Is this the only way to show it? Type a[sup]2[/sup]...
:smack: Why didn’t I think of that? (I tried using the “code” tags but they didn’t work the way I wanted them to.)
Not necessarily, if you apply limits to it.
Limits have nothing to do with this. It’s an algebraic equation, not a calculus problem.
While the original problem does not have a defined solution, you can point out that there is a unique limiting point to the solution. That is to say,
if y = 8/(x-3)-(x+5)/(x-3) for all x <> 3, then
lim[sub]y->0[/sub] x = 3
Which is just a mathematical way of explaining the behavior of monica’s calculator graph.
On preview, I see that ultrafilter has dismissed the idea of limits on non-calculus problems. I’ll politely disagree with him to say that limits are useful anytime you have a discontinuity in a function (which clearly is the case here), not just in calculus.
The question was not “what is the behavior of this function as x -> 3?”. The question was “what are the solutions of this equation?”. Whether you use limits or not, the equation in the OP has no solution.
btw, 8/(x - 3) - (x + 5)/(x - 3) = (8 - x - 5)/(x - 3) = (3 - x)/(x - 3) = -1. y doesn’t go to zero.
I’m not even sure, in this instance, what you mean by the limit as y approaches zero. If you plug in values close to zero for y, what values do you get for x? They’re all undefined, so that limit doesn’t exist.
Even the limit as y approaches -1 doesn’t exist. Values of y close to -1 also give undefined values for x.
[QUOTEFranticMad, if you think undefined should equal undefined, then do you also think that the squareroot of an apple is equal to 8/0? **[/QUOTE]
I had no opinion on whether mathematics should allow undefined entities to be equal. I trust mathematicians to make the right choice for purposes of internal consistency or whatever.
I would love to insert the obligatory and irrelevant reference to Godel here, but maybe someone else can.
I do prefer a banana to the squareroot of an apple, as one can be eaten and the other – well I don’t know what you’d do with the the root of an apple. Hence I would deem them not to be equal, but I don’t pretend that this is a mathematical answer.
Just for an alternative viewpoint, in Matlab, with x = 3, (8/(x-3)) == ((x+5)/(x-3)) evaluates as true (although it does give divide-by-zero warnings). This is because Matlab uses the IEEE floating-point standard 754 for representing numbers, which has rules for operations involving infinities and indeterminate results (Not a Numbers). For example from this site:
Operation Result
n / ±Infinity 0
±Infinity x ±Infinity ±Infinity
±nonzero / 0 ±Infinity
Infinity + Infinity Infinity
±0 / ±0 NaN
Infinity - Infinity NaN
±Infinity / ±Infinity NaN
±Infinity x 0 NaN
I couldn’t find right away what the standard says about comparisons involving Infinities or NaNs. I’ll keep looking. In MatLab, NaNs always compare as unequal, even with themselves.
Of course, the OP was taking a math class, not a programming class, so naturally his teacher was correct.
BTW, If the IEEE standard 754 bothers you, know that it is currently undergoing a revision! Feel free to voice your opinions on this standard to the IEEE. As it says at the link, “Meetings are open and participation is free of charge”.
Right, ZenBeam, it evaluates to true because NaN == NaN. I believe you’d get the same result if you evaluated 1/0 == 2/0.
Of course, any problem that you use limits to solve is ipso facto in the realm of calculus. And your approach will give you some very unsatisfying results. For instance:
What is the value of f(0) if f(x) is defined as 1 for x not equal to 0, and 0 for x = 0.
The “unique limiting point to the solution” is 1, when the actual solution is clearly 0.
No, it evaluates to 8/0 == 8/0 which in turn evaluates to Inf == Inf, which is true (in Matlab, anyway). 1/0 == 2/0 also returns true in Matlab. However, NaN == NaN is always false, so for example, if x = 0/0, x == x evaluates to false. Here’s a site which gives the rules for comparisons with Infs and NaNs, although I’d rather see a more authoritative source say that Inf == Inf is true under IEEE 754, and not just Microsoft’s interpretation of it.
Ah I see. I didn’t read that closely, but that makes sense too. Cool.
Just out of curiosity, isn’t there such a thing as -0 in IEEE? What’s the value of n / -0?
I don’t suppose this would be a good time to mention square root of -1, and how it is imaginary but indispensible for correlational statistics, signal processing…no, okay, I’ll sit down.
Yes, there’s a -0. n / -0 = -Inf if n > 0, +Inf if n < 0, NaN if n = 0 or n = NaN.
Okay. Then I guess 8 / 0 == 8 / -0 would probably return false, even though we think of 0 and -0 as the same. Wacky stuff…