|February 23rd, 2017, 04:39 PM||#1|
Joined: Feb 2017
Assume $\displaystyle a,b,c,d$ are positive and $\displaystyle a>b$ as well as
$\displaystyle c<d $
Can we add the the two the following way?
$\displaystyle a>b$ --- (1)
$\displaystyle d>c$ --- (2)
so $\displaystyle a+d>b+c$.
My teacher said it is wrong, but i am not able to find a counterexample.
Appreciate any help.
|February 23rd, 2017, 04:48 PM||#2|
Joined: Jul 2011
I hate to say a teacher is wrong, but in this case ... I agree with you.
You sure about the starting inequalities? Is there a "problem" to do that you haven't stated completely?
|February 23rd, 2017, 04:54 PM||#3|
Joined: Feb 2017
Thank you for your answer skeeter. Yes, the starting inequalities are correct, it is just a question I had in general, not a specific problem.
|February 26th, 2017, 02:43 AM||#4|
Joined: Apr 2014
So, we travel in ℝ₊
a > b ⇒ a = b + k
d > c ⇒ d = c + m
a + d = b+k+c+m ⇒ a + d = (b+c) +(k + m) ⇒ a + d > b + c
|March 23rd, 2017, 10:05 AM||#5|
Joined: Mar 2017
Maybe your teacher say that: it is not equivalent!
a>b AND d>c so a+d>b+c,
but the counter case is not true!
For example: 2+7>3+4, with a=2; b=3; d=7 and c=4.
|high school math, inequalities, inequality|
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