Witch 93: Base Balls-Up

This is the first of three WitCHes on VCAA’s Specialist Mathematics Exam 1 Sample Questions. We spent a solid hour writing a critique of this one, and another half hour removing the bad language. Given the topic is new, however, and will be not so familiar to many teachers, we decided the question required more prominence, and proper discussion. The question was first flagged by commenter Michael, and the sample questions have been discussed generally on this post and this post.

UPDATE (14/02/23)

Part (a) is simply meaningless, as summed up perfectly in edderiofer’s comment, below.

26 Replies to “Witch 93: Base Balls-Up”

  1. Unfortunately, setting n_0=1000 works too. This question may make sense on the surface, but it gets more mindless the more you think about it. Why must it be n_0 =5? Clearly, 2^n > n^2 for n\geq1000. Maybe this could be better written as: “deduce the smallest n for which the inequality holds”. How would you go about showing n_0 is 5? Show that it doesn’t work for n_0 = 4, but for n_0 = 5? Would it be reasonable for n_0 to be assumed natural too, or could you set n_0 = 4.0000000001, and have the expression still be true, because n is natural? The question assumes a lot, which could set the precedent that students may have to “deduce” arbitrary details of no consequence.

      1. I can see a potential issue then: if you set n=0 or n=1, 2^0 > 0^2 and 2^1 > 1^2, but it seems like blind faith to be able to show that n_0 =5 is the miracle number. I don’t see a logical argument for showing n_0 =5. But is this an issue?

  2. There is no statement that n_0 is a natural number.

    This surely is a bad thing…?

    Also, since the question doesn’t use the quantifier “for all” n greater than a set value, I would suggest simply showing it was true for n=6 might be enough… I’m sure I am wrong in this though.

    The IB had a question similar to this one not that long ago. Marty – I’ll send it through and you can do what you see appropriate.

      1. The two expressions are equal for n=4, so n=4 does not satisfy the inequality.

        From there, an induction proof is not really needed, since one function has a bigger gradient than the other.

        BUT, if we have to stick to natural numbers… when n=5 one expression is clearly larger than the other, so let us try to prove that for all integers n greater than or equal to 5, 2^{n} is greater than n^{2}.

        The subscript notation seems pointless.

  3. Not sure I can read Marty’s mind.
    But the question reveals disturbingly sloppy ignorance of the nature of proof by induction.

    There are *two* different objects lurking in line 1:
    there is an “inequality P(n): 2^n > n^2” (where n is left “free”, and the quantifier is *not* included)
    there is a (sloppily worded) compound statement: “P(n) is true for all n >=n(0)”

    Unless students understand the difference between
    – a single inequality with a free variable (which may be true or false when we give the variable a particular value), and
    – a family of *infinitely many* statements, all compressed into a *single compound* statement
    they are bound to be confused (and so is the teacher and the examiner and the curriculum authority).

    It is relatively easy for students to understand a single statement – even with a free variable n.
    And they can see how to check whether it is true for any particular value of the variable n.

    It is much less clear how to prove *infinitely many* statements *all at once*.

    The first rule is
    – to state things clearly;
    – to adopt a template for handling such remarkable mathematical statements (and their proofs).

    I refer to Chapter 6 of https://www.openbookpublishers.com/books/10.11647/obp.0168 (p.283 ff).

    1. Thanks, Tony. Yes, it’s the first line that is the killer. I’m less worried by VCAA’s disturbingly sloppy ignorance of induction than I am by their disturbingly sloppy ignorance of sentence composition and meaning.

      1. Ah yes forgot to specify: I mean if they’d written what (I think) they meant, eg. ‘show that setting n_0=5 makes this statement true’. It is funny that they wrote the converse of what they meant, when their exam reports will presumably contain many high-and-mighty ‘students failed to distinguish between a statement and its converse’ lines.

        1. Quickly:

          1) Yes, obviously proving the inequality holds for all n ≥ 5 is what (b) is asking.

          2) I don’t think VCAA intended to ask this in (a), to the extent one can conjecture about such muddled intention. One would think they’d be aiming for the base case in (a), which amounts to proving the inequality for n = 5. Yes, what they wrote is a million miles from that. But in terms of setting up a standard induction question, what else might they mean?

          3) What they wrote in (a) is not the converse of what they intended, or anything. They did not write a statement that can be either proved or disproved.

  4. Ah, yes it would make more sense if they were asking for the base case. That question (prove the inequality for n=5) really is indecipherable from part a. Looking at their webinar (ew) they seem to also want students to show that n_0 can’t be smaller than 5 (given n_0 has to be an integer, which is also not specified).

    I’d still maintain that, IF they had intended to ask ‘show that, if n_0=5, this statement is true’, (which, granted, they probably didn’t) then by instead asking ‘show that, if this statement is true, then n_0=5’, they’ve used the converse of what they meant.

    (When you have time to answer) what do you mean by the statement not being able to be proved/disproved? Are you referring to the ‘2^n>n^2 for all n>n0’ statement? Or is the issue with asking to solve for n0 when there are infinitely many values that would make that statement true? (or something else entirely?)

    (and this is aps, but I prematurely clicked enter)

    1. Thanks, aps. I adjusted your name on the comment.

      In regard to your question, you want to prove or disprove the statement in (a), which is a claim about n0. What do you know about n0 in order to prove or disprove this statement?

      1. Hm. So, is the issue that we don’t know anything about n0, and thus we can’t prove ‘2^n>n^2 for all n>=n0’?

        Would the statement be provable if it instead said something like ‘there exists n0 in N such that, for all n>=n0, 2^n>n^2’?

        1. Thanks, aps. See also edderiofer’s comment below.

          In response to your first question: yes. We are invited to “consider” and inequality involving n0, but we are given no indication of the truth of this inequality.

          In response to your second question: sort of yes.

          A proof of your rephrasing would involve finding any n0 that works. That doesn’t quite fit in with an answer of n0=5. But you could tweak your suggested statement to make it work.

          The point is, the preamble in (a) must declare something to be true. Otherwise, the car simply won’t start.

          1. Yeah, that makes sense. In the webinar, VCAA include the clause ‘for the inequality statement to be true’, which suggests that they made the webinar based on the sample questions and picked up on that issue (although didn’t edit the sample).

            I looked at edderiofer’s comment, it’s a good summary. There are just so many layers of wrongness in this question. In my experience, VCAA are very bad at saying what they mean (this goes double for non-maths subjects), and trying to figure out what they want is the student’s problem. It gives me very little hope for a logic component in Specialist, where precision is critical.

  5. Delete “for n>=n0” in the first line
    2nd line: Show (actually: inspect)that 5 is the smallest value of n for which the inequality is TRUE.
    Then, naturally, use Mathematical induction that it holds for all n greater than 5.

    I think one can “consider inequality” , any inequality, but does it imply it is true???
    Must admit that after the 1st reading i took it as a clumsy way of saying that it start holding for n=5.

    1. That would be it.

      Consider the statement,

      “Scott Morrison is an honest man”.

      Does being provided this statement to consider somehow imply that the statement is true?

    2. “Show (actually: inspect) that 5 is the smallest value of n for which the inequality is true” – maybe not:
      n=0 : inequality is true
      n=1 : inequality is true
      n=2, 3 or 4 : inequality is false
      n=5 : inequality is true.

      So maybe just a simple “Show 2^n>n^2 when n=5.”

      1. I think Bancek’s point is n =5 is the smallest value for which the inequality is true from then on. In any case, it’s a very minor detail. Nothing in the saving or condemning of the question hinges upon it.

  6. Right, lemme summarise what other people have stated here, since this discussion is getting a bit messy.

    * There is no statement of whether/when the inequality should be *true*, and it certainly does not state anything like “2^n > n^2 is true for all n ≥ n_0”; it only states that the inequality should be *considered* over this range. So there is not enough information to prove that n_0 = 5, because we’re given zero constraints on n_0.

    * There is no statement that n_0 needs to be a natural number. Not necessarily an oversight, but definitely feels like one.

    * If a) were intended to say “Show n_0 = 5 is the unique value of n_0 such that the inequality holds for all such values of n”, then a) is impossible because n_0 can in fact be any number greater than 4.

    * If a) were intended to say “Show that when n_0 = 5, the inequality holds for all such values of n”, or “Show that n_0 = 5 is the smallest integer value of n_0 such that the inequality holds for all such values of n”, then part a) requires first doing part b).

    * If a) were intended to say “Show that the inequality holds for n = 5”, this doesn’t at all resemble what a) actually says. Indeed, if this really is the intended reading, then n_0 doesn’t even factor into the question at all, and the first sentence should really just say “Consider the inequality 2^n > n^2.”, possibly with a “where n ∈ ℕ” tacked on. The question-writer seems to have conflated the notation “n_0” for the base case of the induction with “n_0” for the lower bound of an arbitrary range over which to consider the inequality.

    Honestly, a) is an absolute dumpster fire of bad wording. I don’t think I’ve seen a question this badly-worded featured on this blog for a while. Whoever wrote and reviewed this did a terrible job.

    1. I’m not sure if, after reporting on the Hindenburg crashing, people came up and said “Great commentary, Joe”. But, great commentary, edderiofer. Perfect summary.

  7. An induction problem for SM students:

    Prove that for all positive integers n,

    \displaystyle{\frac{1}{2}.\frac{3}{4} ... \frac{2n-1}{2n} < \frac{1}{\sqrt{3n}}}.

    From "101 Problems in algebra from the training of the USA IMO team", p. 10.

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