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September 16th, 2015, 04:52 AM  #11  
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271  Quote:
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  
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271  Quote:
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 = {(x3) *(x+5)} / (x3), 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 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra  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 twosided 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 ${x 3 \over x3} = 1$. 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 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra  
September 16th, 2015, 07:45 PM  #18 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  