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From the table of values of $\frac{2n}{n-4}$ it seems to me that the expression never evaluates to an integer for all natural numbers $n$ greater than $12$. How should I go about proving it?

Anonymous
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3 Answers3

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Division algorithm: write $$\frac{2n}{n-4} = 2 + \frac{8}{n-4}.$$ The fraction is between $0$ and $1$ strictly for large $n > 12$, as needed.

Evan Chen
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$\frac {2n}{n-4} = \frac {2n-8}{n-4} + \frac 8{n-4} = 2 + \frac 8{n-4}$. But if $n > 12$ then $n -4 > 8$ and $\frac 8{n-4}$ is not an integer.

.... or....

For sufficiently large $n$ (i.e. if $n > 4$) we have $\frac {2n}{n-4} > \frac {2n}n = 2$.

But if $n > 12$ then $4 < \frac 13n$ and $n-4 > (n - \frac 13 n)=\frac 23n$ and $\frac {2n}{n-4} < \frac {2n}{\frac 23 n}=3$ so

So if $n > 12$ then $2<\frac {2n}{n-4}< 3$.

...... OR....

If $n = 12$ we have $\frac {2n}{n-4} = \frac {24}8 = 3$. $\lim_{n\to \infty} \frac {2n}{n-4} = 2$ and .... well, I was going to say as $\frac {2n}{n-4}$ is clearly decreasing ... but showing that $\frac {2(n+1)}{(n+1)-4} < \frac {2n}{n-4}$ is not as obvious as my blase statement warrants.

... but....

Note: for $n > 4$ then $2 < \frac {2n}{n-4} < 3 \iff 2(n-4) < 2n < 3(n-4)$ and if you expand that out you get $2n -8 < 2n < 3n -12\implies -8< 0 < n-12 \implies 4< 12 < n$.

fleablood
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Suppose $(n-4)$ divides $2n$. Then $k(n-4)=2n$ for some positive integer $k$. Then $n=\frac{4k}{k-2}$. This makes no sense for $k=1,2$. At $k=3$ it gives $n=12$. Then as $k$ increases, $n$ decreases (tending to $4$). So this can only happen for $n\leq12$.

So "$(n-4)$ divides $2n$" implies $n\leq12$. The contrapositive is: If $n>12$, then $(n-4)$ does not divide $2n$.

2'5 9'2
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