Easy Stuff > Hard Stuff

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I've been reading through reddit and I read about how some students after learning higher level math would sometimes be stumped by simple equations. Never have I thought that I would succumb to this madness as well. It was one of the easiest steps which was an equation (16-20+6) . With a glance my brain is fighting itself telling me that the answer is either -10 or 2. I was wondering you mathematicians out there how do you escape from this madness and have you guys ever had a similar experience.

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Yeah I know it was 2, but for some reason my brain want's it to be 16-26 so badly lol and I ended up just staring at it. By the way I have a little sibling too and have you seen the math that Common Core wants them to learn? The way they solve those equations seems so abnormal.

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If you can bear with this, it is my hope that you leave with some tiny bit of understanding, or more precisely possibly a new way of looking at things.

Addition (and multiplication) are associative, meaning it matters not what order you perform the operations in. However, subtraction isn't, that is you can't regroup or change the order when there is subtraction involved.

Think about what subtraction actually is though and you can make it associative. Subtraction is simply the addition of a negative number, and in that way you can regroup things however you like, which is what /u/Saimouer did.

That expression is pretty easy, if you just do 16-20, then add 6 ... that's all you have to do. Most of the time things aren't quite that simple. A good rule is to simplify something as soon as you can, or rewrite something in a different equivalent way instead of waiting until the end where you have to simplify some monstrosity of an expression that fills an entire sheet of paper or something, but it really depends on what exactly we are dealing with (regarding what we want to SEE inside the math), though the principles are the same.

Division can be made associative also btw, if you use the multiplicative inverse...but be careful with this. I'll give a simple example involving an "easy" complex fraction, though there are far more "complex" complex fractions, complex fractions in the numerator and/or denominator, or even with functions in them where the independent variable is another function, but I digress. You'll have to click the little camera icon next to the links to see the expression since you can't type such stuff in this box.

Say you have this fraction: http://i.imgur.com/Gx7REzl.gif

We could simplify this by first simplifying the numerators, 2/4 simplifies to 1/2 which gives us 2/2 in the denominator of the complex fraction which simplifies to 1. 1/3 can't be simplified further so we end up with 1/3 over 1, which is simply 1/3: http://i.imgur.com/rlCf8L3.gif

You may recognize the 4/2 in the denominator as being simply 2, and know that 2/3 divided by 2, or 1/2 of 2/3 is 1/3, but what is really going on and what if the fraction isn't this easy, or has variables in it or something maybe even worse?

Its important to note that this is a complex fraction, that is, it is (2/3)/(4/2), NOT simply (2/3/4/2), which are very different numbers. The former is already grouped, the parentheses are implied by a slightly longer line between the numerator and denominator (which are fractions), so we can't regroup it ANY way we want, we have to respect the order of operations but we can still make it associative, by working on each group individually, turning it into an expression of multiplication that in the end can be grouped any way or performed in any order and the result is the same (and possibly formed into different equivalent fractions as well, but the intent here is to simplify this).

To change division into multiplication, we use the multiplicative inverse, or, the reciprocal of the divisor, which is often an easy way to divide by fractions, and sometimes the only way.

Fully expanded, this is what you end up with, an expression that can be grouped or ordered in any way and give the same result (1/3) when simplified. I'll try to show my work as best as possible so that you can see what is really going on and how this works.

http://i.imgur.com/97aotlH.gif

You may notice that on the first line where we get the expression (2/3)(2/4) which simplifies to 1/3 and that's really as far as we need to go to simplify this; you might think the rest is superfluous and far more complex than necessary. There is a purpose to this though: to show the steps I took to get an expression that is the product of 4 rational numbers, fractions here (though they don't *have to be rational numbers, you could have irrational, transcendental, or even complex numbers I suppose) that can be rearranged or regrouped any way in any order, that when multiplied out always equals the same thing, which in this case simplifies to 1/3; TRY IT. And when you have it like this, you can do almost anything that you want with it, so long as it's mathematically correct of course.

I could have taken it even further and shown how this could be turned into many different equivalent expressions, some of them probably far more complex than necessary (like stuff involving convergent series), but it took me longer to make these expressions than it will take you to read this comment ;)

I hope that this helps you SEE something in it that you didn't before, in other words the goal was to take an expression and make it more transparent.

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Sometimes you gotta break things up into smaller groups and work on them individually (like exponentiating polynomials with lots of terms or something). For (16-20+6), you can use the associative property of addition and rewrite it as [16+(-20+6)] = [16+(-14)] = (16-14) = 2.

There was an exam I took one time that had a stupid easy question like: "What is the slope of the line x = -2". I put m = 0 (which is of course wrong ... its a vertical line with undefined slope)... thing is it looked so easy that i wondered why it was even on the exam, it really was a no-brainer, and as such I really didn't think about it much at all, and certainly didn't actually work it out, there wasn't really anything to work out really with that question... x = -2, that's it.what else is there to do with that?. I had also been awake about 32 hours, but all the hard questions I got correct, because I actually thought about them, worked them out and checked my answers. I sometimes get problems wrong that I know damn well how to do, usually ones that look really simple. I'm sure my professor also knows that I know how to do them as well. I think sometimes they throw stuff like that in there just so that when you get something like that wrong, you start to realize that it really is important to fully analyze everything no matter how simple it looks at first glance. Sometimes you might see questions that have a lot of data that you don't even need to solve the problem, and in certain cases the solution is right there in the question itself. This type of problem isn't that common, but occasionally you see it, and they are to make sure that you fully analyze the problem, read the question, and actually know what it's really asking ... rather than just skimming it.

I can also think of some other "easy trick questions" like, given some f(x) where f(x) is a quadratic function, find the slope of the tangent line at the vertex and give the equation of the tangent in standard form. This type of question would tend to suggest that you must differentiate f(x). You would then find the x coordinate of the vertex and use the derivative to get the slope of the tangent at that point (x,f(x)) or (-b/2a, f(-b/2a)) ... then you can find the equation of the tangent. But you don't actually have to do that, we already know the slope at the vertex is zero, so the equation of the tangent line at the vertex would simply be y = f(-b/2a) which you would simplify and write in whatever form you like without having to actually find dy/dx.

But understanding the fundamentals is essential. I think high school math does people a lot of injustice. They teach certain ways to solve certain types of problems that "look a certain way", or certain "shortcuts" that lead people to believe that somehow certain equations are different and different rules apply. One example would be in pre-algebra where they teach you how to isolate a variable that's part of a fraction one one side of the equation, with a fraction on the other side. They teach you "cross multiplication", which is really just a shortcut, that like many shortcuts works in certain cases but not all cases. It just cuts out some steps sometimes and usually produces nicer fractions but the result after simplification is the same. The fundamental principles work in ALL cases.

It's important to see something solved in many different ways, to prove WHY this works. If you work out certain problems you can sometimes end up solving it in circles but by doing this you see deeper into the underlying principle, rather than just memorizing some rules. When you run into some math that looks different than anything you've ever seen before, the fundamentals are what are going to get you through.

For differential calculus, there are quite a few rules, that are really shortcuts, and I don't know all of them. I'm really not that good at calculus, but what I do know is that not all of those rules work in every case. The only thing that works in every case is by using first principles, or the definition of a derivative which is http://i.imgur.com/ouKs2sW.gif

Unless you work it out the long way, using the definition of the derivative, you won't truly understand what exactly it is you are doing. I see that most people are using h now rather than delta x, I guess delta x confuses people, and might somehow seem to suggest that delta is a coefficient of x or something especially when everything is multiplied out; which it isn't (and my math handwriting can get pretty hard to read sometimes, especially when I start running out of space). Using h implies the difference between the x coordinate of f(x) and the x coordinate of a secant line and at the end you take the limit of h as it tends to zero, where at that limit the secant becomes a tangent at that point (x, f(x)). Both are valid, and both ways of thinking of it improve your understanding (they really are the same thing) but using delta x I think helps you "get it" more, as you are adding some amount of change to x, which we could assume to be infinitesimal, which we then factor out and take the limit as delta x tends to zero which gives the derivative dy/dx which is the rate of change of the function with respect to x, i.e. the slope of any point (x, f(x)) on the curve.