Mathematics for Teaching Using cognitive conflict to teach solving inequalities

## Using cognitive conflict to teach solving inequalities

One way to teach and assess students understanding of math concepts and procedures is to create a cognitive conflict. Here is one way you can create cognitive conflict in solving inequalities:

To solve the inequality x – 7 > 5, the process usually involve adding 7 to both sides of the inequality.

This process uses the principle a > b then a + c > c. There is no change in the inequality sign since the same number is added to both side.

Now, what if we add 7 to the left side of the inequality and 6 to the right side?

The process uses this principle: If a > b, cd then a + c > d. Should this create a change in the inequality sign? Certainly not. There should be no change in the inequality sign when a bigger (smaller) number is added to the bigger (smaller) number side.  Both of these processes create a cognitive conflict and will be a good opportunity for your class to discuss what solving inequality means and, compare the processes of solving equations and inequalities. Comparing and contrasting procedures is also a good strategy to developing conceptual understanding.

For those interested to learn more about inequalities I recommend this book:Introduction to Inequalities (New Mathematical Library)

## 3 thoughts on “Using cognitive conflict to teach solving inequalities”

1. Interesting idea, Erlina.
I’d recommend that elementary teachers start students along this track by having extended discussions of inequalities like your example, which older elementary students should be able to cope with.
My only concern would be that teachers can find the time for such productive discussions, given the crowded math curriculum and increasing attention being paid to testing as a way to measure teaching quality.

2. Jorgen Harmse says:

I like the idea of using cognitive conflict as a step to better understanding. In proof-oriented classes I like to give a few arguments for Orwell’s Lemma (2+2=5) and similar statements because knowing how to prove things includes being able to stare at some impressive-looking pseudo-mathematics and spot the errors. Perhaps I should have framed my question about ‘unlearning’ (over-generalisation of properties of positive integer arithmetic) in these terms.

I agree with Robert’s remark about reversible vs. irreversible processes. I’m not sure what he means by sidestepping. (I couldn’t see the URL on his links.) In solving equations, it is good practice to substitute the answer back into the original equation, which eliminates ‘extraneous’ solutions. In solving inequalities I prefer (as Robert seems to) to make each step obviously reversible. If this is impractical then the best procedure is to try to prove the reverse implication. The next best approach (and perhaps a step to understanding the issue) is to test borderline cases. For example, let x = 11+10^(-1000). Then x>11 but x-7 = 4+10^(-1000) < 5. Setting x=11 and noticing that x-7<5 should also be a clue (but may feed the pre-calculus misconception that all functions are continuous).

3. Robert Morewood says:

This might help address a major deficit in the education of many students regarding reverisible versus irreversible processes. There is a difference between:

x – 7 > 5 implies x > 11

and

x – 7 > 5 if and only if x > 12

A few commonly used procedures in equation solving, especially squaring both sides, are not reversible, leading to “extraneous” solutions. The “allowed” procedures for proving identities in the jurisdictions with which I am familiar amount to an attempt to sidestep this issue without addressing it.