# If I have the conditions $c_1<c_2<a$ and $2c_2>a+c_1$, do I have a contradiction?

The conditions $$c_1 and $$2c_2>a+c_1$$ are equivalent to

\begin{cases} \begin{align*} 2c_2&>a+c_1\\ c_2&>c_1\\ a&>c_2. \end{align*} \end{cases}

Subtracting the second inequality from the first and then subtracting 2 times the third one from the first one I get

\begin{cases} \begin{align*} 0&>0\\ c_2&>c_1\\ a&>c_2, \end{align*} \end{cases}

which is a contradiction, right? Or do inequalities not work this way? In my teacher's suggested solution to the problem that this relates to, he seems to think the inequalities do not contradict each other.

You cannot subtract inequalities like that. For example, $$1>0$$ is a true inequality, and $$2>0$$ is also a true inequality, but if I subtract the two, I get $$-1>0$$, which is clearly false.

What you can do wit inequalities is you can add them, i.e. if $$a>b$$ and $$c>d$$ then $$a+c>b+d$$. You can multiply the inequalities by a positive number, for so if $$a>b$$ and $$x>0$$, then $$ax>bx$$. Multiplying by a negative number inverts the inequality sign.

You also have transitivity in inequalities, i.e. if $$a>b$$ and $$b>c$$, then $$a>c$$.

If you have a particular set of inequalities, in general, it is not trivial to see whether the inequalities have any solution, however, if the expressions in the inequalities are all linear, then you could get some help from linear programming

Try $$c_1 = 0, c_2 = 2, a = 3$$.

You can't subtract inequalities like that.

• Ok, if I am asked to find a contradiction in inequalities, how do I go about it? – Chisq Oct 24 '18 at 13:46

You do not have a contradiction. Let $$c_1=1$$, $$c_2=3$$, and $$a=4$$. Then all three inequalities hold.

The first inequality implies that $$c_2$$ is greater than the average of $$c_1$$ and $$a$$. So, pick any $$a>c_1$$ and then choose $$c_2$$ between the average and $$a$$. Any example constructed in this way will satisfy the inequalities.

$$c_1

Add $$a$$:

$$c_1+a

This is fine.

Note : $$2c_2 , so you are 'replacing' $$c_2 +a$$ by something smaller. May not work.

Try $$c_1=1/4$$, $$c_2=1/2$$, $$a=1$$.

Then : $$a+c_1 =5/4$$; $$2c_2=1$$.

Now choose a bigger, and leave c_1,c_2 as above.