## Great Moments In Mathematical Invention

I was in Australia this last week, working with some teachers at MYSA on You Pour, I Choose. It’s a task that asks which of two glasses has more soda and involves, among other skills, a fairly straightforward application of volume.

A teacher in the workshop called me over. “I’m not a math teacher,” she told me, and then pointed to the person next to her who had calculated the formula for volume of a cylinder.

“But that seems like more work than you need to do,” she said. “We don’t care about the exact amount. We care which one has more. With both glasses, we multiply by pi and square the radius. So all you really need to do is multiply the radius by the height for both glasses and compare the result. That’ll tell you which one has more.”

This was a rather stunning suggestion, made all the more impressive by the fact that this woman doesn’t immerse herself in numbers and variables for a living like the rest of us.

I have two questions:

• Is she right? She’s certainly right in this case. Both the volume formula and her shortcut indicate the left glass has more soda.
• As her teacher, what do you do next?

I’ll update this post tomorrow.

2013 May 29. I knew just telling her, “That’s wrong.” would be unsatisfying because, explicitly, she said, “Am I wrong?” but, implicitly, she was saying, “If I’m wrong, then make me believe it.” I knew the current problem wasn’t helping me out because her shortcut actually worked.

I knew this woman had dropped me off deep in the woods of “constructing and critiquing arguments” but I didn’t know yet what I was going to do about it.

“Whoa,” I said. “Does that work? If that works that’s going to save us a lot of time going forward. Let me bring your idea to the group and see what everybody thinks.”

In the meantime I stewed over a counterexample. It took me more than a minute to think of one because a) I was kind of adrenalized by the whole exchange, and b) I don’t do this on a daily basis anymore so my counterexample-finding muscle has become doughy and underused.

I posed her shortcut to the group and said, “What do you think? Does this work?” I gave them time to think and debate about it. Someone came back and said, “No, it doesn’t work. Imagine two cylinders with different heights and a radius of one.”

Awesome, right? This particular counterexample doesn’t disprove the rule. The square of one is also one so her rule works here also.

Eventually someone suggested two examples where the product of the radius and height were the same but where the radius and the height were different in each cylinder. The shortcut says they should have the volume. The formula for volume says they’re different.

Final note: students are often asked to prove conjectures that are either a) totally obvious (“the sum of two even numbers is even” in high school) or b) totally abstract (“prove the slopes of two perpendicular lines are negative reciprocals” in middle school). It’s rare to find a conjecture that is both easily understood by the class and not obviously correct or incorrect. I’m filing this one away.

I'm Dan and this is my blog. I'm a former high school math teacher and current head of teaching at Desmos. He / him. More here.

1. #### Patrick Honner

May 28, 2013 - 8:55 am -

I might ask the student to prove or disprove the following statement: Squaring makes all numbers bigger.

2. #### Justin Lanier

May 28, 2013 - 8:56 am -

Briefly:

1) In general, she’s right about the pi, but wrong about the exponent on r.

2) This is easy. I’d have her share her shortcut with her peers during our whole-class discussion of the problem. That of course could end up heading in a lot of different directions.

May 28, 2013 - 8:57 am -

As her teacher, I’d draw/find the following two glasses:

That’s all I’d do. “Be Less Helpful” She should be able to figure out if this works.

If you can actually find two glasses that have the same r*h product but different volumes, then fill the one with larger volume and pour it into the other glass. Enjoy the overflow.

4. #### Simon

May 28, 2013 - 8:58 am -

Due to the squaring of the radius, this won’t always work. The next step I might take is use measurements that should come out equal using her model and then mix them up and see if this still holds. I.e. h=3. R=5 for one and reverse those numbers for the other and then ask why there are two different results.

5. #### Andrew Shauver

May 28, 2013 - 9:05 am -

Perhaps if you asked what “r squared” means. If you could lead the student to recognize “r” as a factor twice as many times as “h”, then perhaps the student can recognize the different roles of radius and height.

Perhaps, rewriting pi*r*r*h might show that removing the exponent is a bigger change than the student thought.

6. #### Bob Lochel

May 28, 2013 - 9:09 am -

Interesting discussion. The earlier replies have given some examples as specific conditions where the conjecture can be disproved. But I would then want students to try to explain conditions under which the conjecture IS true. When does it fall apart? Why?

Just whipped together a Desmos that could be used for individual exploration: https://www.desmos.com/calculator/gq0m1xv159

Let me know if it works.

I like Patrick’s idea: there is a general rule here which seems intuitive, but is far more complex when you dig deeper. I want the students to dig, and come up with the argument.

7. #### Nick Skaggs

May 28, 2013 - 9:14 am -

She is right in this example but that shortcut would not work every time. I would ask her to consider a different example. What if the left glass had a diameter of 4 cm and was filled to a height of 3 cm and the glass on the right had a diameter of 8 cm and was filled to a height of 1 cm?

8. #### Andy Mitchell

May 28, 2013 - 9:23 am -

What a fascinating response!

My first reaction is to say, yes, in the real world this shortcut will work.

Then I’d give the person with this solution a challenge: find me an example where this shortcut doesn’t work. I would reassure them that although in most cases this solution is sufficient, there is a special range of radii, all bunched together, that will yield a wrong result when using this shortcut.

9. #### Don

May 28, 2013 - 9:24 am -

What’s great about her suggestion is the ways it reflects an understanding of being able to alter both sides of an equivalence. Less great that she doesn’t spot which one of her two-part changes doesn’t pass muster because it’s non-linear, but an excellent opportunity to explore what pi*r^2 really -means-, as #5 points out.

I’d explore that and ask for other possible simplifications. One that comes immediately to my mind is whether we could just throw away h for computational purposes if h1 and h2 are both smaller than r1^2 and r2^2.

I am more tickled by this the more I think of it; it reminds me of computing O() in computer science classes.

10. #### Dan Pearcy

May 28, 2013 - 9:26 am -

This is a great blog post Dan – getting people to engage with how they deal with a misconception. Often this integral part of teaching and learning isn’t discussed and should be more often I think.

Question 1: Of course, no. Dividing by pi works but dividing by r doesn’t.

Question 2: I’ll try to answer the way that I originally thought. Of course it depends on the student but this was my original internal response.

Me: “That’s a really interesting idea. It’d be great if it worked because it would reduce workload and increase efficiency. I wonder if it does always work or whether it just worked in this case? – Have you tried it out with other examples?”

I’ll come back to you in 5 and you can tell me if you’re onto something.

Come back in 5 mins and they’ve either found a counter-example or haven’t. If they haven’t found one…

Me: I wonder whether this example would work (E.g. First Cylinder: radius=2, height=3, Second Cylinder: radius=3, height=2)

Great effort for trying something new. I wonder though why it doesn’t work? (Amount of time given here dependent on student).

If student can’t see why. Possibly open discussion to rest of class because it is a big misconception. (Don’t make clear who student is in case they don’t want everyone to know).

Me: “Everyone, can you just stop what you’re doing [explain situation]. Can anyone help out with this?…Class discussion.

Finish Discussion with something positive which addresses the key issues here: i.e. “It’s great to see people thinking creatively about these problems. What two things can we take away from that to learn for next time?”

1) Is there a counter-example?

2) You can’t divide two sides of an equation by different numbers (because r is different for the cylinders – if r is the same for both cylinders then it would be okay).

11. #### Dan Anderson

May 28, 2013 - 10:13 am -

To make it more clear to the student, I’d write the formulas with different variables. pi * (r_1)^2 * h_1 and pi * (r_2) * h_2.

Or to keep pushing the rock down the hill, I’d have them cross out the other heights too. After all they’re the same letter.

12. #### Mike Lawler

May 28, 2013 - 10:16 am -

As the person asking the question isn’t a math teacher, I’m not sure that I would have jumped right into the r^2 part.

Her shortcut also assumes if x > y, then C * x > C * y for any constant C. It hopefully wouldn’t take too long to help her understand that we need some restrictions on the constant C for this assumption to be true.

Realizing that there are some limitations on when this assumption is true might help her better understand that there are some limitations on her assumption about r^2, too.

13. #### Jim P

May 28, 2013 - 10:19 am -

Quick response:

Is she right?

Yes and no. Yes for this case, No in general. We can probably play with the set of numbers where her solutions would work – fun for us to do, less fun for her.

What do you do next?

“That is awesome” (because it seriously is awesome that she’s invested and inspired to think about it) “I really want to know when both of them would be the same?”

Give her the two cups, liquids, let her go nuts.

Thoughts: it’s not always about the right answer or not. I am okay with the exploration, discovery, and fun be the outcome.

14. #### Chris Goedde

May 28, 2013 - 11:12 am -

This would be a great entry on the math mistakes blog. I think some of the answers above are making assumptions about the student’s thinking that aren’t valid, because they are assuming the student understands the meaning of exponentiation, and realizes that she is dividing by r. But I don’t think that’s what the student thinks she is doing. I think the student is just canceling all the numbers in the equation; she’s treating exponentiation like multiplication. What’s interesting to me is that the student is mistranslating the mathematics. If I were the teacher, I would probably ask her to explain the volume formula in words, and to explain why she thinks her simplification works.

15. #### Justin Lanier

May 28, 2013 - 11:27 am -

I want to add that she’s right in a *big* way–that “constants” that affect quantities don’t matter when comparing two quantities’ relative sizes. She’s half-right and half-wrong in *small* ways–multiplying by pi *is* such a constant, but the 2 in the formula isn’t a constant, even though it looks like one. It’s actually a mask for a variable (another copy of r), and variables of course matter for relative sizes.

I think most of the above answers are focusing too much on how she is wrong, when in fact her insight is majorly correct.

Also, many of the teaching suggestions above are geared toward getting her to see she’s wrong as expediently as possible, rather than promoting investigation, encouraging classroom discourse, or celebrating and extending her insight.

How much of our jobs should be about “fixing” student thinking, and how much about encouraging student thinking?

16. #### Chris Goedde

May 28, 2013 - 11:36 am -

It would also be interesting to ask the student what she thinks the quantity r*h actually represents, if anything.

17. #### David Rubenstein

May 28, 2013 - 11:51 am -

The examples of getting the students to experiment with other setups to test the idea is a great one. This also provides a great opportunity to reinforce the idea that a rule/theorem is only valid until a non-example is found.

May 28, 2013 - 1:07 pm -

#17 Justin Lanier, Amen! You nailed it. And, Dan, I suggest you leave out the measurement captions for doing so would raise more thinking and conjecturing possibilities on the part of the students. Just a thought. Thank you

Sidi Naajee

19. #### Emily

May 28, 2013 - 1:14 pm -

As a teacher, I see an understanding of the overall objective AND a fundamental misunderstanding of what it means to “square.” She knows the r is important, but (I assume) doesn’t realize that the r is multiplied by itself.

*Note: there is the possibility that she intuitively knows this works BECAUSE the model shows that r >1. She may not have come to the same shortcut if 0< r <1.

Real-world, I would individually praise her for her thinking and ask her to model it her thinking on the board for the whole class (when the activity was done). Her "shortcut" demonstrates a true consideration of the question being asked, a process that I think is valuable to the class. I'd then ask the class your Q1. We'd find that it works for this example. At that point, either A) she discovers the exceptions to her "rule" in the process of teaching it, and we discuss it; B) another student notes the exceptions and we discuss it; or C) I pose the question "Is this a rule we can always use to compare relative volume?" and we discuss it, with varying heights and/or radii. (at some point, I might make the "variation" of radii just different units. Convert cm to m, and suddenly this shortcut doesn't work)

The whole point, it seems to me, of activities like these 3-acts is not for students to practice finding volume, but to begin to think of volume as a function of its parts, which is exactly what she does.

20. #### Justin Lanier

May 28, 2013 - 2:40 pm -

Let it be noted that whether the radii are greater or smaller than one has no bearing on whether or not her shortcut works. It can both work and fail to work in instances with any combination of those factors.

21. #### David

May 28, 2013 - 3:45 pm -

Being somewhat new to this game (teaching, that is) please forgive me if my suggestion branches out into text book style questions. Having read my response too many times, I can’t tell.

Is this not a delightful case of letting the algebra getting in the way of the understanding.

How about a Dan Meyer style, timing the filling of a cylinder with water problem, with the extension:

How long would it take to fill if the diameter of the tank was halved?

If you halve the height instead, how would the radius of the cylinder have to be altered for it to take exactly the same time to fill?

Better still, get them to make some cylinders with different heights and diameters that would all take the same amount of time to fill with water – they could use those old OHP plastic sleeves and a flat surface for a low budget version

22. #### Kevin Lawrence

May 28, 2013 - 10:19 pm -

1) She is correct in this case but it doesn’t work in every case. I’m sure counterexamples are easy to find (spare me, it’s late).

2) As a teacher, I see where this idea may be coming from a student standpoint. If we look at the area of a circle A=pi*r^2 (which is a portion of the volume of a cylinder formula), we know that comparing areas of circles is dependent on the radius only. She may have known that a circle with the largest radius will always have the largest area. The hiccup in using that logic on this is that there is another variable for the volume of the cylinders (height) that does not depend on the area of the circular bases. Simply pouring more liquid into one of the cylinders may drive that point home.

One other thing that I think is interesting is by looking at her updated formula of r*h for her comparison, that represents the area of half of the rectangular cross-sectional area of the cylinder when cut with a vertical plane through the diameter of the cylinder. If I pick on the left glass in the picture above, her “area” formula would be 2.75 cm*10 cm. If we rotated the rectangle around one of the side lengths of 10 cm, I think it would be easy to see that it would recreate the shape (and therefore volume) of the original left glass. But what if you took that same rectangular area and rotated around one of the 2.75 cm sides (h=2.75 cm and r=10 cm)? It definitely wouldn’t create the same cylinder. But would that cylinder have the same volume as the original? The answer to this is “No”, so the product of height and radius doesn’t match to a unique volume. For this reason, simply finding the product of the radius and the height of the cylinders is insufficient in making any generalizations about comparing the volumes of the cylinders.

23. #### Dan Pearcy

May 28, 2013 - 11:34 pm -

This has turned into a great discussion and something that could create a shift in my approach to addressing misconceptions.

As has been mentioned previosuly, many people (including myself) focused on what is wrong – not what is right. Getting her to investigate the scope of her discovery seems to be a great first step.

Furthermore, a good point was made regarding the fact that she is treating the exponent as a constant. This is for me, the key misconception in her skill-based knowledge which must be addressed.

24. #### Justin Lanier

May 29, 2013 - 12:15 am -

@DanPearcy, they way you phrased that just now about the exponent gave me the further thought that if the formula in question had been V = pi*(r*h)^2, crossing out the pi’s and 2’s and comparing would be just fine. What “doing the same to both” means is clearly a pretty nuanced question. Neat.

25. #### Dan Pearcy

May 29, 2013 - 12:31 am -

@Justin Awesone point! Dividing by pi and the square rooting both sides would be fine. Need to invest time into helping students understand how to balance equations!

26. #### Nils Davis

May 29, 2013 - 9:47 am -

The big problem with her simplification, and #17 touches on it, is that while pi is indeed a constant and can be cancelled, the relation you’re looking at is really two equations:

V1 = pi * r1**2 * h1
V2 = pi * r2**2 * h2

It has nothing to do with squaring or less than one, greater than one, or anything else. She’s simply cancelling out factors that are unequal (r1 r2), and that doesn’t work.

You could get to this problem via a question – “how is cancelling out the **2 different from simply cancelling out h?” which then leads you into perhaps a deeper understanding of what exponentiation means.

27. #### BK

May 29, 2013 - 9:51 am -

Have the student take 2 identical pieces of paper and ask what is the area and how they figured it out (9×12 would work well).
Roll each piece up to create a tube, one the long way and one the short way.
Fill one of them with something (cereal, sand, base-10 cubes, etc). If you pour what you have filled from that tube into the other tube will it fill it? Pour it in the other tube and find out.
Ask the student figure out why the 2 tubes made with the same size piece of paper with the same area don’t hold the same amount of stuff.

28. #### James Key

May 29, 2013 - 9:57 am -

I like the fact that she recognizes that we can ignore the factor of pi if all we want to do is figure out which one has more. (Note also that the conversion factor from cubic cm to ounces is not important here, either.) As Dan has pointed out many times in many places, *figuring out what is essential and what can be ignored* is a critical step in mathematical reasoning.

As for the part about “r*h versus r^2*h,” the comments have really got me interested. My first thought is that it is *really hard* to teach students that what is needed here is to *design an experiment* to put her theory to the test. If we want to teach students to think and reason independently, there are two things here:

1) It needs to *occur* to the student to design an experiment. In my experience, many/most students never get this far.

2) The student needs to be *capable* of designing an appropriate experiment.

I went right to work on designing a suitable experiment, and I will admit that I found it *hard to do.* Took me a couple of minutes to even decide what I was aiming for. Took me another couple of minutes to find parameters that met the conditions I was aiming at. Then I read the comments and people made it seem so easy…

I used R1 = 10, H1 = 25, and R2 = 9, H2 = 30.

One point I will share on my thinking: is it better to construct two cylinders where her method shows the volumes are *equal,” whereas the correct method shows one is larger? Or is it even more powerful to show that her method shows that glass 1 is bigger, whereas the correct method shows glass 2 is bigger? Or maybe it doesn’t matter. Does it?

One last point, based on the comments: it’s one thing to show that her method produces a different result from the standard method. This shows that it’s “wrong” in *relative terms,* i.e. assuming the standard method is right. But it’s another thing altogether to prove that it’s wrong on *absolute terms.* I like the suggestion in the comments to get some actual glasses and liquids and let her/the class get to work on them.

29. #### Richard Vahrman

May 29, 2013 - 10:51 am -

Easy proof that the conjecture is wrong. Take one of the examples e.g. the 7 and the 3. If the method is right then height 7 and base 3 would be the same as height 3 and base 7. Doing the maths with the square shows this not to be the case.

30. #### James Conron

May 29, 2013 - 8:18 pm -

Certainly a true “teachable moment”, one of those moments when being in a classroom with a teacher and other students really makes a difference.

Sure, you can include a scenario-based word problem in a textbook based on this, but who’s going to internalize more, the student who discovers that their *own* conjecture fails or the student who reads about *somebody else’s* conjecture failing?

31. #### Tim Erickson

May 29, 2013 - 8:53 pm -

Lovely. Bringing it to the class is often a reliable fallback! Also: I particularly liked your comment on counterexample-muscle atrophy. SO true.

32. #### Sneha

May 30, 2013 - 5:29 am -

Amazing coincidence: one of my students in M.A.Ed just developed a teaching resource based on this very concept. He ask students to roll a sheet of A4 paper into a cylinder – first along its length and then along its breadth.
Question: Are the volumes of the two cylinders formed the same?
So basically when we look for products of r and h which are the same, we are looking for the same lateral surface area (same sheet of A4 paper)
It was amazing how many people thought that the volumes would be the same.

33. #### l hodge

May 30, 2013 - 6:27 am -

I wonder whether this person was thinking about the formulas, or the actual cylinders, or both when they made the suggestion.

One issue is the fact that the correct volume formula would not always correspond with the suggested radius * height as far as choosing the larger volume goes.

A different issue might be an incomplete sense of what volume is and and intuitive sense of how changes in radius & height (or other dimensions) affect volume.

34. #### Mike Caputo

May 30, 2013 - 2:00 pm -

Though there are differences, when you first show that the volume for a pyramid or cone with the same base and height as it’s companion prism, students easily guess that the formula should be 1/2Bh instead of 1/3Bh because they examine only the two dimensional profile of the objects. It’s a starting place for discussion but never the eventual answer.

35. #### Chris

May 31, 2013 - 5:30 pm -

A great question from the lady. The fact that her model was wrong is completely irrelevant at first. She was actively looking for new ways to deal with the situation rather than following the algorithms provided.

With my science teacher hat on (rather than my maths teacher hat) I encourage students to value models based on usefulness rather than “correctness”. In this case the easier arithmetic is certainly a vote in favour of her model if it yields useful results.

Using a modeling framework, I’d ask the followup questions:
– Does it always work?
– If not, when does it work?

Most models are useful as long as we understand their limits. This model clearly works for the provided example. The fact that it doesn’t work all the time doesn’t stop the potential for it to be useful in a range of situations that may well be easily definable.

Trigonometry ratios are shortcuts that are “wrong” in the sense that they don’t work for all triangles. We use them because we understand their limits (right triangles only).

I’m stealing this one for my own classroom. The exploration potential is amazing – thanks :-)

36. #### Bob Hansen

June 2, 2013 - 10:43 am -

The only reason any one questioned this teacher’s conjecture is that the algebra looked wrong. As a teacher, assuming I wanted to help her not be wrong, I would proceed to teach her algebra.

And thinking Tuesday comes after Wednesday is a misconception. This was a mistake in reasoning, namely algebraic reasoning.

Without being there I would suspect that she already knew which was larger (through the discussion that was going on) and then arithmetically saw that r*h resulted in the same conclusion as r^2*h, in this case. Showing that she is wrong with a counterexample will at least teach her that she doesn’t know algebra.:)

37. #### Dan Meyer

June 2, 2013 - 3:10 pm -

Showing that she is wrong with a counterexample will at least teach her that she doesnâ€™t know algebra.:)

A useful first step before you “proceed to teach her algebra.”

38. #### bBob Hansen

June 2, 2013 - 3:50 pm -

“A useful first step before you â€œproceed to teach her algebra.â€

Agreed

39. #### Matt Switzer

June 3, 2013 - 7:45 am -

Was she right?

I think that the answer to this question has been well addressed.

What to do next?

An important idea that arises from her hypothesis is variant and invariant quantities. She correctly identified pi as an invariant quantity but may have also treated the radius as invariant or assumed that if r1 > r2 then r1^2>r2^2.

I tended to not answer “is this right” questions. Instead, I reposed the question as a hypothesis to be tested. She has made a hypothesis about this problem that can then be posed to the rest of the class. Her hypothesis works for this particular problem. The question is whether it is true in all cases. If not, then for what cases is it true and why? This extends the opportunity for the participants to problem solve, model, engage in proof, number sense, operation sense, and do mathematics.

40. #### Mimi Yang (@untilnextstop)

June 3, 2013 - 12:12 pm -

You can use the same two glasses from the picture (if they’re available in real life), and pour in h = 7, diameter = 5.5 into the tall/thin glass, and h = 5.5, diameter = 7 into the other. Not only are the r*h products the same, but the visual “rectangle” cross sections are congruent as well. Then, pour the two liquids out into equally shaped containers to test to see whether they’re actually equal. (I would have the kids vote first, before you run the test.)

Then ask the class to figure out why the volumes are different, via calculations.

“Do increases in both r and h impact the volume at the same rate? When you double r but keep h the same, what happens to V? When you double h but keep r the same, what happens to V? Does the square matter?!”

I’d start with concrete experiment –> table of numerical, sequential example values –> let the kids draw algebraic abstraction/generalization.

41. #### Anders Muszta

June 4, 2013 - 2:15 am -

The I’m-not-a-math-teacher wants to know if knowing that

D1*H1 > D2*H2

is the same thing as knowing that Volume 1 > Volume 2.
_________________________

To compare the volumes she can look at their ratios.

(Volume 1)/(Volume 2) = (D1/D2) * {(D1*H1)/(D2*H2)}.

She knows that the ratio in curly brackets is greater than 1 — for example it could be 3 — and she wants to know if the volume ratio is also going to be greater than 1.
_________________________

If the diameter ratio is greater than 1, then yes, the volume ratio will definitely be greater than 1. But if the diameter ratio is less than 1, then the volume ratio could be less than 1; for example, 3*1.1 is definitely greater than 1, and 3*0.9 is greater than 1, but 3*0.2 is less than 1.

42. #### Ming

June 6, 2013 - 4:58 am -

hi Dan
thanks for the post; always fascinating to hear you reflecting about simple/complex math challenges. Our school board has been moving toward inquiry-based math program; students are presented with a challenge (much like this one) and set to work. I found many instances that i battle with students and try to persuade them to ‘unlearn’ the way they view math; much like what your TED talk in 2010 suggested: formula eager and shortcut oriented students who dont’ see math for more than.. well, a required credit and a passing mark.

I was actually finding myself agreeing with the shortcut (sounds perfectly logical!) then spent the last 2 minutes going over in my head what might have been wrong. The key, to me, obviously falls on the ‘exponent’; and how the base can be drastically ‘increased’ once it’s doubled. Then i began to wonder: will my grade 8s be able to spot the errors in this rationale? To be honest: I am not confident that they will see the flaw in this statement. (actually, most of them, once the formula is derived, will happily accept ‘THE’ one way of doing this type of question without going further. I see this not their fault; but ours.)

I have to say that I really was struck by what you said during your TED talk about ‘conversation’ and ‘math’; too often, in school systems, we worry about covering the ‘math’ and forget about the ‘conversation’.

My challenges, and not sure if it’s shared by other educators out there, is that in an inquiry based learning environment, if the learners (collectively) do not see the flaws in above mention ‘shortcuts’ and are content to accept whatever is given, how do we move forward from here? would love to hear what you would do; such is my challenge with my kids and my staff.

Cheers!