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Crazy Horse 11 said:Something divided by itself always = 1.
Not true.
x^2 - 1 is no different.
Therefore for example when x=1 then 0/0=1:
(x^2 - 1) / (x^2 - 1) = 1. This may be undefined but it is proved true by the fundamental mentioned above.
Prove that 0 / 0 = 1 is not true.
It's essentially a definition. The division "x/y" is defined for all real (and complex numbers) except for y=0. We have left "x/0" undefined by definition.
The reason is simple. What would we define it as?? You could argue that
[tex]\frac{x^2 - 1}{x^2 - 1}=1[/tex]
for all ##x## except ##1## and ##-1##. So you could say that it makes sense to let this equality hold also for ##1## and ##-1##. That way, we could define ##0/0=1##.
The reason we don't do this, is because we could also argue that for ##x\notin \{1,-1\}## that
[tex]\frac{2(x^2-1)}{x^2-1}=2[/tex]
So we want this to hold also for ##1## and ##-1##. But that forces ##0/0=2##.
In the same way, we could make a perfectly fine argument that ##0/0=r## for any number r! So there is no definition of ##0/0## which unambiguously extends the laws of arithmetic. This is why we define ##0/0## to be undefined.