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Out of curiosity, I typed the equation y = 2 + 0(1/(x-2)) into wolfram alpha to find that it simplified it to y = 2. Is this correct? I assumed there would be a hole or some lack of continuity at x = 2 since I didn't think 0 * undefined = 0. Is wolfram correct?

Henry
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    Yes, there is a hole, because $0 \cdot \frac{1}{0}$ is undefined. – MathematicsStudent1122 May 30 '16 at 16:43
  • Hmmm... 0undefined isn't defined but I think 0f(x) = 0 where f(x) is occasionally undefined is legitimate. I think wolfram is correct. It's debatable. But it ultimately has to do with precedence of order of operations. I see the argument that 1/x-2 is undefined at x =2 so you can't do a blanket 0*y = 0 except... can't we? We're rather stuck in an infinite regress if we must consider all potential functions at all potential values in every statement. After all "1 is the additive identity" means 1 = f(x)/f(x) for all f(x) doesn't it? Hence 1 is always undefined? – fleablood May 30 '16 at 16:53
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    On the other hand, there are formalisms specifically made for cases where one doesn't want to care about such single missing points -- for example, in the field of rational functions in $x$ and $y$, $2+0(\frac{1}{x-2})$ does indeed equal $2$. And there's no way for WA to know you're not working in such a context. – hmakholm left over Monica May 30 '16 at 16:53

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