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Okay.
So dividing by zero makes no sense.
r * 24 = 0
24/0 = 0
r = 0
0 * 24 = 0
Thus, is true?
I get what I've heard for some explanations, but that's just weak.
I want a technical explanation :3
EDIT:
I got 24/0 = 0 like this....:
If you split 24 up into 0 groups (i.e. take it away), then you are left with nothing.
24 / 0 is undefined, not zero
Why is this in tech forum?
Yea, I've heard.
That isn't an explanation.
what exactly are you asking?
SPECIFICALLY, why is x/0 'undefined'?
Not, "Because it is undefined".
The mistake you are making is assuming that splitting 24 into 0 groups is theoretically equivalent to "taking it away." Taking the 24 away would be subtraction (24 - 24 = 0). Splitting 24 into 0 groups, on the other hand, is not possible.
Why not?
After all, we can add 5 to itself zero times and get zero. So why wouldn't such abstractions apply here?
well, it's not undefined
it's infinity
machines leave a divide by zero as undefined because you can't represent it
Division is the ratio between 2 numbers
R = a / b
12 / 4 = 3
3 = 12 / 4
the ratio between 4 & 12 is 3
put simply, 12 is 3 times larger than 4
for ever decreasing values of b, the ratio increases
12 / 4 = 3
12 / 1 = 12
12 / 0.1 = 120
12 / 0.01 = 1200
12 / 0.000000001 = 12000000000
the closer the denominator gets to zero, the larger the result
if the denominator is zero, the result is infinitely large
Okay, I see. Thank you.
But then, it isn't undefined, it is ifinity, so why doesn't it work? Infinity is represented as 'infinity'.
Any operation on it results in itself.
But then, that doesn't work because
r * 24 = 0
r = inf
inf * 24 = inf
r HAS to be 0.
Should I do it? Yeah, okay. There is a way to essentially divide by zero, it's the basis of calculus. Calculus is largely about finding inventive ways to divide by zero.
Thats not rigor.Quote:
the closer the denominator gets to zero, the larger the result
if the denominator is zero, the result is infinitely large
Basically if you have y=1/x, as x tends to 0, y tends to infinty.
Ironically, speaking you can actually work with that using nonstandard analysis.
I have a book on abstract algebra, it touches this subject. It says something about this algebra shows thats why division by zero is not allowed, however I couldn't understand the book after about three pages.
In calculus you never divide by zero, actually if you're forced to do that then you have to concluded it has no limit. In calculus you work with infintely small quantities, not zero.Quote:
There is a way to essentially divide by zero, it's the basis of calculus. Calculus is largely about finding inventive ways to divide by zero.
Er, not sure what you mean, Wendy...
Ninja:
Heh.
My Uncle's taking a calculus course. Looks hella abstract }_}
infinity is not a number
you cannot do calculations on it
infinity is a representation of a value so large (or small) you can't define it
how long, in metres, is the universe?
it just keeps going, and going
you can fly around in a space ship all your life, placing rulers end-to-end and always need more rulers
so, how long is it?
Algebra can not divide by zero, you can only do that through calculus. I assume what you are talking about is polynomials that have discrete breaks. That's not quiet the same as division by zero.Quote:
Originally Posted by wendylove
Calc isn't that hard, you just have to get used to it.
Ynot is correct, infinity is a concept, not a number.
It is still represented as infinity.Quote:
Originally Posted by Ynot
Any operation on it results in itself.
What I don't understand, is this.
r * 24 = 0
r = 0
This is true, because 0*x = 0
So then:
r * 24 = 0
r = 24/0
r = 0
If the previous statement is true, then why is this not true?
Calculus can be done by blind monkies with broken fingers. Analysis is abstract, calculus is just doing a technique.Quote:
My Uncle's taking a calculus course. Looks hella abstract }_}
Do set theory and come back here saying that. Infinity is a number, Cantor said so.Quote:
infinity is not a number
This means nothing, and its not rigor. Numbers are abstract, this proves nothing.Quote:
you can fly around in a space ship all your life, placing rulers end-to-end and always need more rulers
so, how long is it?
Just because something can be used for mathematical functions does not make it a number. i is not a number, but i^2 is. Quaternions aren't numbers either.
0 is a weird number in math. You can take a defined equation like the one that you mentioned, move it around and make the variable be different things
r * 24 = 0 : r = 0
0 / r = 24 : r = undefined
r = 24 / 0 : r = infinity
I didn't say it was a number...Quote:
Originally Posted by ninja9578
And that makes no sense.
In order to find that
r * 24 = 0
so then
r = 0
You would have to say
r = 24/0
r = 0
So.
And wouldn't 0/r = 24 be false anyhow?
Take nothing, split it up, and you still have nothing.
Here's a quick lesson in calculus. Say you want the slope of a line at a single point in the line y = f(x). The calculus way to do that is rise over run, but in this case, run is zero because it's just a point.
We need to find some way to solve for a point, so lets use the algebraic way to solve
slope = rise / run
f(x)' = dy / dx
Chose any x, and add a little bit to it. We'll call that little bit ∆x. That ∆x is out run now, but remember we want that run to be zero, so the limit of ∆x is zero.
f(x)' = lim ∆x -> 0 (dy / ∆x)
Now lets change dy. We know that f(x) gives us y so f(x + ∆x) - f(x) will gives us dy.
f(x)' = lim ∆x->0 (f(x + ∆x) - f(x) / ∆x)
Now we need to know what the function f(x) is. For simplicity, lets use f(x) = x^2. So...
f(x)' = lim ∆x->0 ((x + ∆x)^2 - x^2 / ∆x)
Now lets simplify.
f(x)' = lim ∆x->0(2x∆x + ∆x^2)/∆x
Now we have a ∆x on the bottom and in each function in the top. This allows us to cancel them out. This is what calculus is all about, keeping that zero a variable as long as possible and finding a way to cancel it out. However you'll notice when we simplify, there still is a ∆x on the top because on of them is square:
f(x)' = lim ∆x->0(2x + ∆x)
But now the limit of ∆x is zero, so replace ∆x with zero.
f(x)' = 2x, which is the derivative of x^2
Yes, that's why that variation of that equation is undefined.Quote:
Originally Posted by A Roxxor
You fail MAJORLY at basic re-arranging of equations, and everyone else in this thread fails to varying degrees as well.Quote:
Originally Posted by A Roxxor
r*24 = 0
divide both sides by 24:
r = 0/24
r=0
If you can't understand that then I suggest you go back and redo third grade.
As for the undefined/infinity discussion:
A quantity x/0, x real, is undefined. To see why, look at the left and right limits of the function y=1/x as x approaches 0. The left limit will give you -infinity, and the right limit gives +infinity. Thus, the limit does not exist, and 1/0 is undefined in the reals. It's not a rigorous proof, it just demonstrates why 1/0 is not infinity.
I suggest you ALL read this:
http://en.wikipedia.org/wiki/Divide_by_zero
Uhm, there are multiple ways to rearrange the same equation.
All are rearrangements of the same equationQuote:
Originally Posted by ninja9578
a * b = c can be
b = c / a or a = c / b
FYI: This thread is moved to the lounge as this is not a discussion on implementing division by zero.
0/0 could technically be any number
'infinity' is only commonly cited because the function 1/x approaches it from the right
Think about this
5*0 = 0
5 = 0/0
20* 0 = 0
20 = 0/0
135*0 = 0
135 = 0/0
According to the above statements, 5 = 20 = 135 which is why we say 0/0 is undefined, it equals any number.
???Quote:
Originally Posted by drewmandan
Well yea.Quote:
Originally Posted by arby
'r' could be any number in
r*0=0
That's the point.
We know that.
But, in r*x=0 (Where 'x' is any real number) then r = 0
So then, r = x/0
Which was my original question and has failed to have been answered.
there are 4 types of functions: real or imaginary, defined or undefined
y = x : real and defined
y = x^2 / (x - 5) : real but undefined at x = 5
y = √x : imaginary for x < 0, but defined
y = √x / (x + 5) : imaginary and undefined at x = -5
I know that.
What I am asking is why there are different answers to the same equation at given steps.
r*24=0
This is only true when r = 0
Yet,
r = 24/0 is undefined. But it is a step in the previous equation.
WHY?
No it isn't, it would be r = 0/24.
:doh:
Brainfart.
That's what I said ten posts ago...Quote:
Originally Posted by DuB
:takethatfoo:
Most of this topic has hinged on the original poster's algebra mistake.
If anyone thinks 24/0 = 0 then prove it. Correctly.
In the meantime, read http://en.wikipedia.org/wiki/Division_by_zero first section, please.
There is no substance to this topic, we shall let it die.