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Looking for counterexample in inequality proof

  1. Aug 9, 2014 #1
    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.

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


    ∴ (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!
  2. jcsd
  3. Aug 9, 2014 #2
    Here's my favorite:

    [tex]1 = 2~\Rightarrow~1\cdot 0 = 2\cdot 0~\Rightarrow~0 = 0[/tex]
  4. Aug 9, 2014 #3
    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.
  5. Aug 10, 2014 #4


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    That example is OK because each step is reversible. To illustrate a broken usage you need to introduce an irreversible step.
    (a - b)2≥c2≥0
    ∴ (a - b)2≥0, which is true, ∴ the initial result must also be true.
    Or just
    Hence a2≥0, which is true...
  6. Aug 10, 2014 #5


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    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: Aug 10, 2014
  7. Aug 10, 2014 #6


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    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 [itex](a^2+b^2)(c^2+d^2) \geq (ac+bd)^2[/itex] (Cauchy-schwarz in two variables) then you could try to argue as follows:

    [tex](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[/tex] which is true, but this is obviously not a proof.
  8. Aug 10, 2014 #7

    Stephen Tashi

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    That is "a" correct response provided the theory [itex] (a-b)^2 \ge 0 [/itex] 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.

    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.

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