0

Consider $$y= \frac{(x^2-9)}{x-3}$$ For y to be meaningful x must not equal 3 , So, 3 is not included within the domain of the function y (Call it f(x) if you will)

It follows that, $$yx-3y = x^2 -9 $$ Plug in 3 to x out of curiosity, I find, $$ 0=0 $$ A perfectly valid relation! How is this possible?

This problem is even more predominant in partial fraction decomposition. Have a look at this example I found on the web, Partial fraction decomposition x cannot equal -1 or 2. How is it giving us the right answers then? What is the math behind this? When I do these kind of problems it feels like something's is happening mathematically, hidden from my view

All help is appreciated

SNB
  • 31
  • Still new here. Mathjax takes such a long time ;) – SNB Mar 04 '17 at 15:34
  • This is not the same question as the duplicate and the answer to what I believe you are asking is not there. What you appear to be asking is how can you get a true statement by plugging in numbers that you are not allowed to, that should be non-sensical. Consider the statement: if France is in England then France is in Europe. Just because the first statement is false, it does not follow my consequent is false. It is the same here, just because 3 is not a part of the domain does not imply true statements cannot arise. There is a removable singularity at this point – danwalkerdev Mar 21 '17 at 13:29
  • @Dan90 My question exactly, Thanks. What do we call such relations? ( where if one statement is false, the statements arising from it are not necessarily false) is there a specific name, so I may search it? – SNB Mar 26 '17 at 16:48
  • @Dan90 Can you turn your comment into an answer? If so, I will choose it. – SNB Mar 26 '17 at 16:51
  • Unfortunately I can't because the question is marked as duplicate. I have flagged it for attention but nothing has happened yet. The relation is simply based on the nature of implication ($\implies$). As in my example, I can form a statement where the precedent is false but it doesn't mean the consequent is false. For your example, if you plot the graphs you'll notice they are the same everywhere except at 0. But you can specify a value at 0 which makes it continuous on $\mathbb{R}$, which is kind of why the substitution gives you the right answer – danwalkerdev Apr 01 '17 at 10:32

3 Answers3

0

we have $$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3$$ only for $$x\ne 3$$

Dr. Sonnhard Graubner
  • 95,283
  • Yes, but consider the second statement above yx -3y = x^2 -9. I know x cannot equal 3, but if you plug it in just to see what happens, 0= 0 which is perfectly valid, How is this possible? Pls read my question again – SNB Mar 04 '17 at 15:45
0

By definition, we have

$$y=\frac{x^2-9}{x-3}$$

and at $x=3$,

$$y=\frac00=\text{undefined}$$

If your definition had instead been

$$y(x-3)=x^2-9$$

which is not the same as before, we would then note that at $x=3$,

$$y(0)=0\implies y=\text{anything}$$

which is completely different.

Simply Beautiful Art
  • 74,685
0

When you multiply the right hand side by $x-3$, you get: $$\frac{x^2-9}{x-3}\cdot (x-3) \neq x^2-9$$

since those two functions are not equal for ALL values of $x$ (in particular $x=3$).

What we can say is that for all $x \neq 3$, the following equality holds, $$\frac{x^2-9}{x-3}\cdot (x-3) = x^2-9.$$

So it is only with the stipulation $x \neq 3$ that we can make the simplication necessary to continue the problem and that is a stipulation you need to carry out throughout the entire problem.

benguin
  • 3,846