Undefined And Indeterminate

Ajay Mohan · 2015-01-19 20:41:47

Hello people!

I have recently seen a video on khan academy about why we can't define a number to any number divided by 0 ( except 0 ) and why 0/0 is indeterminate.

Video Link!

I have found something wrong with this video. I posted a query but never got a reply.

Problem I :
At 1:00 , he writes an identity which is x*y/y = x.
The problem is that he hasn't specified the values which 'y' can take.

Case I : If y can be any real number including zero , then he indirectly implies that 0/0 is always one , which
contradicts when he says 0/0 is indeterminate.

Case II : If y can be any real number except zero , he cannot use this identity at 3:00 where he uses this identity for
y = 0.

After being confused , I came to MIF website and while explaining why x/0 is undefined while x not equal to zero , they consider the case when x=0 and say it is undefined.

-----> Is it possible to prove that x/0 is undefined when x is not equal to zero without bring 0/0?

Problem II :

Even in the last part of video , when he is trying to prove that 0/0 is indeterminate , he transfers the zero from denominator in LHS to numerator in RHS which would work on normal numbers. A brief working tells that we have to assume 0/0 is one in the process.

Process : 0/0 = k
0*(0/0) = k*0
0*1 = k*0
0 = k*0

----->Also , Can someone say if my idea regarding why 0/0 is indeterminate?

Let 0/0 = k . Now , we can multiply both sides by any real number including zero.
(0*x)/0=0/0. But the RHS has changed which is k*x. Therefore , it can have many values and it is indeterminate.

Last edited by Ajay Mohan (2015-01-20 00:36:04)

Bob · 2015-01-19 22:54:42

hi Ajay Mohan

If a times b = c, then you can define division as c/b = a.

But a problem occurs when b = 0.

c/0 = a requires that a times 0 = c and this has no solutions in real numbers. One way to avoid this problem is to say that division by 0 is excluded from the definition.

And as a times 0 = 0 and d times 0 = 0, this appears to suggest that 0/0 = a = d, where a and d are not necessarily the same. So 0/0 is declared to be indeterminate.

That's how I think of it.

Bob

Ajay Mohan · 2015-01-20 00:28:55

Hello Bob!

The step ( given below ) which you have done from multiplication to division requires that you divide both sides by 'b' , right?

bob bundy wrote:

If a times b = c, then you can define division as c/b = a.

This is the process I posted in my question and for this to be valid , b/b has to be 1.
The process I am talking about :

Ajay Mohan wrote:

Process : 0/0 = k
0*(0/0) = k*0
0*1 = k*0
0 = k*0

The problem is by doing this process we assume that 0/0=1 and then prove that 0/0 is indeterminate. So , isn't it wrong?

Thanks for your reply and please forgive me if I am annoying.

Bob · 2015-01-20 01:51:44

hi Ajay Mohan

I don't think the Khan Academy guy is assuming that y/y = 1. He is saying that division is the opposite process to multiplication and then exploring a 'what if' to show that special rules are required for division by zero. The trouble is, we all know that his assumption is going to lead to a false conclusion; when you assume something is true that is actually false all sorts of strange results can happen. That's why mathematicians shy away from allowing division by 0. The problem arises because the additive identity (zero) causes all multiplications by it to be zero. It's a property of the algebra of 'field' theory and you cannot escape it.

In the end it comes down to this: in any mathematical theory you can make up any axioms you like. They don't have to make sense. But they do have to be consistent. You cannot be able to prove, with your axioms, that 23 = 47 for example. If you allow division by zero that is easy to prove:

23 x 0 = 47 x 0 (= 0)

divide by zero

23 = 47

Whoops.

I'm confused about whether you are then offering an improved proof. If you are, please post just that again and I'll give it some thought.

Bob

Ajay Mohan · 2015-01-20 02:33:00

Hello Bob!

bob bundy wrote:

I don't think the Khan Academy guy is assuming that y/y = 1.

If he didn't assume y/y = 1 then how does he get x * (y/y) = x ?

bob bundy wrote:

I'm confused about whether you are then offering an improved proof. If you are, please post just that again and I'll give it some thought.

The improved proof I was offering was this ( The proof which I am giving is to prove that 0/0 is indeterminate ):

Let 0/0 = k. (k is any real number)

Now , we will multiply both sides by any real number x ( which can also be zero )

LHS = (0*x)/0 = 0/0. But RHS = k*x = any other real number. Therefore 0/0 is indeterminate , i.e ., it can be any
real number.

I am now suddenly getting a doubt.

Shouldn't 0/0 stay undefined?

If 0/0 = a (any real number). Also 0/0 = b (any other real number) . Therefore a = b , but a =/= b!

It is violating the equality sign!

Please clarify me regarding this. I think it is better for 0/0 to stay undefined to remove these confusions!
Thanks for your reply!

By the way I liked your argument/proof on why x/0 ( x not equal to zero ) should stay undefined. I am now satisfied
on why x/0 should stay undefined.

Bob · 2015-01-20 04:22:35

hi

Well I don't mind that proof, but I doubt it would pass the number theory rigorousness test.

Here's what may be wrong with it:

(0/0)= k => (0/0).x = kx ok so far.

But now you want (0/0).x = (0.x)/0 Can you switch about the brackets like this?

Let's continue.

So (0.x)/0 = kx => 0/0 = kx => kx = k => x = 1 which is a contradiction.

So one of your steps / assumptions is wrong. Maybe it is that 0/0 is indeterminate. I'd accept that.

Have you done any calculus yet ?

The differential calculus is founded in the idea that we can, for certain functions, find a value for 0/0 .

So a lot of mathematicians are going to be very unhappy if 0/0 is undefined.

Bob

Ajay Mohan · 2015-01-20 06:04:26

Hello Bob!

bob bundy wrote:

But now you want (0/0).x = (0.x)/0 Can you switch about the brackets like this?

Yes! Right?

bob bundy wrote:

So (0.x)/0 = kx => 0/0 = kx => kx = k => x = 1 which is a contradiction.
So one of your steps / assumptions is wrong. Maybe it is that 0/0 is indeterminate. I'd accept that.

Okay , Let x be any real number other than 1.

bob bundy wrote:

Have you done any calculus yet ?
The differential calculus is founded in the idea that we can, for certain functions, find a value for 0/0 .
So a lot of mathematicians are going to be very unhappy if 0/0 is undefined.

Yes , I have studied calculus. But I have got a bad teacher ( I think so ). He says that (->0)/(->0) is indeterminate
( Shouldn't it be 1 always? ).

Bob · 2015-01-20 06:33:11

He says that (->0)/(->0) is indeterminate. Shouldn't it be 1 always? .

I would say no to both. Look here:

http://www.mathsisfun.com/calculus/slop … point.html

Bob

Math Is Fun Forum

#1 2015-01-19 20:41:47

Undefined And Indeterminate

#2 2015-01-19 22:54:42

Re: Undefined And Indeterminate

#3 2015-01-20 00:28:55

Re: Undefined And Indeterminate

#4 2015-01-20 01:51:44

Re: Undefined And Indeterminate

#5 2015-01-20 02:33:00

Re: Undefined And Indeterminate

#6 2015-01-20 04:22:35

Re: Undefined And Indeterminate

#7 2015-01-20 06:04:26

Re: Undefined And Indeterminate

#8 2015-01-20 06:33:11

Re: Undefined And Indeterminate

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