Hello everyone. In this video, we're going to talk about the following question.
This is going to be an example of nested quantifiers and negating them.
So we're going to take - I haven't done too many examples of applications
of what we've learned, so I'm going to do an application in calculus today,
and it's going to be with limits. "Express the limit as x goes to a of f of x
does not equal L in terms of predicate logic."
So first thing, what is predicate logic? Predicate logic, it means
all those symbols that we've been using in the course, so
"for all", "there exists", "implication", "and", "or", "not" etc.
To get this started…
so this is a "not equals to" question,
but in this case it's far easier to write down the equality
and then go from there.
First we express this
in English.
Now let me mention something about this problem before I go too far;
if you don't know the calculus behind this, that's okay.
There's still something to learn here, but if you'd like you can zoom ahead though to the…
you can zoom ahead a little bit and then just look at the negating of the nested quantifiers
and the actual writing out
of the English sentence; those are the two most important things.
Okay so what is this in English?
So for every
epsilon greater than 0 there exists
a delta greater than 0
such that
if
x minus a is less than delta
for every...for each
real number x, then
f of x minus L is less than epsilon, absolute values…
or magnitude.
Okay.
So here's the sentence, so this is what the
equality limit means in English, "For every epsilon greater than 0 there exists a delta greater than 0
such that if" - so “if”, implication, you should be thinking -
"the absolute value of x minus a is less than delta for each real number x,
then the absolute value of f of x minus L is less than epsilon."
I've worded this a little bit funny.
This should be for every x that satisfies this.
I've kind of worded it backwards, but that's okay we'll word it correctly in predicate logic.
Okay so let's do this now, so let's turn this into a...
let's turn this into a predicate logic statement, okay?
So the first thing, "for every epsilon greater than 0"
that's "for all" epsilon greater than 0.
Let me write this…
now some people won't do this. Some people will do…
so "for all"
you might see this notation.
Just want to mention a couple of notations.
You might see this notation which means for all epsilon in the positive reals.
You might see this notation,
this also means for all epsilon in the positive reals.
I'm not going to specify the domain here just
because I haven't done an example of this in class and I want to do one.
But you can write this in many different ways, so I'm going to write it as "for all epsilon greater than 0
there exists a delta greater than 0…"
that's the next part, there exists a delta greater than 0,
"…such that if
this absolute value statement holds for each real number x…"
So if this statement holds for each real number x
for each real number x where this statement holds is really what this should say maybe I'll just change it now...
so let's see this...
it's also a good example of how English can be really weird, right?
So let's change it, so, "for every epsilon greater than 0 there exists a delta greater than 0
such that for each real number x where this holds then…"
So this is a hidden 'if statement' so it's good that I did this just so that you get another example.
So since I'm saying for each real number x that's the same as "for all"
so let's do that, "…for all x in
the real numbers"
and then, like I said, it's an implication so it's going to be x minus a is less than delta…
so, "…for every x minus a less than delta
we get that
f of x minus [L] is less than epsilon."
So remember intuitively what this says, right, this says that as you get closer to…
so...
f of x gets infinitely closer to L
so long as I can pick - as x approaches a - so long as I can pick a neighbourhood near a,
where as long as I'm close enough to a, then the function is close enough to L.
Okay? That's really the intuitive definition here, but that's not relevant for algebra.
So here's our predicate logic statement for the equality.
Now we have to negate this, okay?
So let's negate this...I'm just going to copy and paste this...
So here we go. Here I'm negating this, okay?
Again the way to think about it is that, if I'm negating a 'for all' statement then
it should be a 'there exists' statement, and if I'm negating a
'there exists' statement it should be a 'for all' statement.
So let's pound through this, and then we get to this implication part at the end,
so let's look at that in a second. So...
Okay so let's do the first three things since that's
easy. Again, we're going to take…
so remember 'for all' becomes 'there exists'
'there exists' becomes 'for all',
and 'for all' becomes 'there exists'.
So let's see what we got.
"There exists an epsilon greater than 0 for all delta greater than 0,
there exists an x inside R…", notice that the domains don't change,
and then we're going to get the negation of the implication…
I started with an 'equals' and I really wanted equivalence...
so let's change the equals…there we go.
Okay, so negated the first three quantifiers and now I have the negation of an implication.
What's the negation of an implication? Well I did this in a previous video but
I pull up the old paintbrush on my little napkin here,
'not' A implies B
that's equivalent to 'not' - an implication, remember, is 'not' A 'or' B, right? Either
A is false and the statement is true, or B is true and the statement's true.
I take this and distribute the negation in there.
I'm going to get A and 'not' B.
That's great. So the negation of an implication is A and 'not' B, so let's just take A and 'not' B here.
A is this
and 'not' B is the negation of that, so
f of x minus L will be greater than or equal to epsilon, okay?
So, let's do that.
So I'm going to copy a previous line and make a couple of changes and that should be it.
Okay so all the qualifiers stayed the same then 'not' - remember so it's A and
'not' B, so f of x minus L
is greater than or equal to epsilon.
And here we have it. So here's an example of
the following question. I scrolled down a little bit, so express
the limit as x goes to a of f of x does not equal L in terms of predicate logic.
So how did we do this? Well first we expressed the
equality in terms of English,
then in terms of predicate logic, and then we negated it.
Sometimes it's hard to just go straight from question to answer.
Sometimes there's a bunch of intermediate steps that will help you get there.
I hope this helps with understanding nested quantifiers and with negating them.
Again, maybe I'll make one more mention now since I'm here,
so again,
for all epsilon greater than 0, so think about it like a...
like an adversary game, right? So your adversary picks an epsilon
greater than 0, you get to pick a delta that may or may not depend on epsilon,
then your adversary says, "Okay well for every x such that you're inside
this nice little area here,
so you're inside this nice neighbourhood around a,
then you must have your function sufficiently close to your limit."
If this is to be equal.
You could think about this game all the time, right, so you can think of the
'for all' is sort of like the enemy picks,
an epsilon that you don't know, you can think of your 'there exists' as you get a little bit of control
as to thinking about which delta you want.
Alright so hopefully this video helps. Thank you very much for listening.