# Graphing y = x vs. y = x^2 / x **URL:** \ **Category:** Factual Questions\ **Created:** [March 25, 2005, 6:16pm UTC](https://boards.straightdope.com/t/graphing-y-x-vs-y-x-2-x/296231 "2005-03-25T18:16:27Z")\ **Posts on this page:** 7\ **Page:** 2
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [March 25, 2005, 11:13pm UTC](https://boards.straightdope.com/t/graphing-y-x-vs-y-x-2-x/296231/21 "2005-03-25T23:13:21Z")
> [@ultrafilter](#): > > > [@David Simmons](#): > > > > You can use L’Hopital’s rule to get a value to use at x = 0, and that value is 0 > > 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). If you’ll read my post real carefully you will note that I didn’t say that the function was equal to 0 at x = 0. I merely said you can get a value to use there which, as **crisk** said, fills in that little hole. And I’ll let the Mathcad folks defend their work. ---
**Author:** ![ReuvenB](https://avatars.discourse-cdn.com/v4/letter/r/9fc348/32.png) [@ReuvenB](https://boards.straightdope.com/u/ReuvenB)\ **Post date:** [March 25, 2005, 11:52pm UTC](https://boards.straightdope.com/t/graphing-y-x-vs-y-x-2-x/296231/22 "2005-03-25T23:52:52Z")
> [@Jon the Geek](#): > > 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? This would be incorrect, seeing as speed (or technically, velocity, which is what you’re measuring anyway) is defined as (displacement)/(time). If time=0, then velocity is undefined. ---
**Author:** ![Jon\_the\_Geek](https://avatars.discourse-cdn.com/v4/letter/j/bbce88/32.png) [@Jon\_the\_Geek](https://boards.straightdope.com/u/Jon_the_Geek)\ **Post date:** [March 25, 2005, 11:58pm UTC](https://boards.straightdope.com/t/graphing-y-x-vs-y-x-2-x/296231/23 "2005-03-25T23:58:43Z")
Yeah, my bad. At time = 0, all I can calculate is the instantaneous velocity, which is 0… but is not arrived at by exactly the equation given (since a derivative is involved, we’ve eliminated the problem of the undefined point, right?). I think that answers my question, then; there is a time when reducing y = x^2/x to y = x will cause you to get an incorrect result in a physical science, and I can imagine that there are likely others… but it doesn’t mean we screw up every time we cancel variables in an equation, so long as we know that the case where it goes to zero is _actually_ a special case, and likely undefined. When I first realize the difference earlier today, I was worrying that perhaps I’d found a crack in everything I understand about physical laws, since they might not be as useful for generalizing as I thought… ---
**Author:** ![TJdude825](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@TJdude825](https://boards.straightdope.com/u/TJdude825)\ **Post date:** [March 26, 2005, 3:04am UTC](https://boards.straightdope.com/t/graphing-y-x-vs-y-x-2-x/296231/24 "2005-03-26T03:04:56Z")
> [@chrisk](#): > > … 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!! … Just for the record, the proof is: 1. Assume that x and y are both equal to 1. x=y 1. Multiply both sides by x. x[sup]2[/sup] = xy 1. Subtract y[sup]2[/sup] from both sides x[sup]2[/sup] - y[sup]2[/sup] = xy - y[sup]2[/sup] 1. Factor both sides, using the Difference of Squares on the left, and the common factor y on the right. (x+y)(x-y) = y(x-y) 1. Divide both sides by (x-y). x+y = y 1. Substitute 1 for x, and 1 for y. 1+1 = 1 1. Substitute 2 for 1+1 2=1 If you’ve been following the thread you can probably tell that Step 5 is illegal because x-y = 1-1 = 0. We are dividing by zero in that step and, as **chrisk** says, this causes the various infinities to fall down on our heads. ---
**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\ **Post date:** [March 26, 2005, 4:09am UTC](https://boards.straightdope.com/t/graphing-y-x-vs-y-x-2-x/296231/25 "2005-03-26T04:09:31Z")
> [@bostonpete](#): > > 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. For this function, it’s not really. In general, though, what we’ve got is a case of two different functions _with different domains_ that agree when restricted to any subset of the intersection of their domains. ---
**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\ **Post date:** [March 26, 2005, 4:17am UTC](https://boards.straightdope.com/t/graphing-y-x-vs-y-x-2-x/296231/26 "2005-03-26T04:17:01Z")
> [@pulykamell](#): > > 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. Actually, here’s a place where xy=x[sup]2[/sup] is a different notion than y=x[sup]2[/sup]/x: algebraic geometry. The algebraic set defined by the ideal generated by (xy,x[sup]2[/sup]) is the union of the lines y=x and x=0 in the affine plane. The algebraic set corredsponding to (y,x[sup]2[/sup]/x) is a subvariety locallized away from the line x=0. In particular, the ideal is not in k[x,y], but in k[x,y]sub[/sub]. ---
**Author:** ![erislover](https://avatars.discourse-cdn.com/v4/letter/e/71e660/32.png) [@erislover](https://boards.straightdope.com/u/erislover)\ **Post date:** [March 26, 2005, 3:34pm UTC](https://boards.straightdope.com/t/graphing-y-x-vs-y-x-2-x/296231/27 "2005-03-26T15:34:06Z")
> [@Jon the Geek](#): > > I think that answers my question, then; there is a time when reducing y = x^2/x to y = x will cause you to get an incorrect result in a physical science… The reactance (in ohms) of a capacitor in an electric circuit is given by 1/(2_pi_frequency\*capacitance) and is thus undefined for a constant DC signal. Just to throw another out there. [Previous page](https://boards.straightdope.com/t/graphing-y-x-vs-y-x-2-x/296231.md?page=1)