and occasionally a new concept requires turning left or right slightly, and that's just hard for me to do I think, I don't know if that makes any sense or not. I do believe I've solved my current issues with this set of problems. Are there any resources for those with autism to help in math?
  13. Sep 25, 2013 #12


    Staff: Mentor

    I don't have any advice on resources for autistic people, but something you said about thinking in straight lines hit me. When you are in the process of thinking about what you need to do to solve a particular problem, if you misinterpret what the problem says, or make some incorrect assumptions about it, and then proceed in a straight line, you will almost certainly get the wrong answer.

    What many new students don't realize is that mathematics, unlike normal human conversation, is very terse, and there isn't a lot of redundancy. It's important to pay attention to details - ignoring or missing them will likely cause you to head off in the wrong direction.
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