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Hello everyone. In this video, we're going 
to talk about the following question.

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This is going to be an example of nested
quantifiers and negating them.

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So we're going to take - I haven't 
done too many examples of applications

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of what we've learned, so I'm going 
to do an application in calculus today,

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and it's going to be with limits. "Express 
the limit as x goes to a of f of x 

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does not equal L in 
terms of predicate logic."

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So first thing, what is predicate
logic? Predicate logic, it means

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all those symbols that we've 
been using in the course, so

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"for all", "there exists", "implication", 
"and", "or", "not" etc.

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To get this started…

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so this is a "not equals to" question,

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but in this case it's far easier 
to write down the equality

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and then go from there.

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First we express this

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in English.

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Now let me mention something about 
this problem before I go too far;

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if you don't know the calculus
behind this, that's okay.

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There's still something to learn here, but if 
you'd like you can zoom ahead though to the…

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you can zoom ahead a little bit and then just 
look at the negating of the nested quantifiers

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and the actual writing out

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of the English sentence; those are 
the two most important things.

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Okay so what is this in English?

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So for every

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epsilon greater than 0 there exists

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a delta greater than 0

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such that

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if

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x minus a is less than delta

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for every...for each

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real number x, then

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f of x minus L is less than 
epsilon, absolute values…

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or magnitude.

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Okay.

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So here's the sentence, so this is what the

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equality limit means in English, "For every epsilon 
greater than 0 there exists a delta greater than 0

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such that if" - so “if”, implication, 
you should be thinking -

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"the absolute value of x minus a is less
than delta for each real number x,

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then the absolute value of f of x 
minus L is less than epsilon."

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I've worded this a little bit funny.

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This should be for every x that satisfies this.

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I've kind of worded it backwards, but that's 
okay we'll word it correctly in predicate logic.

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Okay so let's do this now, 
so let's turn this into a...

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let's turn this into a predicate
logic statement, okay?

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So the first thing, "for every 
epsilon greater than 0"

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that's "for all" epsilon greater than 0.

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Let me write this…

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now some people won't do 
this. Some people will do…

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so "for all"

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you might see this notation.

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Just want to mention a
couple of notations.

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You might see this notation which means 
for all epsilon in the positive reals.

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You might see this notation,

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this also means for all 
epsilon in the positive reals.

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I'm not going to specify the domain here just

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because I haven't done an example 
of this in class and I want to do one.

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But you can write this in many different ways, so I'm 
going to write it as "for all epsilon greater than 0

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there exists a delta greater than 0…"

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that's the next part, there 
exists a delta greater than 0,

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"…such that if

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this absolute value statement
holds for each real number x…"

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So if this statement holds 
for each real number x

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for each real number x where this statement holds is 
really what this should say maybe I'll just change it now...

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so let's see this...

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it's also a good example of how 
English can be really weird, right?

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So let's change it, so, "for every epsilon greater 
than 0 there exists a delta greater than 0

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such that for each real number 
x where this holds then…"

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So this is a hidden 'if statement' so it's good 
that I did this just so that you get another example.

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So since I'm saying for each real
number x that's the same as "for all"

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so let's do that, "…for all x in

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the real numbers"

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and then, like I said, it's an implication so it's 
going to be x minus a is less than delta…

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so, "…for every x minus a less than delta

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we get that

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f of x minus [L] is
less than epsilon."

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So remember intuitively what this says, 
right, this says that as you get closer to…

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so...

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f of x gets infinitely closer to L

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so long as I can pick - as x approaches a - so
long as I can pick a neighbourhood near a,

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where as long as I'm close enough to a, 
then the function is close enough to L.

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Okay? That's really the intuitive definition
here, but that's not relevant for algebra.

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So here's our predicate logic
statement for the equality.

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Now we have to negate this, okay?

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So let's negate this...I'm just
going to copy and paste this...

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So here we go. Here I'm negating this, okay?

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Again the way to think about it is that, 
if I'm negating a 'for all' statement then

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it should be a 'there exists' 
statement, and if I'm negating a

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'there exists' statement it
should be a 'for all' statement.

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So let's pound through this, and then we
get to this implication part at the end,

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so let's look at that in a second. So...

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Okay so let's do the first 
three things since that's

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easy. Again, we're going to take…

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so remember 'for all' becomes 'there exists'

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'there exists' becomes 'for all',

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and 'for all' becomes 'there exists'.

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So let's see what we got.

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"There exists an epsilon greater 
than 0 for all delta greater than 0,

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there exists an x inside R…", notice
that the domains don't change,

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and then we're going to get 
the negation of the implication… 

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I started with an 'equals' and I
really wanted equivalence...

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so let's change the
equals…there we go.

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Okay, so negated the first three quantifiers 
and now I have the negation of an implication.

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What's the negation of an implication?
Well I did this in a previous video but

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I pull up the old paintbrush 
on my little napkin here,

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'not' A implies B

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that's equivalent to 'not' - an implication,
remember, is 'not' A 'or' B, right? Either

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A is false and the statement is true, 
or B is true and the statement's true.

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I take this and distribute the negation in there.

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I'm going to get A and 'not' B.

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That's great. So the negation of an implication is A 
and 'not' B, so let's just take A and 'not' B here.

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A is this

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and 'not' B is the negation of that, so

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f of x minus L will be greater 
than or equal to epsilon, okay?

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So, let's do that.

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So I'm going to copy a previous line and make 
a couple of changes and that should be it.

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Okay so all the qualifiers stayed the 
same then 'not' - remember so it's A and

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'not' B, so f of x minus L

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is greater than or equal to epsilon. 

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And here we have it. So here's an example of

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the following question. I scrolled 
down a little bit, so express

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the limit as x goes to a of f of x does 
not equal L in terms of predicate logic.

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So how did we do this? 
Well first we expressed the

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equality in terms of English,

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then in terms of predicate logic, 
and then we negated it.

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Sometimes it's hard to just go 
straight from question to answer.

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Sometimes there's a bunch of intermediate 
steps that will help you get there.

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I hope this helps with understanding 
nested quantifiers and with negating them.

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Again, maybe I'll make one more 
mention now since I'm here,

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so again,

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for all epsilon greater than 
0, so think about it like a...

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like an adversary game, right? So 
your adversary picks an epsilon

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greater than 0, you get to pick a delta 
that may or may not depend on epsilon,

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then your adversary says, "Okay
well for every x such that you're inside

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this nice little area here,

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so you're inside this nice 
neighbourhood around a,

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then you must have your function
sufficiently close to your limit."

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If this is to be equal.

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You could think about this game all
the time, right, so you can think of the

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'for all' is sort of like the enemy picks,

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an epsilon that you don't know, you can think of 
your 'there exists' as you get a little bit of control

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as to thinking about which delta you want.

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Alright so hopefully this video helps.
Thank you very much for listening.