# Witch 96: Root Con All

This is the fourth of our three WitCHes on VCAA’s Specialist Mathematics Exam 1 Sample Questions. Yeah, yeah it’s Douglas Adams-ish, but it wasn’t deliberate. We were undecided on this last one, and it’s not as compelling as the others. But, for similar reasons for the other WitCHes, we changed our mind and decided to post it. (There were other close calls, and we’ll soon update this post with brief comments on all the questions.)

### UPDATE (14/02/23)

This one is interesting. As noted above, it was an afterthought to WitCH it; it is not nearly as bad as the two other contradiction questions considered in the webinar; although there are simple direct proofs, it is not totally absurd to ask this as a proof by contradiction. Still, as commenters have noted, there are definitely aspects to criticise, and there’s plenty to discuss and to clarify.

First of all, let’s give some direct proofs.

Most simply, we have Anonymous’s one-liner:

√3 + √5 > √3 + √3 = 2√3 = √12 > √11.

A clunkier version, which Anonymous also proposed, is to bluntly estimate the roots. For instance,

√3 + √5 > 1.5 + 2 = 7/2,

and then just note that (7/2)2 > 11.

If nothing else, these two proofs point out the importance of choosing numbers in problems with proper care: if VCAA had gone with √13 or similar, such a direct proof would have required either more cleverness or more work.

Our immediate reaction to the problem was a third direct proof, which begins with first squaring the LHS:

(√3 + √5)2 = 3 + 5 + 2√15 = 8 + 2√15.

We then want to see if this is greater than the square of the RHS, and of course that’s easy:

8 + 2√15 > 8 + (2 x 3) = 14 > 11.

Since LHS2 > RHS2, and since both sides are positive, we also have LHS > RHS. Done.

The expected contradiction proof is, in fundamental working, very similar to this third direct proof. Assuming the conclusion is false, we have

√3 + √5 ≤ √11.

Squaring gives

8 + 2√15 ≤ 11,

and we easily get the desired contradiction.

Now, a couple questions. First of all, which is preferable: the third direct proof or the proof by contradiction? It is perhaps a matter of taste, but for us the direct proof is clearly preferable, for two reasons. First of all, direct proofs are always preferable: why enter the fantasyland of contradiction if it is not required? Secondly, the direct proof begins with a quantity and then manipulates and estimates that quantity; this seems to us much more natural than manipulating statements, than manufacturing a chain of inequalities.

The second question concerns the following equivalence, for positive x and y:

All the above proofs use one or the other direction of this inequality, and necessarily so: in the end, we only know something about a square root through squaring it. Can both directions be assumed by students? Is one direction “easier” than the other? Note that the reverse implication is standardly used in solving questions such as Q16 of the sample questions.

As with the induction WitCH, this seems a case where VCAA has not stated the ground rules for the students, and so it is simply impossible for a student to be sure what may or may not be assumed.

## 50 Replies to “Witch 96: Root Con All”

1. Anonymous says:

It’s very easy to prove this directly in a single line:

√3 + √5 > √3 + √3 = 2√3 = √12 > √11

What the hell does VCAA actually want us to use proof by contradiction for? I can’t actually think of any useful way to use proof by contradiction to solve this problem.

In fact, if I were to do this question in an exam, I’d probably write the following:

Line 1: √3 + √5 > √3 + √3 = 2√3 = √12 > √11

Line 2: Let us assume that √3 + √5 ≤ √11.

Line 3: By line 1, this is false.

Line 4: Therefore, by proof of contradiction, √3 + √5 > √11.

I wonder how VCAA would mark that proof “by contradiction.”

1. marty says:

Now that’s funny.

Thanks very much, A. To be honest I didn’t think of directly estimating the roots as you did. But even without that, proof by contradiction seems a silly route.

1. Anonymous says:

Yeah, my first reaction to this question was just to directly compute the roots (to be more precise upper and lower bounds for the roots). But I then realized you could just do the trick I did above.

Never would I ever have thought of using proof by contradiction.

2. Tungsten says:

Assume √3 + √5 ≤ √11
√3 + √5 ≤ √11
3 + 2√15 + 5 ≤ 11
√60 ≤ 3
60 ≤ 9
Contradiction therefore √3 + √5 > √11.

Correct me if there are any mistakes but I think that I would do it that way if I were asked that question. Anyway, I have no particular issue with asking for a specific proof technique although the fact that VCAA seems to do so I think is a separate problem.

1. marty says:

Yes, that’s fine, and clearly what VCAA expected. And it’s kind of ok to ask for this contradiction proof, even if there are easier direct methods. That’s why I hesitated to post this one. But there’s something else that bugs me about the proof you’ve given.

1. Amber says:

Is it something to do with squaring and square rooting both sides of an inequality?

1. marty says:

Thanks, Amber, and yes. Would that be permitted or not? Can it be used in a direct proof?

1. Amber says:

I’ve been thinking about that – it seems a clear potential trap for students – yes if the proof is over natural numbers, or R+? Here no number set is defined but everything is positive so it works but is the issue with only taking the positive square root in that step? There was an example given at the MAV conference recently with an inequality geometric proof involving triangles and pythagoras where the lecturer explicitly stated that lengths needed to be positive so I assumed we needed that line to make the proof work.

1. Red Five says:

What if we first make a zero right hand side? Does that then fix the issue, because we can assume the LHS is positive/negative and hence squaring both sides either will or will not reverse the inequality.

2. Amber says:

oops just realised there is no square rooting only squaring…

2. Tungsten says:

Squaring both sides of the inequality would be permitted I think, but clearly requires a high degree of care. For example, squaring any inequality involving a negative on either side, and the fact that does not imply . In my proof above we do assume that everything we need is positive as Amber mentioned and then the argument is valid. I see no reason why a direct proof would have different permissions but maybe somebody else can give some insight on that.

2. marty says:

Thanks, Amber, and Tungsten and RF. You’ve caught onto the reason I posted this question. I will try to focus your puzzling.

First of all, the negative numbers thing is mostly a red herring. We’re dealing with explicit positives here. So, yes, it may be reasonable or VCAA-mandated to declare the positivity, but that’s just bookkeeping.

Now, sticking to positives,

(a) Does x2 > y2 imply that x > y?

(b) If so, can that form part of a direct proof of VCAA’s sample question?

(c) If so, would that inequality in (a) first have to be proven?

If the answer to all three is “yes”, then i think we have a problem.

1. Red Five says:

implies .

So .

Now, if then so we can conclude

Which means .

1. marty says:

That answers (a).

1. Red Five says:

(b) is beyond my ability to comprehend.

Or rather: I say it can. What VCAA says might be a different matter.

1. marty says:

No it isn’t. Try to give a direct proof. How might you go about it, if you want to avoid directly estimating the roots?

1. marty says:

No. Forget about what VCAA might want or might do. How could you use (a) to give a direct proof of the sample question?

2. marty says:

I got rid of your 0, but there seem to be other typos. Not sure what “assume” is assuming. (And why assume anything? It’s meant to be a direct proof.)

If you tell me the corrections (by email if you wish), I’ll correct.

2. marty says:

Sorry, my (b) was unclear, and probably unmeaningful.

What I’m asking for is an easy direct calculation (no contradiction) that establishes THIS > THAT. Then, that inequality combined with (a) constitutes a proof of the sample question.

1. Red Five says:

I think such an argument IS possible, but the one I sketched out looks really ugly.

The argument Anonymous proposes at the top of this thread is so much nicer to look at…

2. aps says:

are you looking for a direct version of Tungsten’s proof? as in:

(using your (a) and its converse)

(again using (a) and its converse)
True
or do you mean something completely different?

1. Red Five says:

Something like that except starting from the final line and working backwards.

Otherwise, it may be argued that you are starting from the assumption that the inequality is True which VCAA may not like.

1. marty says:

They’re equivalences, so it’s ok. (Which may not stop VCAA being stupid enough to object.)

2. marty says:

Yes, that’s it. Although I don’t like the chain of equivalences. it seems preferable to me to start with LHS2 and then compute. Then employ (a) at the very end.

Now, what about (c)?

1. Tungsten says:

I am not sure what you mean about your preference. It seems that aps’ method, like my contradiction proof, relies on (the converse of the statement in (a)) or similar, which I am not sure that (a) proves. Red Five’s suggestion would use (a) directly. Anyway, on (c) we might as well prove (a) or its converse as appropriate or else accept it as trivial. With a hypothetical VCAA exam, make a guess based on allocated marks maybe.

1. Tungsten says:

For 3 marks in the sample question using my contradiction proof I think it is safest to prove first. Alternatively, it might be sufficient to justify it by explicitly stating that at the relevant steps.

1. marty says:

Thanks, Tungsten, and aps. I’ll update the post soon with my thoughts (which are mostly questions).

Have a look at Q16 in the sample questions. Does that change your thoughts at all?

2. marty says:

No.

Your contradiction proof uses .

aps’s direct equivalence proof uses .

My suggested revamping of aps’s proof uses .

2. aps says:

Here’s my very amateur and probably erroneous take:
It feels to me like there’s a logical/mathematical side to (c) and then a presentation/‘aesthetic’ side?

Logical side: to use (a), we need to know (a); to know (a), we need to prove (a), (or (a) can be an axiom). But this can be said of many assumptions that are made in the proof, such as sqrt(3)*sqrt(5)=sqrt(15): we have to know sqrt(a)*sqrt(b)=sqrt(ab), so we have to prove this as well; same as difference of two squares in Red Five’s proof of (a). And in proving that, we might come across associativity of multiplication; in this strictly logical side we’d have to prove that as well…

Presentation side: this would consider the level of detail/depth that needs to be shown, given we aren’t going to actually write proofs of all the things the logical side requires us to prove. I would imagine the goal here would be to keep the ‘depth’ of the proof relatively consistent, if that makes sense. I think this would be better achieved by including a proof of (a), since if several steps of arithmetic working aren’t trivial enough to leave out, then surely (a) isn’t trivial enough either. I’m sure VCAA would disagree. But it looks a bit like a matter of opinion to me…

1. marty says:

Thanks, aps. See my reply just now to Tungsten.

2. Red Five says:

It is 3 marks. I really don’t think much more than any of the correct proofs presented here could be marked as wrong.

1. marty says:

Direct proofs would be marked wrong. But my point is to be thinking more carefully about what can be assumed when doing SM proofs.

3. Red Five says:

This maybe very wide of the mark, but is the opposite of “greater than” the symbol “less than” or is it “less than or equal to”?

Nit-picky perhaps, but I’m genuinely curious.

1. aps says:

Using my (very limited) knowledge, I would suggest that given ‘P or not(P)’ is a tautology, then not(a > b) would have to be a <= b, since a negation has to be in some sense a complement. I don't know for certain though.

2. marty says:

First of all, never say “opposite” when you mean “negation”. You’re likely to end up working for VCAA, and we wouldn’t want that, would we?

Now, as to your question, aps is correct: the negation of “a > b” is “a ≤ b”.

This following may seem jargonistic, but it may (or may not) help.

To make a statement S is the same as saying ” ‘S’ is true”. That doesn’t mean the statement is true, it’s just describing what is being said.

So, saying

(*) a is greater than b

is the same as saying

(**) The statement ‘a is greater than b’ is true.

The negation of S is the statement ” ‘S’ is false”. So, the negation of (*) is

(***) The statement ‘a is greater than b’ is false.

But what does it mean for ‘a is greater than b’ to be false? It means exactly that a is less than or equal to b.

In symbols, the negation of

****) a > b

is

*****) a ≤ b

1. Red Five says:

Perfect. Thanks.

1. Banacek Spaces says:

I went with AM GM inequality. Also needed is the fact that x>y implies that x^2>y^2
as well as sqrt(x) > sqrt(y)
Just by recalling that y=x^2 and y= sqrt(x) are increasing functions for positive real x.

1. Banacek Spaces says:

Only sqrt things are actually needed.

4. Terry Mills says:

Re: Q1 in Specialist Mathematics Exam 1 Sample Questions.

In the Victorian Curriculum, N = {0,1,2,….}. This seems to make a difference.

1. Red Five says:

And some documents published recently use both N={0,1,2,3…} and N={1,2,3,…} in the same document…

I’m not sure we can say for certain that 0 is a natural number as defined by the Victorian curriculum.

Which is a source of much frustration.

2. marty says:

That’s hilarious. But, in practice, I think it’s a red herring.

The VCAA glossary for K-10 includes 0 in the naturals. Everything I’ve seen in VCE, however, suggests that n = 1, so I don’t think there’s any practical ambiguity in VCE. I cannot see that the VCE study design defines N, but all the texts I’ve seen, as well VCAA’s support materials, start N with 1.

5. aps says:

I had a look at Q16. I suppose the similarity is that one solution involves assuming that the x-values for min(f(x)) and min(sqrt(f(x))) are the same, which (I think? correct me if I’m wrong) is implied by x sqrt(x) < sqrt(y); and then the question we’re asking would be, should students show that x sqrt(x) < sqrt(y), or otherwise show that the x-values for min(f(x)) and min(sqrt(f(x))) are equal, as part of this solution?

For this question I would say no, because it’s a ‘find’ question rather than a ‘prove’ question. I do think they should state their ‘x-values-are-equal-for-min(f(x))-and-min(sqrt(f(x)))’ assumption somewhere, because it doesn’t seem quite trivial enough to be left implicit in a 4-mark question. But again, that’s just a matter of personal opinion…

I think that’s the real problem, that it’s unclear just how much detail needs to be shown in questions like 5 and 16; VCAA isn’t really clear on what exactly is required. I wonder if different assessors will mark these questions differently based on their personal interpretations of how much detail is necessary (which I suspect already happens a fair bit in show-that questions!). I suspect there will be Issues when it comes to marking proof questions…

1. marty says:

Thanks, aps, and yes, that’s my point.

Forget the “find” and “prove” language thing. The point is, students routinely find a minimum distance and its location by finding instead the minimum distance2. This is done, correctly, without comment, but implicitly relies upon

(*) .

That’s fine. But are the same students in the same subject then supposed to know in a “proof” question, that (*) needs to be proven if used?

1. aps says:

Looking at the webinar (ew) VCAA expects them to state but not prove (*), so to answer that question, seemingly not. It is weird that (*) can be assumed, but (referencing WitCH 94) odd*odd=odd and similar facts can’t. And then, in the ln(5)-is-irrational webinar proof, it’s assumed that 5 can’t divide e^a, which imo isn’t at all trivial! Maybe VCAA should release a guidebook or something…

1. marty says:

A VCAA guidebook? Wrongway Feldman comes to mind.

Thanks, aps. You’re referring to slide 10 of the contradiction webinar? That assumes only

(**) .

I’m not sure VCAA would accept the reverse inequality so readily.

In case you’re in doubt, VCAA’s log 5 proof is utter nonsense.

1. aps says:

Ah, that’s true about Slide 10. Still, I suspect the slide will give many teachers the impression that
(***) x sqrt(x) y^2, then (x+y)(x-y)>0. (1 mark)
b) Hence, show that x^2>y^2 => x>y. (1 mark)
c) Using the statement from (b), show that sqrt(3) + sqrt(5) > sqrt(11) (3 marks)

…yeah.

1. aps says:

Oh no, my comment seems to have gotten corrupted somehow. Maybe I accidentally deleted a chunk of it..? (I tried to edit but it didn’t seem to work).

Anyway, my point was to be that VCAA probably wouldn’t have faith in Specialist students’ ability to prove
(***) x <y —> sqrt(x) < sqrt(y)
on their own, looking at Red Five’s proof and Banacek Spaces’ increasing function argument. So I was trying to write an example of the scaffolded question they might write if they wanted (***) proved as part of question 5. 5a) was:

Show that, if x^2<y^2, then (x+y)(x-y) < 0 (1 mark).
b) and c) were left intact in the original comment.

1. marty says:

If you want me to edit or delete anything, let me know.

1. aps says:

Yeah. My initial comment said (something like):

*****

Ah, that’s true about slide 10. Still, I suspect the slide will give many teachers the impression that
(***) x < y implies sqrt(x) < sqrt(y)
can also be stated without being proved. And I think students and teachers will be right to feel misled if, on the exam, marks are deducted for not writing a proof of (***). Not that this would stop VCAA from doing it.

What I think WILL stop VCAA from requiring a proof of (***), is that they won’t have faith in students’ ability to provide one. A proof like Red Five’s, or a tight argument about strictly increasing functions, doesn’t feel trivial enough for a Specialist exam. Especially if it’s tucked away inside another question. I’d imagine that, if VCAA wanted a direct proof of question 5 including a proof of (***), they’d ask something like:

5a) Show that, if x^2 < y^2, then (x+y)(x-y) <0,
b) Hence, show that x < y implies sqrt(x) < sqrt(y), for positive x and y (2 marks)
c) Using the statement from part (b), show that sqrt(3) + sqrt(5) > sqrt(11) (2 marks)

*****
If you could edit my initial reply to say this, it would be much appreciated 🙂

6. marty says:

I’ve updated the post with a summary.
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