- 02 Aug '15 16:38

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?*Originally posted by vivify***Why is 0/0 considered indeterminate rather than zero?**

Also, what is infinity divided by zero? Or what about infinity divided by infinity?

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 and 0/0 gives =0 sometimes? Does this make sense?

This is only one reason that we cannot ever, no matter what divide by zero. We just don't know the result. It can be whatever.

Same goes for infinity. First of all, infinity is not a number. Whatever we add, multiply we get the same infinity. So this is not possible to evaluate anything when infinity is used.

(For those of you who want to argue with me about this, I say, I'm talking about exactly =0 and =oo here. Nothing near enough. Only equals.) - 02 Aug '15 17:35 / 3 editsInfinity divided by zero or divided by infinity can have various answers. I took a class where we did that. You need to use calculus to find the answer.

https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

The first thing to understand is that there is no number called infinity. There are only limits as a variable approaches infinity. For example,

x/x = 1

2x/x = 2

What is the limit of x/x as x approaches infinity? 1

What is the limit of 2x/x as x approaches infinity? 2

What is the limit of x/0 as x approaches infinity? No answer, because you cannot divide by 0.

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.

What is the limit of 0/x as x approaches infinity? 0. - 02 Aug '15 17:56

Consider 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.*Originally posted by vivify***Why is 0/0 considered indeterminate rather than zero?**

Also, what is infinity divided by zero? Or what about infinity divided by infinity?

As Fabian points out, "infinity" is not a number so you can't divide it by anything. You can take certain limits where numbers are taken to go to infinity, in which case the result depends on what you are taking the limit of. Examples:

lim x-> inf x/x = 1

lim x-> inf x^2/x = inf

lim x-> inf x/x^2 = 0

lim x-> inf x/0 = not defined

lim x-> inf 0/x = 0 - 02 Aug '15 18:32

Regarding 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'.*Originally posted by vivify***Why is 0/0 considered indeterminate rather than zero?**

Also, what is infinity divided by zero? Or what about infinity divided by infinity?

For your second question, I would like to instead ask my own:

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? - 02 Aug '15 19:14 / 5 edits

Well 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?*Originally posted by FabianFnas***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.

Simply put, wouldn't it be fair to say that x/x=1 isn't actually a rule as much as it is a repeatable result? For example, can I say that x/2 will equal half of of x is a mathematical "rule"? A ² + B² = C ² is mathematical rule, but can x/x=1 be considered one? Because if it's not a "rule" or mathematical law, then there's no contradiction if we say 0/0=1...right?

**Same goes for infinity. First of all, infinity is not a number. Whatever we add, multiply we get the same infinity. So this is not possible to evaluate anything when infinity is used.**

What about π/π? - 02 Aug '15 19:30

What if we replaced infinity with pi?*Originally posted by AThousandYoung*

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.

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. - 02 Aug '15 20:49

pi 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.*Originally posted by vivify***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. - 02 Aug '15 21:00 / 4 edits

I 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*Originally posted by twhitehead***there are twice as many integers as positive integers surely?***undefined*.

To see why, consider this;

For each negative number, you can arbitrarily group it with two positive numbers like this:

-1 grouped with 1 and 2;

-2 grouped with 3 and 4;

-3 grouped with 5 and 6

...

...and so on for infinitum.

Thus, using the above heavily contrived perspective, there would appear to be a ratio of all integers to all positive integers of 3 to 2, not 2 to 1 which many laypeople would intuitively think! Using an unlimited number of similar arguments, you can apparently show the ratio of one to the other to be anything you like! But each conclusion of such a ratio is contradicted by other conclusion using the same kind of argument thus the only thing we can say about the actual ration is that it is undefined i.e. it doesn't really exist.

In fact, you can use a similar (flawed ) argument to apparently show that there are an infinite number of positive than there are integers!

But, consider this grouping with prime numbers so the number of prime numbers keeps doubling each time you move on to the next group down:

1 grouped with 2;

2 grouped with 3 and 5;

3 grouped with 7 and 11 and 13 and 17;

4 grouped with 19 and 23 and 29 and 31 and 37 and 41 and 43 and 47;

...

...and so on for infinitum.

So now, if you consider the limit of that, we apparently shown there are*infinitely*more prime numbers than there are integers! Of course, this is nonsense because the proportion is not defined because you can contrive a similar argument which give rise to a different proportion which means it contradicts. - 02 Aug '15 21:34

No, they are equally many. And to show why is easy:*Originally posted by twhitehead***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?**

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 back 7. If you give me 100 I give you 200, and if you give me -100 I give you back 101. So every a you give me I give back b = 2a if a is positive, and b=2a+1 if b is negative. Can you find an a to give me that I cannot give anything back because the numbers of group B is out, there are no more left? No, certainly not. The numbers of integers in group B is as least the numbers of integers of group A.

We can play the game in reverse. I give you an integer from group B and you give me a number of group A and we'll see if your integers runs out quicker than my integers. It shows that the numbers integer in group A is as least the numbers of integers of group B.

So if #(A) is >= #(B) and at the same time #(B) is >= #(A), then the conclusion must be that #(A) is = (exactly) #(B). They are exactly as many. Q.E.D. - 02 Aug '15 22:16 / 4 edits

How can you have "...is as least the*Originally posted by FabianFnas***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.*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.

In fact, it can be shown by proof by contradiction that if you have two infinite sets of anything, the proportion/ratio of one to the other is always*undefined*as a definable*finite*number although you can, depending on the nature of the two infinite sets you are comparing, rationally talk about different "*orders*of infinity" where you can sometimes prove there are infinitely more elements in one infinite set than the other. - 02 Aug '15 22:31

There are different classes of infinity.*Originally posted by humy***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.

All the integers, and all the positive integers are both examples of 'countable infinities' and are thus equal and equivalent.

You can have varying orders of uncountable infinities which are larger [in fact infinitely so] but all countable infinities

are equally infinite as each other.

As for the infinity over infinity question...

I like to think of it like this [because physics ]

Density is given by mass over volume. Water for example having a density of ~ 1000kg / 1m3

We can imagine making volume infinite, envisaged by an infinite flat universe, and then make the mass in that universe infinite

and then ask what is the density of that universe?

Well you could have a density of 1 atom per cubic light year, and with infinite volume you still have an infinite mass.

You could equally have a density like that of a neutron star, and you still have the same infinite mass.

So what is the density of an infinite universe with infinite mass... anything you like, [except I think 0 ] so it's undefined.