Looking for counterexample in inequality proof

Click For Summary

Discussion Overview

The discussion revolves around the validity of proving inequalities by starting with the desired result and working backwards. Participants explore the implications of this approach, seeking counterexamples where it may fail, particularly in the context of teaching inequality proofs.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that starting with the result and working backwards can lead to correct inequalities but may not provide a valid proof of the original statement.
  • One participant provides an example of a flawed backward proof involving the Cauchy-Schwarz inequality, illustrating that while the steps may lead to a true statement, they do not constitute a proof.
  • Another participant emphasizes the importance of reversibility in each step when working backwards, suggesting that equivalence arrows should be used instead of implication arrows.
  • Some participants express that while working backwards can be a useful technique, it is crucial to ensure that the steps taken are logically sound and reversible.
  • There is a mention of the pedagogical aspect, where instructors may allow backward proofs for efficiency in grading, despite concerns about their rigor.

Areas of Agreement / Disagreement

Participants generally agree that while working backwards can be a helpful technique, it is not always valid and can lead to incorrect proofs if not done carefully. Multiple competing views remain regarding the appropriateness and effectiveness of this method in teaching.

Contextual Notes

Limitations include the need for established assumptions and theorems in the course material, which may affect where a proof can begin. The discussion highlights the balance between logical correctness and practical teaching considerations.

cerealkiller
Messages
7
Reaction score
0
Hi guys,

I have to teach inequality proofs and am looking for an opinion on something.

Lets say I have to prove that a2+b2≥2ab. (a very simple example, but I just want to demonstrate the logic behind the proof that I am questioning)

Now the correct response would be to start with the inequality (a - b)2≥0, then progress to:

a2+b2-2ab≥0
∴ a2+b2≥2ab.

What many students do, as it is generally much easier, is to start with the required result and work backwards:

i.e.

a2+b2≥2ab
a2+b2-2ab≥0
∴ (a - b)2≥0, which is true, ∴ the initial result must also be true.

Can anyone provide an example where starting with the result and working toward a true statement will not work? My colleagues would generally discourage this approach but it would be much more convincing to students if I could show them a situation where it won't work.

Thanks for the help guys!
 
Mathematics news on Phys.org
Here's my favorite:

1 = 2~\Rightarrow~1\cdot 0 = 2\cdot 0~\Rightarrow~0 = 0
 
Thanks! Are there any others that you know of? There are some obvious no-nos, multiplying both sides by 0 being one of them, but I was hoping for one that was maybe a little more subtle.
 
cerealkiller said:
a2+b2≥2ab
a2+b2-2ab≥0
∴ (a - b)2≥0, which is true, ∴ the initial result must also be true.
That example is OK because each step is reversible. To illustrate a broken usage you need to introduce an irreversible step.
a2+b2≥2ab+c2
a2+b2-2ab≥c2
(a - b)2≥c2≥0
∴ (a - b)2≥0, which is true, ∴ the initial result must also be true.
Or just
a≥b
a2≥b2≥0
Hence a2≥0, which is true...
 
When working "backwards" it's as haruspex says important that every step is reversible. So instead of implications arrows, you are looking for equivalence arrows, or simply backwards implication arrows. If you're not using only equivalence arrows, you might want to do at least the steps without equivalence arrows of the proof in the right direction.

I would however not discourage this way of proving inequalities, it's extremely helpful to start out with what you want to prove and try to reversibly find an inequality you can easily prove, and not very instructive to look for some random inequality of which your original one should follow from.
 
Last edited:
cerealkiller said:
Can anyone provide an example where starting with the result and working toward a true statement will not work? My colleagues would generally discourage this approach but it would be much more convincing to students if I could show them a situation where it won't work.

Well, it's kind of easy to produce an example where you have a correct inequality and work your way to a correct one, but not providing a proof of the original inequality. Suppose you want to prove that (a^2+b^2)(c^2+d^2) \geq (ac+bd)^2 (Cauchy-schwarz in two variables) then you could try to argue as follows:

(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2 \Rightarrow (a^2+b^2)(c^2+d^2) + (ac+bd)^2 \geq (ac+bd)^2 \Rightarrow (a^2+b^2)(c^2+d^2) \geq 0 which is true, but this is obviously not a proof.
 
cerealkiller said:
Now the correct response would be to start with the inequality (a - b)2≥0, then progress to:

a2+b2-2ab≥0
∴ a2+b2≥2ab.

That is "a" correct response provided the theory (a-b)^2 \ge 0 has already been covered in the course. Since proofs employ assumptions and theorems previously established, where a proof can begin depends on the organization of the course materials.

What many students do, as it is generally much easier, is to start with the required result and work backwards:

i.e.

a2+b2≥2ab
a2+b2-2ab≥0
∴ (a - b)2≥0, which is true, ∴ the initial result must also be true.

Doing mathematics is compromise between what is logically correct and what humans can do efficiently. Working backwards is a valuable technique. Students who work backwards usually don't write words that create a correct proof. Many instructors allow it anyway.. (Such papers are easier to grade than a formal proof! Concerns about human efficiency include concerns about efficient marking of papers.) Many instructors let students "prove" trig identities and write epsilon-delta "proofs" by working backwards.

Can anyone provide an example where starting with the result and working toward a true statement will not work?



In view of what other posters explained about the need for reversibility, try (a^2 + b^2 - 2ab ) \ge -1
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
44
Views
6K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K