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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?

  • 0
    No. Insufficient data for a meaningful answer.2012-10-10

7 Answers 7

6

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.

  • 2
    @jjjjjj The question states that "$k_1$, $k_2$, $a$, $b$ are all positive numbers"2017-04-08
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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

  • 0
    Still better than no bound ;-)2012-10-10
1

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.

  • 0
    I let $k_2$ be $k_1-f(a,b)-1$. (I just used $x$ as an other name for it.)2012-10-11
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  • Note that your property of summing inequalities for positive numbers can immediately be extended to signed numbers.

Indeed assuming $\begin{cases} k_1\le a\\k_2\le b\end{cases}$ then we have $\begin{cases} a-k_1\ge 0\\b-k_2\ge 0\end{cases}$ which are positive numbers.

Their sum is then positive $(a-k_1)+(b-k_2)\ge 0\iff (a+b)-(k_1+k_2)\ge 0$

Once rearranged it gives $k_1+k_2\le a+b$ independently of the signs of $a,b,k_1,k_2$


  • Now the inequalities $\begin{cases} k_1\le a\\k_2\le b\end{cases}$ with positive numbers are in fact $\begin{cases} 0\le k_1\le a\\0\le k_2\le b\end{cases}$

We can invert the first one to get $\begin{cases} -a\le -k_1\le 0\\0\le k_2\le b\end{cases}$

Now summing all inequalities like we have seen in first point gives: $-a+0\le -k_1+k_2\le b+0$

Or simply $-a\le k_2-k_1\le b$

-1

First, recall that you can express your problem statement in terms of the set of inequalities

$0 and $0.

This naturally leads naturally to $-b Here I would like to stress that the set of strict inequalities follows from the fact that $k_1>0$ and $-k_2<0$.

-1

You can also use an alternative strategy based on the identity $|k_1-k_2|=\sqrt{(k_1-k_2)^2}.$

First of all, we use the fact that $k_1,k_2>0$, $k_1\leq a$ and $k_1\leq b$ to show that

$(k_1-k_2)^2=k_1^2+k_2^2-2k_1k_2 and hence, the strict inequality $ |k_1-k_2|<\sqrt{a^2+b^2}.$

In comparison with the inequality that I've derived previously ($-b), this approach gives rise to a rough estimation for the upper and lower bounds of $k_1-k_2$.

-2

Finding the inequality with k(1) - K(2) as the subject

The simplest solution is usually the best solution----Albert Einstein

Given:

k(1) ≤ a

k(2) ≤ b

Required: To determine if k(1) - K(2) can be made the subject of an inequality

       derived from the above inequalities  

Possible solution:

One will operate on the given system of inequalities.

Generally, subtracting c < d from a < b yields a - d < b - c; but below, one

will apply the permissible operations on a system of inequalities. To add a

system of two inequalities, both inequalities must have the same sense (or

direction). To subtract an inequality, multiply this "subtrahend" inequality by

-1 while reversing the sense of the inequality, and then add to the "minuend

inequality", making sure the two inequalities have the same sense.

Now, one will subtract K(2) ≤ b from k(1) ≤ a).

Step 1: Multiply the "subtrahend" inequality, K(2) ≤ b by -1 and reverse the

     sense to obtain - K(2) ≥ - b......(1) 

Step 2: Rewrite (1) so that it has the same sense as the "minuend" inequality,

    k(1) ≤ a. Then one obtains -b ≤ -k(2) 

Step 3 Add the left sides and add the right sides below

   k(1) ≤ a      -b ≤ -K(2)  --------------- 

k(1) - b ≤ a - k(2) <-----the "difference" inequality,

From the above result, k(1) - k(2) cannot be made the subject of the above

inequality.

Conclusion:

Therefore, explicitly, k(1) - k(2) does not exist.

However, implicitly, k(1) - k(2) ≤ a + b - 2k(2)

Repeating the above procedure, using a, b, c, d

Given

a ≤ b...........(1).

c ≤ d............(2);

determine if a - c can be made the subject of the inequality derived from the above system of inequalities.

Solution

From (2) -c ≥ -d (multiplying by -1 in order to subtract)

     -d ≤ -c....(3) (rewriting in the same sense as (1))  

Adding the left sides of (1) and (3); and adding the right sides of (1) and (3),

one obtains

a - d ≤ b - c...........(4)

Conclusion

Explicitly, in (4), a - c cannot be made the subject of (4),

However, implicitly, a - c ≤ b + d - 2c

  • 0
    [mathjax references](https://math.meta.stackexchange.com/q/5020/306553). Also, it would be great to answer in terms of algebra for generality.2018-07-22