# Subtraction of inequalities

Suppose I have two inequalities: $$k_1 \leq a$$ $$k_2 \leq b$$

where $k_1,k_2,a,b$ are all positive numbers

I know that that summation of them can be written as: $$k_1+k_2 \leq a+b$$

But I want to find $$k_1-k_2 \leq ?$$ or $$? \leq k_1-k_2$$

Is there any way to solve this problem?

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Where did you get stuck? –  draks ... Oct 10 '12 at 19:05
@draks: what should be instead of question mark? –  kotoll Oct 10 '12 at 19:07
What do you think? –  draks ... Oct 10 '12 at 19:07
@draks: If i knew i would not ask –  kotoll Oct 10 '12 at 19:08
The link you gave seems pretty clear to me, what is bothering you? –  Jean-Sébastien Oct 10 '12 at 19:11

First thing you can do is invoke the triangle inequality, i.e. $|x + y| \leq |x| + |y|$. Note that this equality works for negative $x$, $y$ as well. This gives you $$\begin{eqnarray} |k_1 - k_2| &\leq& |k_1| + |k_2| &\leq |a| + |b| \text{ and since }k_1-k_2 \leq |k_1-k_2| \\ k_1 - k_2 &\leq& |k_1| + |k_2| &\leq |a| + |b| \end{eqnarray}$$

For arbitrary $k_1 \leq a$, $k_2 \leq b$, you can't do better than that, but since you stated that $k_1,k_2 \geq 0$, you can. $k_1,k_2$ being positive means that $$\begin{eqnarray} -b &\leq& -k_2 &\leq& k_1 - k_2 &\leq& k_1 &\leq& a\\ -a &\leq& -k_1 &\leq& k_2 - k_1 &\leq& k_2 &\leq& b \end{eqnarray}$$ because subtracting a positive number can only make a number smaller, never bigger.

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Let's look at some example. We know that $$4<8, 2<10,$$ but we don't have $4-2<8-10$. On the other hand, $$3<9,4<8$$ but you cannot say $3-4>9-8$. There is nothing to deduce from substrating inequalities that are of the same side

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No, but you can say that $2 = 4-2 \leq \text{max}\{8,10\} = 10$ –  fgp Oct 10 '12 at 20:03
I could say $6<10, -8<9$ but now $14=6-(-8)\leq\max\{9,10\}=10$ does not hold. I know that $-8-6$ would hold, just saying –  Jean-Sébastien Oct 10 '12 at 20:07
According to the question, $k_1,k_2 \geq 0$. –  fgp Oct 10 '12 at 20:09
oh you're right, but then this becomes a trivial bound :) –  Jean-Sébastien Oct 10 '12 at 20:11
Still better than no bound ;-) –  fgp Oct 10 '12 at 20:17

No. The problem is that if $k_2\le b$, then $-k_2\ge-b$

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For an accurate proof that you cannot say anything about it:

Suppose you would be able to find that $k_1-k_2\leq f(a,b)$ holds for some $f$, and $f(a,b)\in\mathbb{R}$. Then $k_2\geq k_1-f(a,b)$. And because $k_2\leq b$ we have $k_1-f(a,b)\leq b$.

However, we can substitute $k_2=x=k_1-f(a,b)-1$. This value is allowed since $x\leq b$. But $x$ does not satisfy the condition $x\geq k_1-f(a,b)$.

That means that $f(a,b)\not\in\mathbb{R}$, so $f(a,b)=\infty$

It may look like an otiose proof, but i don't think it is.

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$k_1 - k_2 \leq |k_1| + |k_2| \leq |a| + |b| = f(a,b)$ is a counter-example to your "proof". –  fgp Oct 10 '12 at 20:06
Yes it is, but i don't see where the 'proof' goes wrong... –  barto Oct 10 '12 at 20:09
Dunno, but the second part looks pretty confused to me. What is $x$? And into which term are you substituting what value? –  fgp Oct 10 '12 at 20:34
I let $k_2$ be $k_1-f(a,b)-1$. (I just used $x$ as an other name for it.) –  barto Oct 11 '12 at 15:11