If I have an equation y = x^2 / x, I can reduce that to y = x. However, if I graph the original equation, I have an undefined point at x= 0. If I graph the simplified equation, there’s a point at 0,0.
Am I missing something? Are simplified forms of equations sometimes not actually equal to the unsimplified forms? Is there a real example (physics, chemistry, economics, whatever) where an equation loses some meaning if you simplify it?
0/0 is undefined. So x[sup]2[/sup]/x isn’t the same as x for x=0.
0/0 is more accurately described as “indeterminate”, a subtle but important distinction.
You can use L’Hopital’s rule to get a value to use at x = 0, and that value is 0.
The derivative of x[sup]2[/sup] is 2x. The derivative of x is 1. The value of
2x/1 at x = 0 is 0.
My Mathcad has no trouble plotting them both as a straight line at 45[sup]o[/sup] going from the third quadrant to the first quadrant through 0 for x from -25 to +25.
Just to confirm what emarkp and parkov have said… I learned that the technical simplication of y = x ^ 2 / x is y = x, x /= 0 or somthing like that. (Can’t remember how you signify ‘where x is unequal to zero’)
Basically, 0 ^ 2 / 0 is complicated enough that it’s best not to worry about defining it, though there are some arguments (possibly like l’hopital’s rule) that can assign it a value, or you can ‘fill in’ the function so that it’s continuous over that small missing hole.
Formally, though, whenever you’re doing simplifications in algebra that multiply each side by the unknown, you need to take care of the case where what you’re multiplying by is zero and exclude it. For one thing, if you don’t take that precaution, there are sample proofs out there that can demonstrate that 2 = 1 
and if we don’t stop that kind of thing, the entire structure of mathematics might collapse!!
After all, if 2 = 1, then 3 = 2 and 1 = 0 and 1 = 0.5, and basically you can create a unity between any two rational numbers if you work hard enough at the algebraic rules given the 2 = 1 premise. Once all rational numbers are equal to each other, then the irrationals and the complexes go straight to hell… (is the root of -1 still imaginary if -1 equals 2?)
And finally, slowly but with utterly devastating crashes, the various infinties will fall down on our heads.
[sub]Sorry for the hijack there. I’ve always wanted to use a variant on that spiel. ;)[/sub]
This is a tricky question! Clearly the two functions are mathematically different but I can’t think of an application of math where this difference is significant.
That’s not correct. The limit of the function at 0 is 0, but the function’s value at 0 is indeterminate. Your Mathcad’s implementation is incorrect if it’s plotting the point (0, 0).
btw, this is what we math folk refer to as a removable singularity, because you can make the function continuous at 0 by simply defining f(0) = 0.
The rule for simplifying fractions by cancelling says something like
ac a
___ = _ , assuming that c is not equal to zero.
bc b
So, when you cancel c, you are assuming that c is not equal to zero. So, you could say that (ac/bc) = a/b except that c is not allowed to equal zero in the second fraction (because otherwise the two fractions would not be the same thing when c = 0).
So, x^2/x = x if we decide that x can not be zero.
So the function y = x is different from the function y = x (where x can not be zero) = (x^2/x), so yes, you do get different functions when you simplify, unless you pay careful attention to the domain of the resulting function (or unless, by chance, the factor that you cancel is repeated in the denominator).
When I first taught algebra, we got to use a book where we could emphasize things like x^2/x being different from x, but unfortunately after the first section on simplifying fractions, the rest of the book never continued the practice of being aware of the difference. Where I teach algebra now, the book we use just assumes that anything in the denominator is never zero anyway… 
Stupid spaces didn’t stay where they were supposed to.
That should say the equivalent of
(ac/bc) = a/b, assuming that c is not equal to zero.
Yeah, that’s what I was trying to find. I’m guessing that any time this would come up in the real world, the limit at x = 0 (0) would be the “solution” at x = 0, but I’m wondering if there’s a time that isn’t the case.
Or, put another way, is there a time when the technical rule (that you’re adding the stipulation “where x != 0”) must be included, or something breaks.
I thought of an example where the distinction between x and x^2/x yields different answer:
xy=x^2
divide both sides by x:
y=x^2/x
simplify:
y=x
If you go back to the original equation:
xy=x^2, you’ll see that while y=x satisfies most cases, there is one case where it does not. Namely, when x=0. When x=0, y can have any value and satisfy the equation.
:smack:
Actually, my post really isn’t much more than a restatement of the original problem, now is it?
Not including that stipulation is always wrong.
Yes… but does it break any scientific equations. It’s wrong to a mathematician… but are there times in the physical sciences or economics or something, well, beyond pure math when the equation no longer works when it’s reduced, for that case of x=0?
A valid reduction doesn’t change the value of a function at any point where it’s defined, so no.
And there’s nothing specific about that “always” to mathematicians.
Well, here’s a simple example where the function (like the one you defined above) that has no meaning when x = 0, while it’s perfectly meaningful for any x>0. Suppose you hold an object out your window and drop it. Then the distance (in feet) that the object has fallen after x seconds is s(x)=16x[sup]2[/sup]. If you want to know the average speed of the object after x seconds, that would be given by the function f(x) = s(x)/x. This function will give a value of 16x for all x>0, but is clearly undefined for x=0, and that makes a lot of sense, right? You can’t average over 0 seconds. Note that I defined f(x) to be the average speed, not the instantaneous speed, which would be 32x.
Yeah, um, but…
average speed = 16x^2/x = 16x (if I carelessly reduce the equation without putting in the “where x != 0” warning). At 0 seconds, the average speed is 16 * 0 = 0 ft/s. So now the simplified form of the equation works (yielding a correct answer), but the unsimplified doesn’t?
At 0 seconds, the average speed is undefined.
Theres a proof that 2 = 1 that rests on the fact that your secretly dividing by 0 when simplifying an equation. Personally, I’ve encountered the same thing maybe 4 or 5 times when trying to simplify something and coming out with absurd answers.