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September 16th, 2015, 04:52 AM   #11
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Quote:
 If it's "indeterminate", it's undefined.
If they are both defined to be the same, how do you determine which one to use?

What is your view on this question

x + y = 15. (x, y in R)

Determine x, y? September 16th, 2015, 05:06 AM #12 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra They aren't the same. Indeterminate is a subset of undefined. Indeterminate only makes sense in the context of limits. In asking the value of $0 \over 0$, the OP is talking about a numerical operation which is undefined (because it involves division by zero). However, it is a ratio to which we can sometimes meaningfully assign a value based on the context. Because the value differs with the context, that value is indeterminate. September 16th, 2015, 08:36 AM   #13
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Quote:
 Because the value differs with the context, that value is indeterminate.
Yes I agree, that iswhat we have already said.

But context in a mathematical sense is more information, boundary conditions, equations whatever.

So consider my previous example

x + y =15 (x,y in R)

Find x and y

Both x and y are clearly defined, but they are undetermined or indeterminate; there are an infinite numebr of possible pairs of x and y that fit the information and definition supplied.

But if I also add x =10 or x = 2y or in the 0/0 format x/y = 2

Then x and y can be determined uniquely.

But they are never undefined.

Let us look further at the expression 0/0

consider Expression = E = {(x-3) *(x+5)} / (x-3), when x = 3.

This leads to E = (0/0) * (3+5) = (0/0) * 8

In this instance many define 0/0 as = 1 so E = 8

But this situation does not work with (8*3) / 0.

The problem is the division by 0, not the ratio of 0 to 0.

That is because, unlike 0/0, X/0 leads to a result that is not in R

Its difficult here because if I use mathml on the other computer I can't see the result.
and I can't use it here.

Last edited by studiot; September 16th, 2015 at 08:38 AM. September 16th, 2015, 09:07 AM   #14
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Quote:
 Originally Posted by studiot In this instance many define 0/0 as = 1 so E = 8
I've never seen anyone do that, and they'd be wrong if they did. The function you give is undefined at $x=3$. But it's a removable discontinuity which simply means that the two-sided limit as $x \to 3$ exists and is finite.

The only reason you get ${0 \over 0} = 1$ in there is because away from $x=3$ you have ${x -3 \over x-3} = 1$.

Last edited by skipjack; September 16th, 2015 at 03:44 PM. September 16th, 2015, 09:25 AM #15 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra $x+y=15$ defines an infinite number of ordered pairs $(x,y)$. In the $x,y$-plane the pairs are points on a line. September 16th, 2015, 10:18 AM #16 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 The fact remains that there is no additional information you can supply that will lead to a result (in R) of division by zero (because it is not defined) This is different from 0/0 where additional information can lead to a result. This is why define and determine are different words, not subsets of one or the other. September 16th, 2015, 02:12 PM   #17
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Quote:
 Originally Posted by studiot not subsets of one or the other.
I agree with everything else you wrote apart from the above. September 16th, 2015, 07:45 PM   #18
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 Originally Posted by ABRAR ?
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