MHB What happens when dividing by a negative in an inequality?

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Dividing by a negative in an inequality requires flipping the inequality signs, which is a crucial step in solving such expressions. The discussion highlights the example of the inequality -10 < 2x + 3 < -1, which simplifies to -13 < 2x < -4, and ultimately to -13/2 < x < -2. Participants emphasize the importance of recognizing this sign change when rearranging inequalities, particularly when dividing by a negative coefficient. One method is noted as more efficient, but the preference for clarity in presentation is also acknowledged. Understanding the implications of dividing by a negative is essential for accurate inequality manipulation.
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-10 < 2x + 3 < -1
-13 < 2x < -4
-13/2 < x < -2
 
-c < ax + b < -d

Easier to work with (after re-arranging):
c > -(ax + b) > d
 
Wilmer said:
-c < ax + b < -d

Easier to work with (after re-arranging):
c > -(ax + b) > d
But when you get to the step c + b > -ax > d + b you have to divide by -a anyway and the >s flip again. I think it's good in that you get that extra step (to stress the point of what happens when you divide by a negative) but greg1313's method is slightly more efficient.

-Dan
 
topsquark said:
But when you get to the step c + b > -ax > d + b you have to divide by -a anyway and the >s flip again. I think it's good in that you get that extra step (to stress the point of what happens when you divide by a negative) but greg1313's method is slightly more efficient.
-Dan
Ya...agree...BUT li'l ole me prefers ? > ? > ? to ? < ? < ?
A bit like reading left to right instead of backwards!
Anyway, not important...
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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