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?
$\endgroup$ 87 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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
$\endgroup$ 5 $\begingroup$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.
$\endgroup$ 4 $\begingroup$- 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 $$-a\le k_2-k_1\le b$$
$\endgroup$ $\begingroup$First, recall that you can express your problem statement in terms of the set of inequalities
$0<k_1 \leq a$ and $0<k_2\leq b$.
This naturally leads naturally to $$-b<k_1-k_2 <a.$$ Here I would like to stress that the set of strict inequalities follows from the fact that $k_1>0$ and $-k_2<0$.
$\endgroup$ $\begingroup$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
$$(k_1-k_2)^2=k_1^2+k_2^2-2k_1k_2<a^2+b^2,$$ and hence, the strict inequality $$ |k_1-k_2|<\sqrt{a^2+b^2}.$$
In comparison with the inequality that I've derived previously ($-b<k_1-k_2<a$), this approach gives rise to a rough estimation for the upper and lower bounds of $k_1-k_2$.
$\endgroup$ $\begingroup$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
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