Basic Mathematics - 2021-08-16

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F. Zer

13:08

Calculus - Spivak:
  Chapter 1:
    11. Find all numbers x for which
      (viii) |x–1|·|x+2| = 3
        Given x ∈ ℝ:
          If |x–1|·|x+2| = 3:
            If x-1 ≥ 0 ∧ x + 2 ≥ 0:
              x ≥ 1
              (x–1)·(x+2) = 3
              x^2+x-5 = 0
              x = (–1–√21)/2 ∨ x = (–1+√21)/2
            x ≥ 1 ⇒ x = (–1–√21)/2 ∨ x = (–1+√21)/2
            If x-1 ≤ 0 ∧ x + 2 ≤ 0:
              (x–1)·(x+2) = 3
              x^2+x-5 = 0
              x = (–1–√21)/2 ∨ x = (–1+√21)/2
            x ≤ –2 ⇒ x = (–1–√21)/2 ∨ x = (–1+√21)/2

@user21820 I am struggling to complete the formal proof of this exercise. Could you help me discovering where it went wrong ?

user21820

13:33

@F.Zer I don't understand what you're asking. Except for your lack of justifications, what do you mean by "wrong"?

@shintuku Please don't use "^". Either write "and" or "∧". Otherwise it's very confusing.

Anyway what you're writing is hardly a proof in my system. If want to work within the system, you have to work within the system. Deviating only makes both you and me unsure of whether you actually know how to do a formal proof.

@F.Zer: By the way, I'm not sure why you cannot justify the last "⊥". It's exactly the same reasoning as in the first two cases. Are you also unable to justify where you got the conclusion in each of those?

To figure out what the final conclusion should be, you just need to collect all the possibilities that you have found across all the cases and weed out those that don't actually work.

F. Zer

14:05

@user21820 Thank you for taking a look. My intermediate conclusions "x ≥ 1 ⇒ " and "x ≤ –2" are incorrect since x should be strictly greater than 1 and strictly less than –2.

@user21820 Well, the equation "x^2+x+1 = 0" has no solutions in ℝ. So, I do not know how to formally justify the last "⊥".

@user21820 Yes, in the first two cases I can certainly justify the answer. I used the quadratic formula. However, in the last case, since the discriminant is less than 0, there are no solutions (in ℝ).

F. Zer

14:36

@user21820 Well, thinking a bit more, although the book solution says "x > 1 ⇒ ..." and my solution is: "x ≥ 1 ⇒ ...", when x = 1, I can certainly reach a contradiction and infer the same statement...

user21820

15:31

@F.Zer I did not say you could use the quadratic formula. If you cannot prove it, you cannot use it.

@F.Zer What is more concerning is that your ⇒intro was wrong. I didn't notice earlier because I assumed you could copy-paste correctly...

1 hour later…

F. Zer

16:47

@user21820 Ok. I will continue working on it.

F. Zer

17:39

@user21820 I do not currently have the tools to justify quadratic formula (involves square roots) and do not know any other method to solve a quadratic equation.

3 hours later…