I started arguing this in high school, and perhaps I am being stubborn and denying the rebuttals and/or maybe I am just searching for someone who agrees with me. So far the only person to agree with me on this is my best friend.
From what I am taught any number divided by zero is undefined.
x÷0=infinity
And any number divided by itself is 1.
x÷x=1
So what about zero? I approached it in terms of quantity:
If I have 3 marbles in my left hand, and I move 3 of those marbles to my right hand, I moved 1 whole of the marbles (3÷3=1). Likewise, If I have zero marbles in my left hand, and I move zero marbles to my right hand, I moved 1 whole of the marbles (0÷0=1).
Also look at it this way: 0÷0 = 0 × (1/0) = zeroes cancel = 1!
If that’s not enough find :
lim sin(x)
[sup]x-›0 X[/sup]
By direct substitution you get sin(0)÷0 = 0÷0 but it = 1!
I think you’re looking at it wrong. 3/3 means that you’re dividing three marbles into three equal groups: how many are in each group? 1, obviously
0/0 means you’re dividing 0 marbles into 0 equal groups. How many are in each group? Problem is, since there AREN’T any groups, there’s no way to tell how many are in each group.
Assume 0/0 = 1.
Then multiply both sides by x.
Then x*(0/0) = x
So (0*x)/0 = x
So 0/0 = x
But 0/0 = 1
So x =1. The argument applies for any x you choose, so x = 1, for any x.
With your moving marbles example, yes you do move 1 ‘whole’ lot of you marbles. But also you move 2 * the whole (as the whole is 0), and three times the whole, and so on, so you get a similar result as above.
With the 0/0 = 0*(1/0) example, you have assumed that 1/0 has some meaning, which is doesn’t, it is undefined, as matt_mcl points out.
The final example also gives an ambiguity. If you take 2Sin(x) rather than Sin(x), you get the limit to be 2 (assuming your original calculation is correct), but 2Sin(0)/0 = 0/0. as before. The problem is you are assuming that sin(x)/x is continuous for all x.
In the cases where your ‘proofs’ make sense, they don’t lead to just one value of 0/0, which is rather difficult mathematically, as division is supposed to be ‘well defined’ so it gives only one answer.
matt_mcl, I was using a graphical description. When you put y=x÷0 into a calculator, you get a vertical line.
DanielWithrow, thanks for the clarification.
One rebuttal I have heard is “Type it into a calculator and see what you get.” Well, of coarse I am going to get undefined because the programming of the calculator check division entries to see if entry1 is divided by zero. If entry1 IS divided by zero it throws the error ‘undefined’! So the calculator flaw can be blamed on the programmer.
If you think of division as the inverse of multiplication, then for example, 15 ÷ 5 = 3 because you have to multiply 5 by 3 to get 15. Then 0 ÷ 0 could be anything, because any number times 0 = 0. (So, in terms of the OP’s marbles example, yeah, you moved 1 “whole” of the marbles, but you also moved 2 wholes of them, and also 3 wholes, and also one half of them, and so on and so on.)
If you want to talk about limits, in calculus, 0/0 is an example of what’s called an indeterminate form, meaning that just knowing that the numerator and denominator each approach zero doesn’t determine what, if anything, the limit of the whole thing is. Pick your favorite number N. When x = 0, N sin x = 0, and x = 0, but the limit of (N sin x) ÷ x (as x approaches 0) is N.
Yes, this is how it works. 0/0 is defined only in the sense of a limit, i.e., we can look at the limit of the ratio of two continuous functions f(x)/g(x) as x->0 where f(0) = 0 and g(0) = 0. And, it can indeed be any result depending on the functions; it doesn’t even have to be finite. For example, while sin(x)/x -> 1 as x -> 0, sin(x)/x^2 -> infinity as x -> 0 (actually, it goes to +infinity when you take the limit from above, usually denoted something like x -> 0+ and it goes to -infinity when you take the limit from below (x -> 0-). And the limit of sin(x)/sqrt(x) as x -> 0 is 0.
0 isn’t a number, is it? It’s a symbol or a place holder. Heck, it wasn’t even around when the number system was first invented. It was invented later. It can’t be treated the same as a number in terms of division.
Now take the limit of 1/x as x -> 0. Then take the limit of x[sup]2[/sup]/x as x -> 0. Which one is right?
0/0 is indeterminate. In order that 0/0 = a, it must be the case that 0a = 0. That works for any real number a.
a/b is defined to be the unique number c such that bc = a. If c is not unique, a/b is indeterminate. If c doesn’t exist (as in the case of 1/0), a/b is undefined.
If you take 0 marbles and divide it amongst 0 people, how the hell do you get ONE marble for each person who wasn’t there to get their share?
Therefore, 0/0 = Does a bear shit in the woods? I don’t know, ask the friggin’ bear! (Hint: Undefined!)
Brian - Your assumption is wrong, you wouldn’t have 10 marbles, because you were not in line to get your share! Otherwise, 10 divided by 1 (you Brian, being the one in line) is 10.
How many nickels can you get for a dollar? Well, 100 / 5, right? That makes twenty.
How many nickels can you get for a quarter? 25 / 5, or 5. Right?
Now, how many nickels can you get for NOTHING? The answer is 0 / 5. In other words, zero.
So far, this all makes sense, right? Now, one last question:
How much nothing can you get for a dollar?
You see the dilemma? “How much nothing” is an absurdity, a meaningless concept. But when you try to divide by zero, that’s what you get! You can get an INFINITE amount of nothing for any amount!
No; if you divide(share out) 10 marbles equally into zero groups, how does it work out for you?
If you have no marbles and you divide them out in to no groups, how many marbles are in each group? - the answer is not zero; it is undefined bcause you don’t have any groups to count
If you thought that was fun, try null sometime; it’s a strange beastie.
