02 Aug '15 16:01>
Why is 0/0 considered indeterminate rather than zero?
Also, what is infinity divided by zero? Or what about infinity divided by infinity?
Also, what is infinity divided by zero? Or what about infinity divided by infinity?
Originally posted by vivifyLet say we have x/x. If we evaluate this we always get =1, right? 4/4=1, 137/137=1, and so on. But what happens when we have 0/0? Does that equal =1 as well?
Why is 0/0 considered indeterminate rather than zero?
Also, what is infinity divided by zero? Or what about infinity divided by infinity?
Originally posted by vivifyConsider the division of two numbers a and b: a/b. This is equal to the number c, such that c*b = a. For example, 6/2 = 3, because 3*2 = 6. But what is the number c, such that c*0 = 0? Any number c multiplied by 0 will do; in other words, there is no such unique number c, and hence the division is not defined.
Why is 0/0 considered indeterminate rather than zero?
Also, what is infinity divided by zero? Or what about infinity divided by infinity?
Originally posted by vivifyRegarding your first question, if the other posters were a bit too technical for you, then just think of it this way: when we say x divided by y, we are asking 'how many times does y go into x'. Now with zero divided by zero, you are asking 'how many times does zero go into zero' and the answer is 'as many as you wish'.
Why is 0/0 considered indeterminate rather than zero?
Also, what is infinity divided by zero? Or what about infinity divided by infinity?
Originally posted by FabianFnasWell zero is a different number than all the rest. Just because every other number divided by itself equals one, why should this rule apply to zero? Every number that's either multiplied or divided by zero is zero; so why not simply say this rule trumps the x/x=1 rule in the case of 0/0?
Let say we have x/x. If we evaluate this we always get =1, right? 4/4=1, 137/137=1, and so on. But what happens when we have 0/0? Does that equal =1 as well?
Further, say that we have 0/x. If we evaluate this we always get =0, right? 0/4=0, 0/137=0, and so on. But what happens when we have 0/0? Does that equal =0 as well?
So 0/0 gives =1 sometimes a nnot ever, no matter what divide by zero. We just don't know the result. It can be whatever.
Originally posted by AThousandYoungWhat if we replaced infinity with pi?
What is the limit of oo/x as x approaches infinity? No answer, because you can't have infinity in an equation - it is not a number.
Originally posted by vivifypi has an infinitely long representation in the base ten number system we use, but it is not infinite. It's a little more than 3.
What if we replaced infinity with pi?
And you're right about not dividing a number by zero. I was always taught that any number divided by zero simply equals zero. After reading your post, I entered 9/0, and the Google calculator returned "infinity". 0/9 equals, but not 9/0. So I learned something reading your post.
Originally posted by twhiteheadI am not sure if you meant that as an rhetorical question but, if not, and for the benefit of other readers who don't know the answer or think the answer is yes; actually, there isn't twice as many integers as positive integers! The ratio of one to the other is undefined.
there are twice as many integers as positive integers surely?
Originally posted by twhiteheadNo, they are equally many. And to show why is easy:
There are an infinite number of positive integers, and an infinite number of integers (positive and negative). Can we say that the total number of integers divided by the number of positive integers is two? If we ignore zero, then there are twice as many integers as positive integers surely?
Originally posted by FabianFnasHow can you have "...is as least the numbers of..." if you are talking about infinities, which are not numbers?
No, they are equally many. And to show why is easy:
There are two groups: (A) All integers and (B) Positive integers.
Let's play a game. Give me an integer from group A and call it a, and I will give you back an integer from group B and call it b. And let's see whose integers run out first.
If you give me 3 I give back 6, if you give med -3 I give you ...[text shortened]... (A), then the conclusion must be that #(A) is = (exactly) #(B). They are exactly as many. Q.E.D.
Originally posted by humyThere are different classes of infinity.
How can you have "...is as least the numbers of..." if you are talking about infinities, which are not numbers?
I don't see how.
I think that is partly why I think your conclusion of "They are exactly as many" is incorrect; as I just explained why in my last post, the proportion of one to the other is undefined and this is proven by proof by contradiction.