Well, I won't do all of the examples in, in assignment 7 but let's do a few of
them. Starting with number one prove or disprove
the statement, all birds can fly. Well, we'll, we'll find a counter example.
I mean, that's false, and to show it's false what I need to do is to find the
counter example, and an obvious one is the ostrich.
That's a bird that can't fly so we're going to counter example, okay?
Well now let's look at number 2. Prove or disprove the claim for all x in
r, for all x and y in r, x minus y squared is greater than 0.
That's also false. And again to prove to something that's a
universal quantifier, just like the one above it, it's a universal quantifier, or
in this case 2 universal quantifiers. And to prove that our universally
quantified statement is false. What you need to do is find a counter
example. Well that's one way of doing it at least.
And the most obvious one here, well anything that takes x and y equal will,
will do it. So let's give a specific counter example,
let's take x equals y equals 1. And in that case, x minus y squared equals
0, and 0 is not scrictly based in 0, okay? It came close, right?
If we'd excluded x and y being 0 then we'd have a, then we'd have a positive result,
but uh,it says it's true for all x, y, and r, not a single counter example.
Is all it takes and in this case any pair of equal numbers gives us the gives a
counter-example. Okay, number three prove that between any
two unequal rationals there's a third rational.
So let's let x and y be rationals x less than y, okay?
Then because they're rationals x is p over q.
We can write y is r over s, where p, q, r and s are integers.
And we have to show there's a number between them.
Well the obvious thing is to take the, the mean.
Let's just take x plus y over 2 and if that's rational then, then we'll have
proved the result. Well look, here's the proof that it's
rational, x plus y over 2 is equal to p over q plus r over s over 2, okay?
Which is ps plus qr over qs, all over 2. Which is ps plus qr over 2, over qs, can't
see underneath my hand here, which is rational.
Because it's a quotient of 2 integers, but of course, x is less than x plus y over 2,
is less than y. And we're done, okay?
That's 1, 2, and 3 knocked off very, very quickly.
Let's go on to do, do number 7, the one about square root of 3.
Well just as we did with the square root of 2, we're going to prove it by
contradiction. So we're going to assume square root of 3,
were rational. Okay., then I could write 3 root 3 is p
over q, where p and q are natural numbers. And I can always assume that they have no
common factors. Okay, because if you pick a pair of
integers, a pair of natural numbers that do have a common factor, you can cancel
out. So, you can always express a, a rational
number in that form. Then, if we square that, I get 3 Is p
squared over q squared. So I can multiply across by the q squared
and get 3q squared equals p squared. So 3 divides p squared, but 3 is prime and
if a prime divides p squared, that means 3 divides p.
If a prime divides a pair of numbers multiplied together, it divides one of the
numbers. So 3 divides p, so that means p is of the
form 3 let's call it 3r, okay? So that means p squared equals 9r squared.
So I can take p squared and substituting it back in here to get 3q squared equals p
squared, equals 9r squared. So if I'll forget the middle term now 3q
squared equals 9r squared, let me the factor the 3 out.
I've got q squared equals 3r squared, so that means 3 divides q squared.
But if 3 divides q squared, then just as before with p, that means that 3 divides
q. And now I've got a contradiction, since p
and q have no common factors. And yet, we've just shown that 3's a
common factor. So, there's a contradiction.
Almost exactly the same as the one for square root of 2.
So, if we compare the two to see if the proof for square root of 2.
In the case of square root of 2, we talked about numbers being even.
But if we talk about something like p being even, then that's just another way
of saying 2 divides p. So, really all I've done here is I've
taken the previous proof, the one for root 2 and instead of talking about it in terms
of even or odd, I could recast in terms of whether it's divisible by two or not.
And then the facts, the fact about 2 that we used there, we di-, we said if 2
divides p squared, then it divides p, it used the fact that 2 was prime.
So, I might just as well use 3. So, it's exactly the same proof as before,
except instead of talking about divisible by 2, I'm talking about being divisible by
3. I didn't express it that way before in
terms of talking about even and odd, but even just means that they're a multiple of
two. Okay, well we've, we've moved on.
Let me decide what to do next. I think I'll do number eight that, that
shouldn't take too long we'll, so we'll do number eight next.
Okay, so the converse of a, of an implication of a conditional is
implication in the opposite direction while shop around the antecedent in the.
Consequence but everything remains more the same otherwise unchanged.
So, the, the converse of this is, if the Yuan rises the dollar falls.
We just swapped them around same here. If negative y less than negative x, then x
is less than y, and in the case of c, if 2 triangles have the same area, then they
are congruent. Now in terms of what we do and this is
totally trivial. This is just swapping things around
without doing anything. What I'm really trying to get at is this,
this exercise is in part to contrast it with the contra positive.
And also to, to observe that truth and falsity can change.
In this case, we start out with something that's true, this is a truth to
implication, that's a true implication too.
So sometimes, the converse of a true implication is a true implication.
Let's look at this one. If two triangles are congruent they the
same are? That's true, but this one, if two
triangles have the same area they're congruent.
That one's false, so sometimes the converse of a true statement is true.
And sometimes the converse of a true statement is false.
Now, that's a different situation from, from, from what we found with the, with
the contra positive, but when you swap around the order you can sometimes get
through going to false and you can sometimes reserve truth.
Okay, well that, that's really all there is to that, it was just to sort of give,
give an opportunity to reflect on these things and to recognize that truth and
falsity plays its own game when you're dealing with congruences.
Okay, let's go and do number 11, okay. That was that one about the rational, the
irrational numbers. Well we've already looked at some of these
in the lecture where we've, we've observed that some of them are rational.
Let me see, we observed that this one can be rational.
We observe that this one can be rational and there's an issue about this one being
rational okay. The, the, these were easy, we did those in
the lecture. That was, that was a later issue that
leaves 1, 2, and 5. They're the ones where, where they're
irrational and so we have to prove them this time, okay?
And so the, starting with number one, say what we're trying to show is that it is
irrational, so yes, it is irrational. Okay, and let's see, suppose, suppose that
r plus 3 were rational, okay? That's a plus sign there, then r plus 3
would be of the form p over q, where p and q are integers.
Well then, r would equal p over q minus 3. Which is p minus 3q over q, which is
rational and that's a contradiction. Because r is assumed to be irrational, and
we've shown that if r plus 3 were rational, then r would have to be
rational, and the others are similar. You simply assume the continuity you
express it in terms of p over q. And then you do a tiny bit of manipulation
and you end up showing that the, the number r is rational here.
And the square root of r is that, that, that r is rational in the square root
example. Okay, very straightforward.
Nothing really more to be said about this one, okay?
The only one left now to do on, on assignment 7 is question 12.
So let me just quickly go through question 12.
Well the key facts that you use in, in dealing with all of these are that, if n
is even then, and I actually should, if and only if, then n is 2k for some k, some
integer k. Okay, and this n is odd if and only if n
equals 2k plus 1, for some k. Okay, the even numbers are the ones that
are multiples of 2 and the odd ones are the ones that are 1 more than a multiple
of 2, okay? They're in between the multiples of 2.
So for example, if we do part a if m and n are even then we would have m equals 2k,
for example, we'd have n equals 2l, so m plus n would equal 2k plus 2l, which is 2,
into k plus l, which means it's even. It's 2 times something else do I need to
do 1 more? Well, let me just do 1 more anyway.
Just, just because of I mean, I'm on a roll now.
Let me do a number d, part d. If one of them even, it's 2k and the other
one's odd. It's 2l plus 1, then m plus n equals 2k,
plus 2L plus 1, which equals 2 into k plus l plus 1.
Okay, so it's 2, twice something plus 1. Et cetera for all the others.
Okay, nothing terribly deep about, about these things.
But in terms of, of writing out a proof, it's basically a case of just counting our
odds and evens in this way. Doing, doing a tiny amount of algebra.
Okay, that was a very easy assignment, I think.
Well, it was meant to be an easy assignment.
It's easy for me to say that, right. I've been doing this for years.
If you haven't met this kind of thing before, then it probably isn't an easy
assignment. I've fell into the trap of mathematicians
err of saying something's easy in the same way we talk about trivial, that's a sort
of a term of art that we use in the game. So I everything's easy when you know how
to do it, right? Okay but I was looking ahead to assignment
eight. Which was decidedly difficult for me and
for you guys, okay? Assignment eight, coming next is
difficult.
