1
$\begingroup$

I've heard it claimed that for a system of inequalities such as:
$a<b$
and $ c<d $
Then $a+c<b+d$ is a valid move, but subtraction isn't. And similarly for a system such as $e>f$ and $g<h$ then $e-g>f-h$ is valid and so is $g-e<h-f$, but addition isn't valid.

In short, I've been told that you can only add inequalities when they both have the same inequality sign (I believe the term is "sense") and you can only subtract them when they don't.

However, I have neither seen nor can I derive, a prove that these rules are true.

Does anyone have such a proof?

$\endgroup$
2
$\begingroup$

Given $a\lt b$ and $c\lt d$ then $a+c\lt b+c\lt b+d$

Given $e\gt f$ and $g\lt h$ then $e-g\gt f-g\gt f-h$

$\endgroup$
2
$\begingroup$

Hint: $a<b \iff \exists x>0, \; a+x=b$. Similarly $c<d\iff \exists y>0,\; c+y=d$. Thus we have $a+c+(x+y)=b+d\implies a+c<b+d$ as $x+y>0$.

In similar vein, you can prove all those statements you used.

$\endgroup$
0
$\begingroup$

Do one part at a time. If a< b then a+ c< b+ c. And if c< d then b+ c< b+ d. From both of those, a+ c< b+ d. As for subtraction, you have to use the fact that multiplying both sides of an inequality by -1 reverse the inequality. If a< b then -a> -b. So if a< b, a- c< b- c. And if c< d then c- b< d- b. But to get "b- c" you have to multiply by -1 so b- c> b- d. That does NOT tell where b- d is in relation to a- c.

$\endgroup$
0
$\begingroup$

You'd better never substract (directly) and never manipulate inequalities with different senses. That is too error-prone. It is much safer to stick to these rules:

  • You can add inequalities with the same sense.
  • You can multiply inequalities with the same sense, provided all numbers are positive.
  • If the numbers are negative, you can multiply them, but have to change the sense of the inequality.
  • You can change the signs in an inequality, or multiply both sides by a negative number, but have to change the sense of the inequality.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.