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I was dealing with this problem:

Find $\inf A$ and $\sup A$ if

$$A=\{\frac{2mn+4m+3n+2}{2m-mn+2-n}: m,n\in\Bbb N\}$$

I have to separate this set into 2 sets to make things easier, like this:

$A_1=\{-\frac{2m+3}{m+1}:m\in \Bbb N\}$ and $A_2=\{\frac{8}{2-n}:n\in\Bbb N\}$

So $A=A_1+A_2$. Now I would know how to find $\inf$ and $\sup$, but my question is how did someone come up with $$\frac{2mn+4m+3n+2}{2m-mn+2-n}=-\frac{2m+3}{m+1}+\frac{8}{2-n}$$ I understand that $2m-mn+2-n=(m+1)(2-n)$ but what do you do in the numerator to get something that will cancel out?

3 Answers 3

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$2nm+4m+3n+2=2mn-4m+3n-6+8m+8=-(2m+3)(2-n)+8(m+1)$ Now when u separate the two you get

$$\frac{-(2m+3)(2-n)}{(m+1)(2-n)}+\frac{8(m+1)}{(m+1)(2-n)}$$

from there it's obvious how the last expression is achieved.

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    How did you know you had to expand it like that? – 2017-01-28
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    @Lewis Thing is, the expression in the denominator is easier to spot. Once you do see it, like you did yourself, you try to make one of the two expressions in the numerator as well. E.g. when you try to aim for $m+1$, you obviously need to leave the $2nm+3n$ alone, and try to work something out with others. But generally, learning to to it fast is just a practice-makes-perfect kind of thing in my opinion. – 2017-01-28
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$$\frac{2mn+4m+3n+2}{2m-mn+2-n}=\frac{A}{m+1}+\frac{B}{2-n}$$

$$(2mn+4m+3n+2)=A.(2-n)+B.(m+1)$$

Put m = -1 and calculate A.

Then put n = 2 and calculate B.

You have terms for numerator.

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the denominator can written as $$m(2-n)+2-n=(2-n)(m+1)$$ for the numerator we have $$2m(n+2)+n+2+2n=(n+2)(2m+1)+2n$$ other possibilities for the fraction are $$-{\frac {2\,m+7}{m+1}}+{\frac {-16-4\,m}{ \left( n-2 \right) \left( m +1 \right) }} $$ or $$-2\,{\frac {n}{n-2}}+{\frac {-5\,n-2}{ \left( n-2 \right) \left( m+1 \right) }} $$

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    The OP had this already. The question was the *numerator*! – 2017-01-28