4
$\begingroup$

I have trouble understanding something in a book I am reading.
It says for the following:

$c_1\leq \frac{1}{2} - \frac{3}{n}\leq c_2$ where $c_1,c_2$ are positive constants, $n$ is a positive variable.

It says that

We can make the right-hand inequality hold for any value of $n \geq 1$ by choosing any constant $c_2 \geq \frac{1}{2}$.

But if I start to work with the right hand side, I end up with $c_2\lt\frac{1}{2}$.

Does the book have a typo or am I wrong?
I also have different result for the left-hand side inequality but I leave it out.


UPDATE (after comments):

$\begin{align*} \frac{1}{2}-\frac{3}{n}\leq c_2 &\Longleftrightarrow \frac{n}{2}-3 \leq nc_2\\ &\Longleftrightarrow \frac{n}{2}-nc_2 \leq 3\\ &\Longleftrightarrow n\left(\frac{1}{2}-c_2\right)\leq 3\\ &\Longleftrightarrow n\leq \frac{3}{\frac{1}{2}-c_2}\\ \end{align*}$ So it must be $\frac{1}{2}-c_2 \gt 0$, so $c_2 \lt \frac{1}{2}$ (I think)

  • 1
    @user384706: You are showing that *if* $c_2\lt \frac{1}{2}$, then the inequality will only hold if $n\leq \frac{3}{\frac{1}{2}-c_2}$. Pick any $n$ that is larger than $\frac{3}{\frac{1}{2}-c_2}$, and the inequality will not hold for that $n$. As to the second, you **asume** that $c_2\lt \frac{1}{2}$ when you don't change the inequality sign after dividing by $\frac{1}{2}-c_2$. Then you use the resulting inequality to "conclude" that you must have $c_2\lt\frac{1}{2}$. That means that you show that **if** $c_2\lt\frac{1}{2}$, **then** $c_2\lt\frac{1}{2}$.2011-10-03

2 Answers 2

10

You aren't trying to "solve for $c_2$"; there are many values of $c_2$ that will satisfy the inequality for specific values of $n$. However, only values with $c_2\geq \frac{1}{2}$ will satisfy the inequality for all positive values of $n$.

If $n\geq 1$, then $0\lt \frac{3}{n}$; so $\frac{1}{2}-\frac{3}{n}\lt \frac{1}{2}$, no matter what $n$ is. Thus, if $c_2\geq \frac{1}{2}$, then certainly $\frac{1}{2}-\frac{3}{n}\lt \frac{1}{2}\leq c_2$, so the inequality holds.

That's the assertion being made: that if $c_2\geq \frac{1}{2}$, then the inequality will hold for any $n\geq 1$.

7

The book is correct. Since $n$ is positive, $\frac12-\frac3n$ will always be less than $\frac12$. Thus the inequality will hold as long as $c_2\ge\frac12$. If this isn't clear, please write out how you ended up with $c_2\lt\frac12$.

  • 0
    I have update post2011-10-03