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I have done quite a few exercises, and I have never considered some restrictions, for example:

For m $\neq 0$ , $\dfrac{m+n}{m} - \dfrac{n-m}{m}$

Resolving:

= $\dfrac{m + n - n + m}{m}$ = $\dfrac{2m}{m} = 2$

Well, the answer is correct. I even arrived at the result, regardless of the restriction. What is that restriction telling me? if $m = 0$, the result is erroneous? Or if the result was $0$, is it wrong?

Besides this, do you have practical exercises where to apply these rules is totally fundamental to solve exercises? (Consider that I am a school student yet)

an4s
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ESCM
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    In the context of the question, the condition that $m \neq 0$ is given to prevent division by 0. Even when you resolve the equation to $\dfrac{2m}{m}$, you are still dividing $m$ with $m$ and dividing $0$ by $0$ results in undefined behavior (why? https://math.stackexchange.com/questions/1025690/seeking-elegant-proof-why-0-divided-by-0-does-not-equal-1) – an4s Mar 07 '18 at 23:25

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Dividing any number by zero does not make sense. In this case $m\neq 0$ is necessary for the statement to even have any meaning. In other cases the restrictions may be necessary for certain steps in the resolution.

Edit: one example of this is the following: suppose $xm = 5m$, what is the value of $x$? If $m\neq 0$ you can divide by $m$ and conclude $x=5$, but if $m=0$ not only this step would be wrong but also $x$ can have any value, as $1\times 0 = 5\times 0 = \pi \times 0= \cdots$.

Lazward
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  • So, most restrictions have to do with fractional indeterminacy? – ESCM Mar 07 '18 at 23:26
  • Not really, but in algebraic manipulations they are often common. Another example: "Let $x_0$ be a positive solution of $x^2-a=0$, ..." you need to restrict $a > 0$ for the equation to actually have a positive solution. – Lazward Mar 07 '18 at 23:51
  • Why i "need" to restrict ? i can't work with complex numbers ? – ESCM Mar 08 '18 at 16:46
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    Or is totally needed a real solution ? – ESCM Mar 08 '18 at 16:46