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Test Flight Tutorial 3

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Introduction to Mathematical Thinking

Stanford University

4.8 (1,468 ratings) | 220K Students Enrolled

Learn how to think the way mathematicians do – a powerful cognitive process developed over thousands of years. Mathematical thinking is not the same as doing mathematics – at least not as mathematics is typically presented in our school system. School math typically focuses on learning procedures to solve highly stereotyped problems. Professional mathematicians think a certain way to solve real problems, problems that can arise from the everyday world, or from science, or from within mathematics itself. The key to success in school math is to learn to think inside-the-box. In contrast, a key feature of mathematical thinking is thinking outside-the-box – a valuable ability in today’s world. This course helps to develop that crucial way of thinking.

Skills You'll Learn

Number Theory, Real Analysis, Mathematical Logic, Language

Reviews

4.8 (1,468 ratings)

5 stars
1,237 ratings
4 stars
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1 star
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JJ

Apr 16, 2019

In the last lecture, based on what he previous taught, the professor give us the definition of limitation which is the beginning of the Calculus. I wish I taken this course before university.

HD

Jun 01, 2018

Great course about mathematical thinking, mathematical logic and proofs. I have a much better grasp on doing rigorous proofs after taking this course. Thanks for putting it together for us!

From the lesson

Weeks 9 & 10: Test Flight

Test Flight provides an opportunity to experience an important aspect of "being a mathematician": evaluating real mathematical arguments produced by others. There are three stages. It is important to do them in order, and to not miss any steps. STAGE 1: You complete the Test Flight Problem Set (available as a downloadable PDF with the introductory video), entering your solutions in the Peer Evaluation module. STAGE 2: You complete three Evaluation Exercises, where you evaluate solutions to the Problem Set specially designed to highlight different kinds of errors. The format is just like the weekly Problem Sets, with machine grading. You should view the Tutorial video for each Exercise after you submit your solutions, but BEFORE you start the next Exercise. STAGE 3: You evaluate three Problem Set solutions submitted by other students. (This process is anonymous.) This final stage takes place in the Peer Evaluation module. After you are done peer reviewing, you may want to evaluate your own solution. It can be very informative to see how you rate your own attempt after looking at the work of others.

Test Flight Tutorial 133:28

Test Flight Tutorial 236:52

Test Flight Tutorial 312:35

Taught By

Dr. Keith Devlin
Co-founder and Executive Director

Transcript

We're turning to the third sample problem set solution, let me tell you up front

that all 10 answers are going to end up getting 24 marks.

This is, if you like, a model solution sheet.

I mean, I don't give model solutions because you can't grade this kind of

mathematics by comparing someone's work with a model solution.

You have to look at the work they've done and evaluate it on its own terms.

Just as you can't do this kind of work by looking for

templates, you can't grade it by templates either.

You have to evaluate it in its own terms.

So I don't give out model solution sheets.

On the other hand, this particular video that you're watching will show you

a good solution for each one.

And in most cases, actually in all cases, a very slick solution.

So before we begin, we know that we're going to be able to congratulate

whoever it is that did this solution.

Now wait a minute, that was me.

I did the solution and I can't really congratulate me,

because I'm the one putting the course together.

Okay, let's just make some,

what I'll do is I'll go through and I'll just make some comments.

So the grading is going to be fairly straightforward.

It's going to be 4, 4, 4, 4, 4, 4, 24 and it's going to look the same throughout.

Now this is a very short proof.

It's certainly false.

We observe that for any m, if n is bigger than or

equal to 2, then this thing's bigger than or equal to 13.

So we only need to show that there's no m for such a case, i.e,

no m for such a case, and that's immediate.

Okay, very short, very sweet.

That's all you need.

Full marks.

So I can just go straight in and

put the numbers 4 in everywhere and a total of 24 for this one.

And this is very much like the solution we saw in sample exam solution two,

except in that case, the answer had an elementary mathematical mistake,

here an arithmetical mistake where the student put 8 rather than 10.

But other than that, this is essentially the same argument.

It's written extremely slickly and

everything's there that you need to be there.

I wouldn't normally expect a typical student in the class to produce something

quite as short and succinct as this, but it's absolutely a perfect proof.

All of the reasoning is there, everything you need there, so very nice little proof.

Looking at number 3 now, I'm not going to put in all of the 4s anymore,

I'll just jump straight in and put the 24, a beautiful proof.

Consider the two cases even and odd separately, very elegant.

If it's even, then that guy works out as being odd and

I've actually stated that it's odd.

I admit that I was the one that wrote this out.

If n is odd, same thing, you follow it through and you show that this guy is odd.

And then you wrap this thing up by concluding that in both cases n squared

+ n + 1 is odd.

And the danger with this particular video is that it's going to go by too

quickly for you to sort of take it in.

So what I suggest you do is as we get to each one you freeze the video and

you just take a look at it and you see why it works.

Incidentally, some of the early research that was done at Stanford onto students

doing MOOCs showed that there was a very strong correlation between the number of

times the students used the pause and rewind button and

the success they had in the course in the long run.

And that's for a whole variety of reasons.

One is, of course, it shows that the student's engaged,

if you have to keep stopping and rewinding.

It shows you're engaged, it shows you're reflecting on the work,

it shows you understand when you don't understand and what's going on.

You know when you don't understand.

And you're taking control of your own learning.

So the correlation between using the pause and rewind button and

success in the course is actually quite high.

And indeed one of the advantages that MOOCs have over classroom lectures

is that you can do that.

It's very hard when I'm lecturing in the class to freeze me and rewind me.

In fact, I wouldn't recommend you try to do that, okay?

I certainly hope you wouldn't try anyway.

Okay, let's go on to number 4.

Okay, so for question 4, we know we're going to give 24,

let's just go through it, m is a natural number by the division theorem.

So stating the reason for why this is going to be the case,

there are unique numbers n and r, then I switch to that.

So actually giving a statement of the division theorem.

So m has got to be one of these four guys.

Two of them are even, so that if the number m is odd,

the only possibilities are the other two.

And that's the proof, very slick argument.

Again, you might want to freeze the video at this point and

look at this, but absolutely perfect little proof.

Everything's there that you need to be there.

Number 5, again put down 24.

I think the only comment to make here is the fact that this solution doesn't

state explicitly that q is an integer as I've mentioned before in the grading.

When you're doing number theory you can assume that any variable that's not

explained or defined denotes an integer.

Integers are the default case.

So simply by looking at the context you can get by without saying it's an integer.

It's not wrong to say it's an integer,

you might want to do that just to play extra safe.

But in this context, that's going to be an integer.

If we were trying to make a claim where we were saying something about there had

been a positive integer, then we'd need to say it.

But the default case when you're doing number theory is that

anything that's not specified can be taken to be an integer.

So there's no need to say that it's an integer, this is fine.

Okay, 24, we know that.

I think the only remark I want to make here, and I've made this remark before,

it's perfectly okay to cite previously established results,

just as it's okay to cite standard results.

So it's okay in the context of putting together a piece of

work to say by the answer to the previous question.

It's like using a level when you're proving a theorem.

So perfectly good, very short, very slick, definitely worth full marks.

Well, the way most people prove this is by induction.

This person's done it a different way and this is very nice.

Look what they do. They say let S be the sum we're

interested in.

Then double S, and you multiply it through by 2, so you get a 2 squared there and

that last one becomes a 2n+1.

Then you subtract the first one from the second one, and everything drops out,

except you've got the 2n+1 at the beginning from the end.

And you've got the -2 from there.

So you've got 2n+1- 2, but 2S- S is just S.

So you've got S equals that guy.

S, this is S, remember, = 2n + 1- 2.

Isn't that nice?

That is beautiful.

In fact, this is so nice, this is so elegant, no need for

the power of induction.

I'm going put a smiley sticker on that.

That's a very, very nice proof, very clever.

It's pleasing to see things like that.

Okay, so let me put the 24 down, and let's see how it works.

So epsilon greater than 0 be given, very nice, starting as one should,

by the assumption we can find an N such that this.

Why are we picking an N with this?

Well, because the whole thing is about multiplying by M, and so

in order to make it work out at the end, we're going to start by dividing by M.

Because we're looking ahead, we're going to say we're going to end up

with an M times something, and we want to end up with an epsilon.

Well, how can you take M times something and end up with an episilon?

Well, there's something you multiply it by, it's going to be epsilon over M.

So that where that came from, we get this by looking ahead to this last step.

When you first meet this kind of structure,

these kind of proofs are hard to do.

But when you've seen a few of them, you get pretty good at looking ahead and

saying, yeah, I'm going to have to divide by M or I'm going to have to take

epsilon by two or whatever it is, and that gives you the limit.

So you almost certainly are having difficulty with this at this stage in

the game, but if you spent a few more weeks looking at this kind of argument,

you'd get very good at looking ahead.

It's not guess work, it's literally looking ahead to see what you want to do.

And if you can't look ahead, just work with anything you want and

then go back and adjust it, and you can just go back and change things, okay?

This is about as slick as you can get.

Incidentally in the last sample exam solution,

this one was written out in words.

I think you'll agree with me,

[LAUGH] given that you're still in this course, that it's much clearer

to write it like this in symbols than to write it in terms of words, with saying

things like an is a distance less than epsilon over M of L, and so forth.

To the modern reader, symbols are much cleaner and

easier to understand than a lot of words.

This is a particularly nice proof.

Admittedly, the person doesn't prove this part,

we've talked about that with the previous sample exam solution.

I think it's okay not to do that, it would sort of been nice to have it there,

but it's okay, given that this is so good.

And let's just, because it is obvious, I mean, these intervene,

arguably you should state that.

Okay, but we're pushing the limit of where that's really necessary now when we get to

this kind of sophistication.

And this is sophisticated, it's very slick.

You let those with the intervals, you observe that this is sort of almost,

it's getting very close to saying that, but it's a little bit beyond that.

You're saying it's a set, everything's within 0 to 1.

So everything in the intersection is going to be in there, and

then this is very nice.

If x is anything in there,

we can find a number such that it's going to be excluded from that interval.

This is very nice because that just nails the fact that nothing

can possibly be in the intersection, so the intersection is empty.

So very, very nice economical, elegant proof.

This is what mathematicians mean when they talk about a proof

being beautiful or elegant.

It's just perfect.

Okay, let me put the 24 down.

So this is almost exactly the same as the previous example except

that this is now left closed.

So 0 is an element of all of these things observed here, but now it's

okay to refer to the previous argument, the one for question 9, because that shows

that no element x strictly between 0 and 1 can be in the intersection.

So the intersection consists only of the point 0.

It really is the same argument as before at that point.

We've just included the initial point in there.

So, again, very elegant, very slick.

And that really completes the three sample solutions.

The first one was really a bit of a low end one.

There were all sorts of things wrong with that.

You often get papers like that.

In fact, you may get a fellow student's paper like that.

It's the kind of thing that you often get when students are really beginning

and struggling.

The second sample solution, that was the kind of thing that you would tend to get

from pretty good students in a class where we've got sort of 74, 75%.

That's probably as good as you'd get.

This solution set, the third one where everything was perfect,

now that's something that only a professional is likely to produce.

I mean, it really was a pro's job and

I would be very surprised to see that coming from a student.

It would be most unusual.

But, there we are.

So having done the three sample solutions,

your next task will be to grade three papers from your fellow students or

at least three papers from your fellow students.

Chances are, they're not going to look as slick and as elegant as this.

They're probably going to be a bit longer.

These are all compressed to fit on the slide, carefully chosen and

edited to make it look neat.

The real world is never as neat as that.

So you're probably going to find it challenging in different ways, but

I hope having gone through these three sample solutions

you've got some degree of confidence as to the kind of thing to look for.

It's clear that this is not an exact science.

By having a rubric, we can get some level of precision.

As I said, the professionals tend to agree by and large, but

you're going to get variations of one or two points absolutely, and

sometimes even more than that.

This is qualitative stuff, it's not quantitative, and assigning numbers

to something that's essentially based on judgments is not an easy thing to do.

Okay, well, good luck on grading your fellow students' work.

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