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Lecture 7B - Proofs 2

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Introduction to Mathematical Thinking

Universidad de Stanford

4.8 (1,507 calificaciones) | 220K estudiantes inscritos

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.

Destrezas que aprenderás

Number Theory, Real Analysis, Mathematical Logic, Language

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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!

De la lección

Week 5

This week we take our first look at mathematical proofs, the bedrock of modern mathematics.

Lecture 7A - Proofs 122:22

Lecture 7B - Proofs 224:12

Impartido por:

Dr. Keith Devlin
Co-founder and Executive Director

Transcripción

Let me say a little more about that proof.

It's an example of what mathematicians call proof by contradiction.

Proof by contradiction is a general method that works as follows.

You want to prove some statement phi.

You start by assuming not phi.

You reason until you reach a conclusion that's false.

Often, by deducing both psi and not psi for some statement psi.

For example, in the proof that the square root of two was irrational.

We proved that p and q have no common factors, and yet we knew that p and

q were both even.

So we have a statement and its negation.

Well, not strictly speaking its negation, but close enough for these purposes.

This is not an exact rendering of psi and not psi,

but the point is we've deduced two contradictory things.

It cannot be the case that they have no common factors and that they're both even.

A true assumption cannot lead to a false conclusion.

Hence the assumption not phi must be false.

Remember, we began by assuming not phi.

We reach a false conclusion, a contradiction, a true assumption cannot

lead to a false conclusion, so the assumption must be false.

In other words, phi must be true.

If not phi is false, then phi is true.

We begin by wanting to prove some statement phi, and

we're going to end up with phi.

And in the middle, we reasoned by assuming the contrary.

So here's the contradiction proof going on in here.

There's phi.

There's phi.

In here, we're reasoning with not phi.

We start by assuming not phi and we reach a conclusion,

a false conclusion, and conclude that that's false.

Let me say a little bit more about this.

We can look at proof by contradiction in terms of truth tables.

What can we conclude from a proof of theta yields psi where psi is false?

In the case of proof of root two being irrational,

the theta was the statement square root of two is rational.

And the psi was the false statement, that we never actually wrote out in full,

that p and q are both even and have no common factors.

Which is false because there are no pairs of integers which are both even and

have no common factors.

So in terms of proofs by contradiction,

the theta is the assumption you make at the start of the proof.

And the psi is the false conclusion you reach at the end.

So I'm going to use truth tables to understand how it is that

starting with an assumption and reaching a false conclusion

leads to concluding that the assumption was false.

And since the assumption was counter or contrary to the thing we're trying to

prove, that would amount to proving the theorem.

So I'm going to write down the truth table for theta yield psi, or

theta conditional psi.

Write the truth values down.

T, T, T, F,

F, T, F, F.

And we know what the table looks like here.

It's T, F, T, T.

Well, if we've carried out a proof of this, that means this thing is true.

So we can forget that line.

We're only interested in what happens when this thing is true.

We've carried out a proof that this implication, this conditional, is true, so

it's one of these three.

In this case, however, the psi is false.

Where is the psi false?

Here, psi is true.

Here, psi is true.

Here, psi is false.

So this is the only line in the truth table, Which fits this.

We have this being true, which means it's one of these three, and

we have psi being false.

That's the only possibility.

In other words, theta is false.

So when, in a proof by contradiction, you make an assumption,

a counter-assumption, an assumption counter to what you're trying to prove.

And you carry out some reasoning to deduce a false conclusion,

a contradictory conclusion, not a contradiction.

So you've established in your argument that this thing is true.

This says that the argument is valid.

Then the conclusion from having reached a false conclusion, the big conclusion,

the global conclusion, is that your original assumption is false.

Okay, so a little difficult to explain this in words because we have to keep

talking about truth, and falsity, and assumptions, and conclusions, and

assumptions, and sub-assumptions, and so forth.

But if you go through the original proof that square root of two is irrational

once more.

And then think about it in terms of what I just said a moment ago about

proof by contradiction in general.

And then think about it in terms of this truth table analysis.

And hopefully,

you should be able to understand why proofs by contradiction work.

It's a very clever idea and they're used a lot in mathematics.

And I've just noticed that I've misspelt truth table, so let me put the L in there.

Okay, there we go.

That was a fairly lengthy discussion of such a short argument.

But I know from many years of experience that beginners find the root two

proof hard to really understand.

You may think you understand it, but do you really?

Let's see if you can produce a similar one.

Try to prove that the square root of 3 is irrational.

You should definitely try to do this exercise, but

be prepared to spend some time at it.

Go on, give it a try.

I'll say it again, this course is about the process of thinking,

not about getting results.

You can use the thinking abilities you develop in a course like this to get

results in other courses.

And in other situations in your life.

It doesn't matter if you don't get the root three proof.

You'd have benefited from trying.

Proofs by contradiction, which we used in the root two theorem,

are a common approach because they have a clear starting point.

To obtain a direct proof of some statement phi,

you have to generate an argument that culminates in phi.

But where do you start?

The only way to proceed is to try to argue successively

backwards to see what chain of steps ends with phi.

There are many possible starting points but just one goal.

And you have to end up with that goal, and that can be difficult.

But with proof by contradiction, there is a clear starting point.

And the proof is complete once you have deduced a contradiction,

any contradiction.

With such a wide target area, that's often a much easier task.

The proof by contradiction approach is particularly suited to establishing that

a certain object doesn't exist, for example,

that a particular kind of equation does not have a solution.

You begin by assuming that such an object does exist.

And then you use that assumed object to deduce a false consequence of

a pair of contradictory statements.

The irrationality of the square root of 2 is a good example since that states

the nonexistence of two numbers p and q whose ratio is equal to root 2.

Even though there is no cookie cutter template approach to constructing proofs,

there are some guidelines.

We just met two.

Proof by contradiction is often a good approach

when there's no obvious place to start.

And proof by contradiction is a useful method to prove nonexistent statements.

Of course, you still have to construct a proof.

You've simply replaced another goalpost with an unclear starting point,

by a much wider one with a known starting point.

But like Robert Frost's fork in the trail, that choice can make all the difference.

There are a number of other guidelines.

I'll tell you some, but do bear in mind that these are not templates.

As long as you continue to look for templates to construct proofs,

you're going to encounter significant difficulties.

You have to start each new problem by

analyzing the statement that you want to prove.

What exactly does it say?

What kind of argument might establish that claim?

Let's look at another guideline.

How might you go about proving a conditional statement,

one of the form a implies b?

We want to prove a conditional phi yield psi.

Well we know this is true if phi is false, so we can assume phi is true.

Why do we know it's true if phi is false?

Well that was part of the definition of the conditional that we developed

using truth tables.

So to prove it, we assume phi, and deduce psi.

This, of course, confirms the point I made earlier that despite its

strange definition, its counterintuitive definition perhaps,

the conditional really does capture genuine implication.

Because in all actual practical examples when you try to establish

a conditional, what you do is you assume phi and you deduce psi.

And this is genuine implication.

This says that psi is following from phi.

For example, let x and y be variables for real numbers, and prove the following.

If x and y are rational, then x + y is rational.

Okay, this is not a surprising result.

This is nothing deep.

I’m focusing not on the results, but on the method I used to prove it.

And so I've deliberately chosen an example that's extremely simple so that we

can look at the process of reasoning that's involved in proving a conditional.

So step one, assume x and y are rational.

In that case, there are integers p, q, n, and

m, such that x is p/m and y is q/n.

Wherein that case, x + y is p/m + q/n,

which equals pn + qm / mn.

Hence x + y is rational.

Okay, as I mentioned a moment ago, there's nothing surprising here.

It's almost not a proof at all.

It's just really adding two things together.

But it actually is a proof because it has the right structure of a proof.

Here's what we did.

We began by assuming x and y are rational.

We concluded that the sum is rational.

And in the middle, there was an argument to demonstrate that fact.

The argument was actually fairly typical.

And what we first did was take the assumption, and

then unpack the assumption in terms of some useful information.

And once we've done that, we reasoned with that information to get a conclusion

that, in fact, was the thing we were aiming for.

So we write down the assumption, we carry out some reasoning, and

we reach the conclusion.

All three steps are important.

Declare the assumption, carry out the reasoning in a clear,

understandable fashion, and then state the conclusion when you've reached it.

Remember that proofs have two purposes.

One is to convince yourself, and two is to convince other people.

And you may know what you're doing if you don't mention what your assumption is or

what your conclusion is.

Although I'll guarantee from experience that a week from now you'll forget exactly

what you were doing.

So it really is good, even for your own purposes,

to write down what your assumptions are.

But certainly from a communicative angle it's important to begin by stating

the assumption, then to lay out the reasoning in a simple understandable

fashion, and then to state the conclusion that you've reached.

Okay, well that's the most basic method of proving a conditional.

Let me give you a quiz.

Let r and s be irrational numbers.

Say which of the following are necessarily irrational.

And let me stress that word, necessarily.

Number 1, r + 3.

Number 2, 5 times r.

Number 3, r + s.

Number 4, r times s.

And number 5, square root of r.

Now this is a quiz format where I'm just asking you to select the ones that

you think are necessarily irrational.

Which unlike most of the quizzes, where I expected you to be able to answer very

quickly, I'd like you to think a little bit before you answer each of

these because the focus isn't really on getting the answer right.

I mean, obviously I want you to get the answer right.

I'm sure you do too.

But that's not the focus.

The focus is on the reasoning that you need to carry out in order to get those

answers.

In each case, you're probably going to have to carry out one or

two lines of simple reasoning in order to answer the question.

That's the focus of this particular quiz.

In fact, in the assignment that's coming up in assignment seven,

I'm going to ask you to write out proofs of each of the five answers.

So let me stress that the focus is on the logical reasoning.

Okay, see how you do.

Well, the ones I found necessarily

irrational are 1, 2, and 5.

In each case, you use the fact that a rational number

is one that can be expressed as a quotient of two integers, and

an irrational number is one that cannot be so expressed.

And then you carry out a couple of lines of reasoning

to show that r + 3 has to be irrational.

If r + 3 was rational, then r would be rational.

Likewise you would carry out a couple of lines of reasoning to show that if 5r was

rational and could be expressed as a quotient of two integers, then so could r.

Similarly, you'd carry out a similar argument for the square root of r.

The two that are not necessarily irrational are 3 and 4.

In order to show that r+s is not necessarily irrational,

what you would need to do, find examples of irrationals r and s for

which the sum is rational.

Well how about taking r is the square root of 2 and

s equal to, let's say,

10- the square root of 2.

r+ s is equal to 10, which of course is rational.

It's an integer.

We know that the square root of 2 is irrational.

And a very simple argument shows that 10- the square root of 2 must be irrational.

In fact the argument you use to show that this number is irrational

is a combination of the two arguments you used in parts 1 and 2.

Okay, so, r and s are irrational, but their sum is rational.

In the case of this guy, we could take r=square root of 2 and s=square root of 2.

R and s are both irrational, and yet r times s is 2, which is rational.

Okay, how did you do?

Remember that in assignment seven I'm going to ask you to give

proofs of each of the 5 answers.

Well, conditionals involving quantifiers are sometimes best handled

by proving the contrapositive.

What's the contrapositive?

Well, we met that in assignment four.

So to prove an conditional phi

yields psi, what you can do is

prove not psi yields not phi.

That's the contrapositive of phi yields psi.

You reverse the phi and the psi, and you put negations in front of them.

And in assignment four, if you look back, we proved that those two are equivalent.

And we proved that using truth tables.

So let's look at an example of how this might work.

And the example I'll take is this one.

I want to prove that if the sine of an angle theta is not equal 0,

then for all N in the natural numbers,

theta is not equal to n pi.

Okay, there's a conditional.

I'm in the middle of some argument, we'll imagine.

And I want to prove that if the sine of theta is not 0,

then theta is not a whole number, multiple of pi.

Well the statement's equivalent to not the case that for

all in n in N, theta not equal to n pi implies not sine theta is not 0.

Well there's lots of nots so let's clear them all out and

put this into a positive form.

So, in positive form this is not for all n and N,

we're going to get there exists n in N.

And the not is going to move inside, and

it's going to negate this not.

It's going to wipe than one out.

And then I'm going to have theta =

n pi yields not that sine theta is

not 0 means that sine theta is 0.

So this is the contrapositive of this.

And that's the thing we're trying to prove.

Well we know this.

We know that whenever you've got a whole number multiple of pi, its sine is 0.

So this proves the desired result.

Obviously this is a highly contrived, manufactured example.

Again, the focus is not what I'm actually proving, it's the method I'm proving.

I'm picking simple examples so

that we don't have to worry about the mathematical content.

We can just look at the logical reasoning.

And in this case we start out we want to prove something.

We replace it by the contrapositive.

And then we prove the contrapositive.

In this case, the proof of the contrapositive,

we just pull on well known knowledge about the sine function,

that the sine crosses the x-axis whenever you're at a whole multiple of pi.

Okay, let me tell you about one more thing involving proofs of conditionals.

To prove a biconditional, phi equivalent to psi, we generally construct two proofs,

one of phi yields psi, the other of psi yields phi.

And since the biconditional is just the conjunction of the two conditionals,

that clearly amounts to a proof of the biconditional.

Occasionally, it is easier to prove the two conditionals phi yields psi, and

not phi yields not psi.

And I'm going to leave you to find out why this is enough.

Why does this work?

If you look back at the assignments, you should find a clue as to why it's enough

to prove these two in order to prove the biconditional.

Okay, now I'd like you to complete assignment seven.

As usual,

completing an assignment means at least you should attempt all of the questions.

You may not be able to get them all out, at least not at first.

But remember the benefit from working on the assignments in this course is

actually trying them.

Whether you get them out or not is not that important.

Remember my favorite example of learning to ride a bike.

The fact that you don't succeed in riding a bike for many days or

weeks when you're learning doesn't mean you're not making progress.

The same with learning to swim,

you actually get the benefits during the process when you're trying and failing.

And then suddenly you find you can either ride the bike or you can swim.

And it's the same with this kind of material in this course.

Okay, good luck on assignment seven.

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