Another Theory question. . . .

(n/0)=Chuck Norris

 

Where is the conveyor in this question?

 

P.S. Tiny correction, it's "Caltech".

 

Is that paper published anywhere?

 

lim (x/n)=?
n-->0

 

Well Im going to have to say no....becuase if my trusty calculator cant do then it IMPOSSIBLE....lol.

Ok now I dont wanna a bunch of crap from people, that was meant to be a joke poking fun at the educational system in the US and kids current reliance on calculators.

Mark

 

The question, short as it is, is poorly worded .

"zero" as a text string is different than "0" as a text string or 0 as a numeric value or 0.0 as a floating point representation in a computer. Hence there are multiple answers, which is guaranteed to keep this new thread going for a long time .

P.S. We haven't gotten to the definition of "division" or "possible" yet either...

 

The proper terminology being 'undefined'?

 

I'm not heavy in math theory, but I remember my calc prof. saying something about if you can divide by zero, you can make anything work in a proof. She didn't explain it, and I imagine that could be an entire course in itself. You can't physically represent division by zero, and that is all that I am concerned about.

 

Simple answer?

Sir, step away from the number line and keep your hands visible at all times.

Zero is an absence, not a presence.

If you treat an absence with the same validity as a presence you can push through an answer on paper that has no validity on the real world.

So, you can divide by Zero but you shouldn’t, simply because the ‘answer’ is junk.

Chances of this post being seen as closure to your question:

Zero.

 

Division by zero is considered an "undefined" mathematical operation. But actually we can say quite a lot about such divisions in specific instances.

It's best to represent these things by the limit notation.

lim (a/x) = ?
x-->0

When a is nonzero and real, the limit is either positive or negative infinity. This can be seen by looking at the asymptotes to hyperbolas like y = a/x at the vertical axis. In fact the sign of the infinity is dependent on whether the direction of approach is from the negative side or the positive one. You can represent this like so.

(a > 0)

lim (a/x) = - infinity
x-->0-

and

lim (a/x) = infinity
x--> 0+

where the '+' and '-' signs should be taken as superscripts of the zero.

What about 0/0 ? That is, what is

(a = 0)

lim (a/x) = ?
x --> 0

Now this one really is undefined ...as it stands.

In actuality, if you can define the numerator and denominator as proper functions of a variable like x, you can study how the numerator and denominator "behave" as they approach their respective limits and work out the overall limit of the quotient. The value in fact, can be zero, it can be positive or negative infinity, or it can be just about any real value in between.

In this case, I will represent the limit as

L = lim f(x)/g(x)
f(x) = 0
g(x) = 0

There is a mathematical trick in basic calculus known as L' Hopital's Rule that allows you to evaluate "indeterminate forms" like the above. Even stuff like infinity/infinity can be evaluated, provided the functions involved meet some technical rules.

L' Hopital's Rule (from here on LHR) basically says differentiate the top, differentiate the bottom and take the limit and you'll get the correct result. So if the result after differentiation is easier to evaluate, you're home free.

In mathematical jargon, LHR states :

L = lim f(x)/g(x) = lim f'(x)/g'(x)
f(x) = 0
g(x) = 0

So let's take a look at each of the examples, where the limit L is zero, infinity and a real finite number.

Examples of each :

A) L = lim sin(x)/x = ?
x --> 0

Differentiate top and bottom to get :

L = cos(x)/1 = 1

Here the limit is a finite real (1). Multiply the numerator or denominator (or both) by arbitrary nonzero real values to get any arbitrary nonzero real result.

B) L = lim(x^2)/sin(x) = ?
x --> 0+

Differentiate here and you get

L = lim 2x/cos(x) = 0
x --> 0+

Here the limit is zero.

C) L = lim sin(x)/(x^2) = ?
x --> 0+

Differentiate here and you get

L = lim cos(x)/2x = + infinity
x --> 0+

(basically, just reverse the limit in B)

So here the limit is infinity. Just invert the signs of the limit to get (-) infinity.

 

She's referring to classic fallacies (euphemistically called paradoxes) like the following :

a = b (a, b are nonzero)

a^2 = b^2

a^2 - b^2 = 0

Factorise the left hand side

(a+b)(a-b) = 0

Divide both sides by (a-b), you get

a + b = 0

a = -b, which contradicts our initial statement of equality. You can go one step further, dividing both sides in the last statement by a (since a = b by initial assumption) and get this beauty :

1 = -1

if you group the terms, another beauty :

2 = 0.

And you've demolished the whole number system.

This sort of error arises because of the attempted division of both sides by (a-b) which is, in fact, zero. You get the indeterminate form 0/0 here, which cannot, in this instance, be assumed to be zero (or hold any specific value for that matter).

 

TurOflat4, you must be the one who wrote the paper for Caltech.
You already have two pages on this thread and I can tell your not finished yet.

You can multiply, divide, subtract, add and theorize about anything you want.

You guys are way too smart for me.

 

No, you're too kind. But as a coincidence, I did actually get admission to Caltech (electrical engineering) on a Singapore govt scholarship. Didn't have to write a paper though, just the SATs and Achievement Tests or something.

This is just basic math, man. I love this stuff. Regret doing Medicine every day.

 

I just noticed this :

I don't think this one quite needs to drag on to 25 pages, do you ?

PS : I happpily stayed out of the other thread. I know how to pick my battles.

PPS : It does fly.

PPPS : So nyaah.

 

Very interesting post. I would think that if anything, 0/0 would be assumed to be 1, as is anything when divided by itsself. Even so, it still doesn't work.

 

"Is an action possible when only influenced by nothing?"

No.

 

Yes it is possible.

The question is x/0 = ?

Case 1: x is not equal zero.

x/0 = + or - infinity

Case 2: x is equal to zero

x/0 = indeterminate

 

wow. All I know is that 1+1=2...haven't solved 1+2 yet.

 

hahaha, so 11th grade.
when you divide by zero, you can prove anything.

see the step: "divide by (a-b)"
(a-b) = 0, if a and b are substituted. so ha! i saw right through that logic


as for the real question at hand.....
(-)infinity does not equal positive infinity therefore

lim x/a DNE (by limit definition, where left and right sides have to equal)
a->0

 

Certainly a fair point. My example was for illustration only, which is why I didn't bother with that much rigor. The one-sided limits (left and right) certainly exist but the function has a jump discontinuity at 0.

Just square the function (or take the absolute value function), and this objection is addressed, the discontinuity is removed.