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### Mathematics and Intuition

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Date: 07/10/2001 at 21:32:46
From: Archer Veledoit
Subject: How to Handle Intelligent Unbelievers

This problem may have more to do with math pedagogy.  It is, however,
a vexing problem, and experience from other teachers may offer some
help.

Certain "puzzlers" in mathematical recreations defy our sense of
experience, leaving you wondering if the answer to a problem can
really be true.

One example is the well-known birthday probability problem, and the
answer that 23 people in a room leads to a 50/50 probability that two
will share the same birthday.

Another is the problem of adding, e.g., "only" one meter to a rope
around the Earth, and determining that the "gap" created between the
lengthened rope and the Earth is about 16 cm. How can it be that
adding such a short length to the rope will result in such a large
gap? Of course, it's easy to show using simple algebra that the result
is a pure value ("amount or rope added"/2pi) independent of any
circumference, so that whether you do it around a superball or around
Jupiter the result will be the same.

My question is how to handle the intelligent non-believers.

I showed this problem to a friend. She had no argument or confusion
over the solution. Her responses were along the lines of, although I
understand the algebra, how do you really know? Has anybody actually
measured it? Aren't there situations in which a mathematical proof
leads to a result that, investigated empirically, proves to be false?
(Of course there probably are, or at least the possibility exists that
there could be.) She just refuses to believe that intuition can be
that wrong, and until somebody actually goes out there and does it and
measures it, she will not be convinced.

One could result to examples. Do it with a pill bottle, then a coffee
cup, then a round table top, ... if the measured results are all the
same, doesn't that suggest that the circumference doesn't matter? Oh,
but those are just a few instances. Such induction can't prove that it
will hold for larger celestial bodies, can it?

How do we respond to those who question that what seems certain might
not be? And is it not a good question, by the way, to ask whether it
(certainty) might not be? Are we really justified in asking others to
toss aside their intuitions in favor of a few sensible jottings on a
piece of paper?

Archer
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Date: 07/11/2001 at 12:00:44
From: Doctor Peterson
Subject: Re: How to Handle Intelligent Unbelievers

Hi, Archer.

We certainly have some experience dealing with this sort of
"unbeliever," as you can see from our FAQ! People have trouble
believing that 0.999... = 1, that -1 * -1 = 1, and so on; our usual
approach is to give them a variety of explanations and hope that one
of them might get through. And it's not just untrained people who have
this problem; mathematicians have stumbled over their intuition in the
past, as to whether numbers can be irrational, whether negative or
imaginary numbers make sense, whether there are as many integers as
rational numbers. We've learned through such experiences not to trust
our intuition.

The first thing we have to recognize is that our intuition can be
wrong. Mathematicians (and the rest of us) need a healthy dose of
humility, because this happens all the time. I suppose one of the
benefits of studying math beyond mere arithmetic is that it can teach
us not to trust our assumptions, or even what seems like sound
reasoning, but to look closely at the logic behind what we believe.
What seems true, not only in math but in all of life, may not be!
That's just part of growing up.

Second, we have to be convinced that math really tells the truth. It's
often been pointed out that airplane pilots flying by instruments can
be deceived by their own senses into thinking the plane is upside-down
in a cloud, when it is really right-side up. If they try to "right"
it, they will be in trouble. So they have to learn that their
instruments really are reliable. Similarly, if we don't have reason to
believe logic, it will not be able to convince us that our intuitive

Of course, one problem here is that math often _doesn't_ tell the
truth - about the real world, that is. Math is based on reasoning from
stated premises (axioms); as long as those are true, and we don't make
mistakes in our reasoning, the results have to be correct. But those
axioms deal with an ideal world, not the real one in which lines are
made of atoms with a finite size, "space" may be curved by gravity,
and so on. So it's easy to find cases where math gives a wrong result
- not because the math itself was wrong, but because it was applied to
an incompletely understood reality, or one that differs in small but
important ways from our assumptions.

The application of math to the real world is based on induction: we
try something repeatedly and see that, yes, our calculations about
circumference do work in the real world, so the assumptions on which
they are based must be accurate. If we measured big enough circles, we
would find that relativity makes it not quite work right; that means
that the world doesn't quite match the Euclidean geometry on which our
calculations are based. But induction does show us that it is a close
enough approximation in normal cases.

Math itself is not inductive, but deductive. Having accepted that the
world is reasonably Euclidean, we have to accept the results of the
deduction that the radius always increases by the same amount. And
that's why math is useful: it makes it possible to find answers
without having to check every possible case.

So I think the best thing to do is not to focus on testing the actual
solution to this problem, but to build confidence in the mathematical
methods by explaining the reasoning in ways that make intuitive sense.
An added benefit is that, in analyzing WHY the math does what it does,
we can gain a better understanding of the whole problem. Let's try

We have a sphere of radius R, with a rope of length 2 pi R around its
circumference; then we add L units to that circumference. What is the
new radius of the rope? The new circumference is (2 pi R + L); the new
radius is that divided by 2 pi, or R + L/(2 pi). This means that the
radius is increased by L/(2 pi), which is independent of R. This seems
a little less surprising, perhaps, than if we solved it with specific
numbers, since we don't have a specific unexpectedly large number to
reject outright; we're forced to look at the algebra. But we can still
question the algebra once we see what it tells us. The next step, of
course, is to check the answer: plug in actual numbers for R and L,
solve for the change in R, and then add that to R to find what the new
circumference will be. It will be L more than the original, showing
that as long as we accept 2 pi R, our answer is right.

Next, if this still seems too weird to be true, we can look into
what's happening. Algebraically, the main point is that 2 pi R is
circumferences:

2 pi (R1 + R2) = 2 pi R1 + 2 pi R2

So if we add something to the radius of a circle, the new
circumference is the sum of the circumference of the original circle,
and the circumference of a circle whose radius is the added amount:

***********
***     |     ***
**        |        **
*          |          *
*           |           *      *****
*            |     R1     *    *  |R2*
*------------+------------*   *---+---*
*            |            *    *  |  *
*           |           *      *****
*          |          *
**        |        **
***     |     ***
***********

***********
****     |     ****
***         |         ***
*            |            *
**             |             **
*               |               *
*               |               *
*                |     R1+R2      *
*----------------+----------------*
*                |                *
*               |               *
*               |               *
**             |             **
*            |            *
***         |         ***
****     |     ****
***********

If you can accept that, the result follows automatically. If you

To convince ourselves of this, we can start with something easier to
understand than a circle, such as a square. Suppose we do the same
(half side) and compare perimeters:

2 R1
+-------------------------+
|                         |
|                         |
|                         |      2 R2
|                         |   +-------+
|                 R1      |   |     R2|
|            +------------+   |   +---+
|                         |   |       |
|                         |   +-------+
|                         |
|                         |
|                         |
+-------------------------+

R2            2 R1             R2
+---+-------------------------+---+
|   |                         |   | R2
+---+-------------------------+---+
|   |                         |   |
|   |                         |   |
|   |                         |   |
|   |                         |   |
|   |                  R1     | R2|
|   |            +------------+---+ 2 R1
|   |                         |   |
|   |                         |   |
|   |                         |   |
|   |                         |   |
|   |                         |   |
+---+-------------------------+---+
|   |                         |   | R2
+---+-------------------------+---+

Here you can see that the additional perimeter in the larger square is
the sum of the perimeter of square R1 and the four corners added,
which is the perimeter of square R2. You can do this with other
regular polygons, and see that it still works. Keep going, and you
have reason to believe the same thing is true for a circle.

Now there's one final way I can see to make the answer seem
reasonable: analyze why the wrong answer seems right, and correct the
underlying misunderstanding. In this case, I think we are used to
proportionality, and figure that adding a relatively small amount to
the circumference should make only a small change in the radius. But
that's exactly what happens! Your 16 cm height is a very small amount
_relative to the radius of the earth_; in fact, as I've shown, it is
proportional to the small change in circumference. It only seems large
because we're focusing on the height above ground, rather than the
distance from the center of the earth.

I don't know that all this effort is really worthwhile just to
convince a friend, but for students it can be important to see math
make sense.

What we're doing here is building a foundation for the unexpected
result. By first making the basics believable, and gradually shoring
up our abstract reasoning with connections to intuitive understanding,
we can make the leap of faith shorter. It may still seem surprising,
but it will start to seem like a natural consequence of things we've
come to know well.

Below I've quoted an answer Dr. Rick wrote recently that touched on
this sort of issue, and I think it may help you. As he points out, one
way to deal with a "non-intuitive" result is to retrain our intuition
so that it agrees with reality. After all, children develop an
intuition gradually, starting with wrong assumptions (things we don't
see don't exist, for example), and having trouble with basic ideas
like conservation of volume (two glasses of milk must be more than one
larger glass, even if they see it being poured from the one to the
others). Intuition is learned. And for that purpose, perhaps an
inductive approach can help - if only to shake up our assumptions and
allow us to accept the mathematical result by seeing our predictions
turn out wrong. But if we refuse to accept what we see, and keep
asking for more evidence, then it's useless to continue with more
examples. Such a person's skepticism goes too far, and until he or she
learns to accept reality, there's not much we can do.

==================================================
Question:

I have heard a problem which I can solve, but I can't really
comprehend it. The "potato problem" goes like this:

You have 100kg (or pounds) of potatos. Like most food it's mainly
made up of water - in this case 99% of its weight is water. The 1%
left is "potatosubstance."

You then have the 100 kg out on a warm day and of course the water
in the potato begins to evaporate. When you take the potatoes
inside, you get the information that now, "only" 98% of the mass of
the whole potato (potato substance + water) is water.

What does all the potato weigh now?

Take a guess before calculating it. Isn't the answer hard to grab?
I would be very greatful if someone could make it more intuitively
understandable (the calculation itself isn't the problem)!

Our naive intuition is that a change from 99% to 98% isn't much, so
the change in weight can't be much, either.

Then we think the problem through. The 100 kg of potatoes contain 1 kg
of "substance." If this 1 kg is 2% of the total after evaporation,
then that total must be

1 kg / 0.02 = 50 kg

To me, the derivation of the equation makes the solution
understandable. What does it mean to make it intuitive? Perhaps it
means to train our intuitions so that next time we encounter a problem
like this, we won't be fooled again. Our intuition isn't going to
solve the problem; we still need to think it through carefully in
order to get a quantitative solution. But at least we can learn not to

One way to train my intuition is to compare the problem with something
I've seen plenty of times in real life. I've painted murals, mixing
the colors I want from poster paint. Often I want a light color, and a
little color goes a long way. I have learned to be very careful not to
add too much color to the white at first. Why? I If I need 1 part blue
to 100 parts white (which is not unreasonable in my experience), and I
put in 2 dabs of blue instead of 1, I need to add another 100 parts of
white to get the color to be what I wanted! I can't tell you how many
times I've ended up with far more paint than I needed, by the time I
got the color right. I've learned that it's better to throw away half
the too-dark mixture, rather than try to save the whole batch by

Do you see how this is the same idea as the potatoes? It's just
reversed. The concentration of potato "substance" is analogous to the
blue paint. In order to *increase* the concentration of the "potato
substance" from 1% to 2%, evaporation must *remove* half the water
(analogous to the white paint).

What we learn from the two examples is this: To make a small change in
the concentration of a minor component of a mixture requires a large
change in the dominant component of the mixture.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
==================================================

Thanks for an interesting question. I'd like to hear back from you if
anything actually helps in this specific situation, or if you have
further ideas in this area.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
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