The Algebraic and Order Properties of the Real Numbers

Video Series: Real Analysis for Undergraduates (Complete In-Depth Tutorials)

(D4M) — Here is the video transcript for this video.

00:00
[Music] hi everyone welcome back in this episode
we’re going to review the basic algebraic properties of the real numbers
and we’re also going to review the basic
order properties of the real numbers and so let’s go ahead and get started
and so this uh episode is part of the series um the series is real analysis for
undergraduates complete in-depth tutorials for beginners
and so yeah let’s let’s do some math so let’s uh start off by asking uh
what is a field and so we’re going to start off by asking this question in
here we’re going to start off with a set so we’re going to let r be the set
and we’re going to have two operations on here defined on here there’s going to
be binary operations they’re going to be called addition and multiplication

00:01
and they’re going to be you know just basically what you’re familiar with
already addition and multiplication and we’ll use a dot for multiplication at
first but then we’ll get rid of it after a while and these are going to be total
functions in other words you’ll be able to add together any two real numbers and
you’ll get an output and that output will be a real number and the same thing
with multiplication and now in order to make this field of
real numbers here we’re going to need some properties here
so the first property is going to be the communicative
property of addition which is going to look like this i’ll write it over here
so a1 will look like this so this is the name of it the cumulative property of
addition but it’s going to be like a plus b equals b plus a
and this is going to be for all a b and r
and again this is called the commutative property of addition it doesn’t matter

00:02
which way you add a plus b or b plus a and the second property is called the
associative property so a2 is the associative property so it looks
like this a plus b plus c is going to be a plus b plus c
and this is for all a b and c and r so this um a b c and r
and yeah that’s the associative property
and so the existence of a zero element a three and this will look like this
so there exists a element in r there exists an element which we’re going to
denote by 0 an element 0 and r such that 0 plus a equals a and a plus 0 equals a

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and then this is for all a and r and so a a statement a 3 here gives us
the existence of a zero element and so we we have these two properties
here that are holding and a four here is um the existence of negative elements
so for all a and r every element has a negative so there exists a
negative a and r such that a plus minus a equals zero and minus a plus a
equals zero and so this is the existence of negative elements here
so this is the name of it but this is the actual mathematical statement here
using these quantifiers here now before we go continue let me just say that

00:04
there are multiple ways of defining the real numbers so
one way is to construct them using simpler sets
um and in fact there’s several ways to construct them using simpler sets you
can you construct the real numbers based upon the rationals or even based upon
the natural numbers um and there are several ways to axiomize them also
um and so this is the approach that we’re taking
um in this episode today is we’re going to give a list of axioms which define
the reals but actually with the caveat we’re actually going to give all the
axioms today today we’re going to start concentrating on the algebraic
properties and the order properties of the reals but it turns out that that’s
actually not enough axioms to get to the real numbers so today we begin our
journey to understanding the real numbers and one approach is to give the
list of axioms for field and then add some additional axioms to

00:05
get the real numbers so notice that these four right here
are all basically um um centered around the notion of addition
we got uh addition is communicative addition is associative
and we got the existence of uh additive identity which is the zero
and we have the existence of negative elements and so now we’re going to kind
of mirror that approach but this time now with multiplication so let’s
mirror that with multiplication so now i’m going to write out these pretty much
the same statements here but now in terms of multiplication so m1 will be
a times b equals b times a and this will be for all b a b
for all real numbers a and b and so that’s the cumulative law
so this is la uh law that you should have seen before somewhere like in
precalculus or calculus or whatever and then we have the um

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associative property for multiplication so a times b times a c
is a times a b times a c and this is for all a b and c real numbers and m three
um so for addition here we had the existing existence of a zero element
which was additive identity so now we have existence of a unit element
and so that’s going to be there exists a one a a special number in
the real numbers a special element of the real numbers one
and it’s not um equal to 0 and such that i’ll just put
there exist a 1 in r such that 1 is not equal to 0 and 1 times a is a
and 8 times 1 is a and this is for all real numbers

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so there exists a special number 1 that works with all real numbers a like this
1 times a is a or a times 1 is a and it’s not and it’s not 0.
all right so that gives the existence of a unit element um number 4 m4
existence of reciprocals so that looks like this for all real numbers a um
also with the caveat that it’s not zero so for all a not zero
i’ll put it like this here for all a not zero in r um there exists an element
so there exists an element which we’re going to denote by 1 over a in r

00:08
such that a times one over a is one and one over a times a is 1 and this is for
all real numbers a for all real numbers a
all right and so that’s the existence of reciprocals right there
and now we’re going to have one that’s going to tie them all together so
i’ll put this down here d here so d is the distributive property of
multiplication over addition and so what we’re going to say is a
times b plus c is going to be equal to a times b plus a times c and
and so you can think about that is distributed from the left but also we
can do from the right so b plus c times a will be b times a plus c times a
and so yeah this is called the distributive property of multiplication
over integers now the reason for uh giving uh the list of axioms for a field

00:09
is uh well there’s a lot of reasons but one of the reasons is
that it’s it’s really nice to isolate and narrow down the
basic properties that you need and then from there you can derive more
complicated properties and if everyone is understanding of what the basic
properties are then it makes it much easier to follow and
and you can have a certain level of abstraction because see these are
actually field axioms and there are a lot of examples of fields for example
the rational numbers are a field the real numbers are a field
you have the complex numbers r field and then you have finite fields and you
could have all kinds of fields this is just four examples um rationals reals
complex finite fields but there are in fact all different kinds of fields and so
if you lay out the axioms the very basic properties of an axioms and then start
to approve theorems based upon these axioms then you know that these the

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theorems that you prove will hold for all these different types of fields
and so there’s one uh reason why abstraction is very useful um and so
another reason of course is if you want to change things up like what if you
want to look at what a field is well then you want to take away one of the
axioms you know what kind of structure does that give you what theorems can you
prove you have to go through your list of theorems and say all right i used a4
on this theorem here this theorem here if i want to remove a4 then i no longer
have those axioms potentially so you’d have to look for an alternate proof or
you would in fact may not be able to have any proof at all so in any case
these are the basic axioms for a field and the real numbers is definitely going
to be an example of a field but in order to get the real numbers the full real
numbers notice i’m using just a regular letter r here
when we’re going to get all the final axioms for the reals we’ll denote it by

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this [Music] boldface type here and so this is just the beginning these are
just the algebraic properties of the reals we don’t have quite the real
numbers yet but the real numbers when they’re fully
defined we’ll use these axioms right here these axioms for field
all right so now what are the basic properties of
the real numbers right so what can we prove
using those axioms that we just listed so we’re going to prove a couple basic
properties now please keep in mind that this is not all the properties that you
can prove there are just lists and lists and lists of theorems you can prove
just based upon those field axioms but we should see a couple right so
here’s going to be our first one if z and a are um in r
um and this is going to be our working definition of the real numbers we’re
just taking the field axiom so far and so we have z plus a equals a if that

00:12
happens then in fact z has to be zero so remember um a um we had the um
existence of a zero so remember so this was a
three so i just wanna make sure that you’re remembering this so a3 says there
exists a zero in r such that um zero plus a equals a
and a plus 0 equals a and this was for all real numbers a or for all elements
in r and so you can see that this was a3 here
and this is a stronger statement because what this is saying is that
if you’re adding something to it any a if you’re adding if you’re adding
something to it and you’re getting back the same number here
then in fact this number right here has to be zero right so we already know zero
plus a equals a right we already know that but what we’re saying is if we put

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anything else here well that something else has to in fact be zero so this is
actually showing that not only does zero exist which is what a3 says but this
right here is saying that actually 0 is unique
and so let’s see the proof of this right here
and so what we’re going to try to prove is z equals 0.
and so what we’re going to start off with is z equals
and then we want to show a bunch of steps in here
and we want to end up with z equals zero
and to to to write a proof out correctly we would need to justify all the steps
in in these you know in this work here and so let’s write a proof let’s say
proof and so i’m going to say z equals and
the first thing i’m going to say is that z is z plus 0 and so this is by
axiom a3 so we know we can write z plus 0

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because this holds for all real numbers z is a real number z is in r
so here we go um so that’s why a3 now we can write this zero out as a
um just take any any real number we already have an a in our problem so we
can write it out as an a plus a minus a and so this is by a4 by a4
and now i’m going to use just uh just uh sorry associativity property z plus a
plus minus a so i’m distributing this um and this is the
i’m sorry associative property for addition which was um a2
and now we’re using the assumption that z plus a is a so that’s why i’m using

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this right here so this right here is an a plus or minus a and so this is why
let me get out of the way here a little bit so this is by
assumption or hypothesis and now we have a plus minus a which we know is zero
um and that was by what was that uh a4 right a4 gave us the
existence of negatives so this will be by a4 here there we go
and so we proved z equals zero so we proved this so using our hypothesis here
which would which was right here and the
axioms we now show that z has to be zero
so this right here is very nice property which we’ll use over and over again in
fact you often use this subconsciously that zero is unique real number there’s
only one zero that has these two properties right here
all right so let’s go on to another one let’s go on to b right here if u

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and b not zero are are in r and u times b is b
then in fact u is one and so this is also going to give us uniqueness
it’s going to give us the uniqueness of the unit element so let’s see a proof of
b here this will be proof of b and so we’re going to show that u equals

  1. so i’m going to start with u and i’m going to do a bunch of work and
    i’m going to get equal to 1. and remember we need to justify
    each of these statements in here and here’s what we can use so far we can use
    the axioms of a field and we can use part a because we already proved part a
    we’re not going to use part a because it’s about zero but if we needed to we
    could we could use part a so that’s the rules of the game right
    so we’re going to start off with u and i’m going to say use u times 1 and so
    that’s by m3 and then i’m going to say that 1 is i’m

00:17
going to substitute in a 1 with a b times 1 over b the inverse of b and
that is so just saying that 1 is equal to this that was m4 so i’ll say by m4
and now i’m going to use associativity u times b times 1 over b so i’m moving um
this parentheses right here to these two right here
and so that’s why associativity which is m2 multiplicative associativity
um and now we’re using the assumption u b equals b which is so this is u times b
if you kind of notice this proof is following pretty much the same format as

00:18
the last proof is proof of proof of a and um this follows logically speaking
because the axioms for addition and the action
multiplication uh kind of mirrored each other
all right and so now this is going to be one and so this is by hypothesis
and this is by um m4 and so what we’ve shown is u is equal to
one and we used our field axioms and our hypothesis here
and so what this is saying is that if if u holds into this if this equation
right here holds for any u as long as b is not zero
right then u in fact has to be one so so we know this works for one but in fact
if it if you put any u here well then u has to be one so what is unique
and so that’s what you can get from this statement right here b
in statement b is that uh the unit is unique so let’s look at another example

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here so here’s another example um if a is an r then a times a zero is zero
um and that seems obvious but you know can we prove it using the field axioms
something that you just know in your bones
anything times zero is zero right how do you know that so let’s try to prove it
so in order to prove this i need to show that a times zero zero now what we’ve
proven in part a is that zero is the only element that has these two properties
that this is a and that this is a so we know that the
axiom says there is a zero and part b says there’s only one thing that
satisfies this so if i want a times zero
to be a zero to be v zero i need to show that a times zero
has you know this property here so here we go proof

00:20
we’re going to say a plus a times 0 and so i’m trying to show a times 0 so a
plus 8 times 0 i want that to be equal to step step step i want that to be equal
to a because i know that if i put 0 here that that is true
and i know that 0 is unique and so if i show that 8 times 0 has this property
also then a times 0 has to be equal to the 0 element
so here’s how we can do this so we’re going to say 8 times 1 so a is equal to
8 times one and so that’s by the property of that um you know
one times a is equal to a and that was property number what if i
look up property number one that was um multiplying by one that was the

00:21
existence of so that was m3 so it was by m3 there
so i’ll just say m3 uh i can squeeze it in by m3 by m3 there all right and now
i’m going to use the distribution uh yeah so a times and then one plus zero
and so that was the by d and now i’m going to use that one plus zero
so we know that 1 plus 0 is in fact 1 and so that’s by the existence of the 0
element which was the a3 by a3 and then once a times one um is a
and eight times one is a is by the existence of one
which was m3 so we’re going to use m3 again and so now i’ve shown that a plus a

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times 0 is a and so that means so i’m going to say by part a it follows that
a times zero equals zero there’s one and only one uh zero
element and this satisfy the property of being zero element so in fact it has to
be equal to zero and so there’s example how you can
improve uniqueness because we prove part a and there’s how there’s an example of
how you can use uniqueness which is uh what we did in
part c there so this is a proof of c all right so now let’s do uh one more d
here we go so for part d here um a is not zero and a and b are real
numbers and we’re going to assume their product is one then in fact
b has to be equal to 1 over a the inverse of a now we already know that a

00:23
times 1 over a is equal to 1 because that’s the existence of inverses
and so what this right here in d is going to show is that if you put
anything else here say any b then that b has to be 1 over a so this
is going to give us the uniqueness of multiplicative inverses
so let’s see a proof of this so uh by the way um as we keep following
through these proofs here you you slowly start to get better at them
and you you know you start to understand um how proofs work and you know
how important they are you’re going to see lots of examples counter examples as
you move along uh something that you think is true
think is very intuitive may not be true i mean you can find counter examples to
things or you can try to find a proof that shows it always works
um yeah and so this is just sort of the beginning

00:24
understanding of the real numbers and how they are going to be working
how course in real analysis is going to work everything’s going to be proof
based so even though we’ll work out some examples
and i’ll show you how how to work out some examples as you as you move along
as we move along um but yeah everything’s going to be about writing
proofs that’s going to be the the main part of real analysis is writing these
proofs so here we go we’re going to show that b has to be equal to 1 over a
so here we go i’m going to start off with b and so let’s say here b is um
we’re doing part d now so proof of d and i’m going to start out i’m going to
show b is equal to a so i’m going to start off with b
so b equals and again we want to come up with 1 over a that’s we want
b to be equal to 1 over a so i gotta show some steps in here and what i can
use is the axioms of of a field that we have so far and we can use a b and c if

00:25
we need to any one of those right and so here we go so i’m going to start off
with b is 1 times b and so that’s by the existence of 1
which was let’s see here what was that that was m3 um and
so now we’re going to say that this 1 here is 1 over a times a and
we can put this in parentheses there so that element times a
that’s for the one and then times b and so this was by m4
the existence of multiplicative inverses and so now what we can do is use um
associativity so i’m going to say 1 over a times and then a times b
and this was by m2 associativity for multiplication and so what we can do now is

00:26
um we have this a times b here it’s a one so i’m going to substitute that in
so it’s times one and so i’m going to say by hypothesis here
and now we know that by m3 that this is just one over a so let’s say by m3 here
and so this shows us that multiplicative inverses are unique if you give me any
non-zero real number then anything i put here has to be equal
to one over a so inverses are unique multiplicative inverses are unique
all right very good and so now let’s look at uh part c part e here
so remember this is just a small taste you can create lots and lots of
properties based upon the real numbers and you can prove all these properties
using the axioms and the more properties you build the more fancy your theorems

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can get and any case these are just basic properties no theorems here
all right so here we go a is equal to 0 or b is equal to zero if your product is
zero right and so what i’m gonna do is i’m gonna say case one um
a is equal to zero all right and so then a equals zero or b equals zero holds
and that’s just by point of logic there that’s really nothing fancy there but if
you know that a is equal to zero then you know a equals zero or b equals zero
holds so case two will be a is not zero right so one of these two
cases hold if a is in fact zero then you know that a is equal to b uh a is equal
to zero or b is equal to zero just from the definition of the or
definition of the or means one of the two holds or or perhaps both old but if
you know one of them holds then you then you know the or holds

00:28
so now let’s say case two a is not zero so in this case i actually need to end
up with then b must be equal to zero so let’s try to show that b equals zero so
then so let’s start off with the b and see what we can get
so i’m going to say that b is equal to 1 times b as we did previously so by m3
and i’m going to say that is since a is non-zero right i’m going to
say this is 1 over a times an a times a b
and so we’re saying 1 over a times a is 1 and so that was by m4
and so now i’m going to use associativity again
and this will be a times b so these three steps are the same that we did in
the last problem here but the the assumption is a little different

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here now so this one a times b was one this one a times b is zero
either way i get an a times a b right here that’s a little dot and so this is by
associativity so by m2 and then now using our assumption right
here this is one over a times zero so by hypothesis
i’ll get out of the way again by hypothesis and then now we know that um
if a is in r then anything times zero is zero so anything times zero is zero
and so what we’ve just shown is b is zero so i’ll say by
part c right here so by by c and so in fact b is actually having to
be zero so if a is not zero then b is zero
or you could be in this other case where a is zero

00:30
all right so in other words we’ve shown that a is zero or b is zero that’s
really all there is to that all right so um next up we’re going to
talk about subtraction and division so let me erase this real quick so um
subtraction and division so we’re going to define subtraction
in terms of the binary operation addition and so this is going to create a new
binary operation a minus b which we call it subtraction and here’s how we
actually define it so i’m using that symbol right
there to to denote the definition right there so we know what addition is
and we know what minus b means and so we can define addition very similarly
so we can define the division of a by b as multiplication
and then using e these inverses right here
these additive inverses the sorry these uh multiplicative inverses and these

00:31
additive inverses and of course this is not defined when b
is not 0 because the inverse is right here is only defined when b is not 0 there
so we can also define powers so for example if i want to say a
squared that’ll be just um a times a or a to the third will be a squared times a
and a to the fourth will be a to the third times a and we can just keep
repeating this so a to the n is a to the n minus one times a
um and so you know we can just define all of these right here
and um you know we’ll talk about induction and uh upcoming soon but you
know that’ll be something that we need to know
and we can have a convention that a to the first power is a
and that a to the zero is one um whenever a is not zero [Music]

00:32
and we can have oops can’t see that um i’ll just put it over here a to the
zero is one when a is not zero and we can also say
a to the minus n will be one over a to the n
so to define this notation right here um you know knowing what something in r to
the n looks like uh we can define that then um and so let’s see here
now let’s talk about um kind of getting everyone on the same
page let’s just make sure everyone is ready and set up for this series
i just want to mention the series again real analysis for undergraduates
complete in-depth tutorials you can find the link below in the description
and so what about other types of numbers right want to make sure that you are

00:33
understanding this notation here so this is going to be the set of natural
numbers here and we’re going to identify something in here a natural number
is going to be equal to one plus one plus one because we know
what one is right we’ve defined one and we’re going to say this is one end times
and so we can talk about what natural numbers are
just by adding up a bunch of ones and the similar similarly we can identify
minus n will be minus one plus minus one and we can do that n times as well
and so we can talk about the natural numbers as being a subset of the real
numbers so i’ll put subset here we can talk about the natural numbers as
being a subset of the real numbers so this will be the notation that we

00:34
eventually use the natural numbers are a subset of the real numbers
and similarly we can talk about the integers so now we will include 0 in here
and we’ll include all the negatives right so actually should have z here right
so we can talk about the the negative integers here and in
this way we can make the identification that z is a subset of the uh
real numbers also so then we have two more uh subsets here
right so you know we can write it like this right here starting with 1 2 [Music]
and so on and z would be something like 0 plus or minus 1 plus or
minus 2 and so on and so you know we just got this
identification here you can identify every anything in in in the any integer
with a real number by by taking that approach there and then we can also talk

00:35
about the rationals so something is a rational number let’s say r is a
rational number means r is equal to a over b so it’s a quotient
where a and b are integers so a and b are integers and
we’re going to assume that b is not zero there so
and we can identify this with a times one over b
and that that would be the rational numbers now we’re going to talk a lot
about the rational numbers and we’re going to talk about uh the difference
between the rational numbers and the reals
so one of the things we’ll talk about is repeating decimals and non-repeating
decimals but right now what i what i want to mention is that
there are two things that are important for us to understand first of all every
rational number has a can be simplified in other words we can

00:36
cancel out and make these numbers in lowest terms for example if i take 21
over 7 i can put that in lowest terms so this is certainly a rational number and
we can identify this with a real number and we can say this 7 21 here is 7 times
3 and so the lowest terms would be 3 over 1 or we would just write it as a 3.
and so you can always take out the uh lowest terms
and you can you know just simplify it right now the irrational so that that’s
one thing that’s important to understand is you can always uh
reduce any rational number to its lowest
terms and the second thing to understand is that there are some rational uh
irrational numbers uh that are real numbers right
and so the last one we want to put together is the real numbers here and
we’re still working on defining them but we’re going to say that every real

00:37
number is either rational or irrational but first off we actually need to
understand that there are some rational numbers that um there are some
numbers in the real numbers that are not rational numbers
in other words um just because i call this the irrational numbers and by the
way not everybody uses this notation here so irrational numbers
been around a long time but you may ask why why do we even have them
are there any irrational numbers at all and the real numbers will make up these
two all these sets here these these will all
be subsets of the real numbers but you want to think about the real numbers as
a combination of rational numbers and irrational numbers
and they’re distinct uh distinct sense there
so um let’s work to understand that here so this is our um are there really any
irrational numbers that’s the question now uh why wouldn’t you just have

00:38
rational numbers and that’s really all that you need
or all that you have why why would we you know think that there are any
rational numbers at all irrational numbers at all so um to understand this proof
um we need to think about even and odd and
and reducing rational numbers and lowest terms so that’s some of the things we
need to pay attention to here is rational numbers so rational numbers can
be written as a over b and a over b can be written in lowest terms you can
always simplify that and a and b are integers
and so you have to rely upon the fact that a has to be even or odd and b has
to be even or odd and b cannot be zero of course and you also have to know that
um every integer it has to be even or odd but it cannot be both
so once you understand those basic facts about rational numbers then we’re ready
to understand that there are there are some irrational numbers there are some

00:39
in fact there are actually a lot of them but that’s a discussion for a different
day let’s just work out the fact that there are some irrational numbers so
what we’re going to prove is there does not exist a rational number whose square
is 2. in other words the square root of two is
irrational number so let’s prove the square root of two exists
and so here’s how such a proof would work let’s work this out proof
and so i’m going to assume that assume the contrary that r is equal to p over q
where p and q are in lowest terms p and q are in lowest terms
in other words we’ve simplified it so i’m assuming that r
its square is actually equal to 2 and it’s a rational number where p and q

00:40
are in lowest terms and are integers so we’re assuming p and q are integers
are integers with q not 0. so i’m assuming i have a rational number and
its square is 2. okay so we’re assuming all of this here
so we’re assuming that r is equal to p over q p and q are in lowest terms and
our integers and q is not zero so in other words we’re assuming that r is
rational number and its square is two in other words assuming the contrary we’re
assuming there does exist a rational number whose square is 2.
okay so what’s going to be so wrong with that well you know what could go wrong
with that so what i’m going to do is i’m going to square both sides here
and so let’s just work it out over here so we’re going to say r squared is

00:41
p squared over q squared p squared over q squared
and we’re assuming that r squared is 2. so in other words p squared is
2 times q squared and you know what could be so wrong with
that what why isn’t there a rational number why aren’t there some p’s and q’s
where this equation is true well let’s see here so what this right
here says that p squared is even because p squared is
two times another integer so p squared is even p squared is even
p squared is even now what i’m really interested in though is what about p
is p even or is p odd so let’s look out what if p is and this is just going to
be some scratch work here i’ll put it in brackets like that p is um

00:42
odd so p is 2k plus 1 and so this is just some scratch work
right here that i’m going to do here p is um p is odd
well what is p squared then p squared is you know square this right here that’s
4k squared plus 4k plus 1. and notice that p squared right here
is odd so this is odd and so this little scratch work right
here helps us realize that if p squared is odd i’m sorry if p is odd then p
squared is odd but when our argument up here back to
our proof p squared is in fact even so we know that p cannot be odd because
if p is odd p squared is odd right so knowing that p squared is even and this
equation is true right here now we actually know that p is even

00:43
p cannot be odd if p is odd then p squared is odd
right so this tells us that p squared is even so in fact p has to be even
all right so if p is even so p looks like say 2 m plus 1 for some integer m
for some integer m all right and so what does that say what
does this equation right here say now so this will be
i’ll square this right here so this would be 4 m squared plus 4 m plus 1
and this will be 2. q squared so p is even even

00:44
so p is equal to 2m for some integer um let me bump this over here for some
integer m here we go p is even all right and so
what does it look like then so this would be 4m squared is 2q squared and
if i cancel the twos from both sides i get 2m squared is q squared
okay and so this just is kind of now the same argument that we just had
this is saying that q squared is even so q squared is even
and by the same argument that we just made up a minute ago if if p squared is
even we showed that p must be even so again if q squared is even then q is even
and so what happens is now we have a contradiction
p cannot be even and q cannot be even because that would violate the fact that

00:45
they’re in lowest terms right here so this contradiction right here that p
is even and q is even contradicts that they’re in lowest terms so this
contradiction right here um implies that p and q do not exist p and q
do not exist so p and q do not exist in other words
there’s no rational number whose square is two
and so this is the reasoning right here if you suppose that a rational number
is square is two you’re going to run into these arguments right here you’re
going to run into this contradiction right here that you cannot have it in
lowest terms but we already know that you can write any rational number in
lowest terms so p and q do not do not exist in other
words there is no rational number r all right so
that uh does it for this episode right here

00:46
um i look forward to seeing you in the upcoming episodes and we’re going to get
to some more um actually i take that back we did not do the second part yet
yeah so we need to talk about the order ordering properties yet as i promised
and so let’s do that real quick and sorry about that um yeah so if you have
any questions uh so far let me know in the comments below and i’ll get to them
um talk about uh what are some other theorems or some other basic properties
and then let’s go on and talk about the ordering on on the reels
all right so to get the ordering on the reels there’s a couple of ways to go
about it um we’re going to take the approach that
there’s going to be the subset non-empty subset of r
called the positive real numbers and so this is going to be an assumption
this is going to remember at the beginning i said that

00:47
we had the field axioms but there were a lot of fields that have those axioms so
we’re going to add another assumption right here to the set r here
that’s going to allow us to hone in more uh
in terms of what fields right so not all fields have this property now there are
more fields that have this property um than the reals and so i’ll talk about
that and so we’re going to need even more assumptions so right now we’re
going to go from the algebraic properties to ordering properties which
is what we’re going to do right now and then we’re going to come up with some
some more properties and we’ll finally get to the full definition of the real
numbers so right now we’re just going to
be concentrating on the ordering part of the real numbers so we’re going to
assume there’s some subset and we’re going to call this subset the positive
real numbers and it’s going to satisfy the following properties if any two real
numbers are in p in other words read this as if a a and b
are positive then so is the sum and the second property is

00:48
if a and b are positive then a times b is also positive
and then the last condition is any real number
they’re exact holds exactly one of these three cases either that number is
positive or the number is zero or their number is negative
and so this defines these three properties define what i mean by
positive real numbers you have to have all three conditions to hold to define
what the positive real numbers are now you might be asking wait a minute
there’s no ordering there well that’s true but we can use these three
properties here to define an ordering on the r on the set r and so i’ll do that
here in a second well let me just throw a couple of names
out so this property number three is is actually called trichotomy trichotomy

00:49
it’s called the trichotomy property and then we have the set of real numbers
minus a such that a’s in p in other words the additive inverses of
all the positive numbers this make up this makes up the set of negative numbers
and if something is in p the set of positive numbers you may also actually
also call that a strictly positive number because we know that everything
is going to be either strictly positive strictly negative or it’s going to be
actually 0. and so yeah let’s now define what i mean by the ordering
so i’m going to take any two real numbers anything any any two elements in r
and i’m going to say that a minus b is positive the the difference between the
two and i’m going to write this that a is greater than the b
or we could write it like this b is less than a
so if a is a minus b is positive that means a is bigger than b

00:50
that’s the intuition you write want to have right if a minus b is positive
or if it’s zero if the difference is zero then we’re going to write greater
than or equal to b and similarly we can reverse the
inequality sign there and so these are the two basic
definitions here that we need to define these relations here we have greater
than and symmetrically we have the less than
and we have the greater than or equal to it
all right so now let’s prove some basic properties of these numbers here
and so part a let’s go with a proof here proof proof of a
and so if a is greater than b and b is greater than c
then a is greater than c and that seems very intuitive to you but let’s see how
we can prove it so by assumption a is greater than b

00:51
that implies that a minus b is in p and we’re also going to assume that b is
greater than c and that implies that b minus c is
greater than p so we know both of these are true because we know both of these
are true but how can we show this right here how
can we show that a minus c is in p that’s we need right there that’s we
need to prove and so to show that a minus season p
we need to um you know come up with the fact that it’s going to be
all right so here we go we’re going to say a minus c is equal to
a and then we’re going to use a minus b and then a plus b
in fact i’ll just say here you know following our axioms that we’ve done
before i’ll replace that a with a zero here
and then i’ll just say minus c here for this 0 here i’m going to put in here

00:52
a minus b plus a b with parentheses here so we can really break this down
piece by piece now i’m going to use distributive
property here i’m going to say a plus minus b
with parentheses here and then i’m going to use uh b and then a minus c here
so skipping a couple steps here this would just be a minus b plus b minus c here
and so we know this is in p and this is np
and remember p has the property that the sum of any two things in p must also be
in p so and this tells us that a minus c is in p because a minus b is in p
and b minus c is in p in other words i wrote a minus c to know that it’s in p
i wrote it down as a sum of two things that are already in p

00:53
and let me get other way a little bit so i know this is in p by assumption and i
know this is in p by assumption so now i know that a minus c is in p
and so therefore or i’ll just say thus a is greater than c so a minus c’s and p
because those are so yeah if a minus season p then a is greater than c that’s
it all right let’s go into the next one here so proof of part b here proof of b
and so b says if a is greater than b then if we add a c to both sides
then you know well we can just do that that’s all it’s saying right here if a
is greater than b all right and so what we need to know is that a plus c m minus
b plus c is in p in other words to get this conclusion here

00:54
i need to come up with this right here and if you look at this this has got a
c minus a c in it all right so we’re ready to start now
so um i got to come up with this right here right so we’re going to say here
you know a plus c minus b plus c is in p and this is equal to
a minus b and that’s in p right so you know just just distributing
the minus sign through there that’s in p um
that’s in p right there so i have the difference right here is in p and so
you know thus um a plus c is greater than b plus c um so

00:55
i’ll just start off by saying if a minus b is in p then
because we should have this hypothesis written somewhere in our proof right so
if this right here holds right so this right here holds then this right here
which is a minus b right here right so that’s a minus b and that’s c minus c so
you know you can show all the little steps right there to get that and so
you know that this difference right here and so we
have a plus c is greater than b plus c right there
and so that’s it for b right there what about if we multiply both sides by c
so this is adding both sides of c and what if we multiply both sides by c
so we already know that if c is positive then the inequality is not going to
change the sign right there so here we go for part c right here proof of c

00:56
so if a minus so if a is greater than b then a minus b is in p and
we know c is greater than zero since c is greater than zero
i’ll just say since c is greater than 0 we have c is also in p so then we have
c times a minus c times b which if we factor or you know [Music] distribute law
axiom d that’s a minus c right here and this is in p why is this in p
because season p right uh i forgot to put the c because c is in p
and a minus b is in p and p is closed under multiplication

00:57
remember if you take any two things in p then its product is also in p
so now we’ve shown that this difference right here is in p
so once we’ve shown this differences in p and so c a is greater than c b
all right and so now as you would might imagine what happens if c is negative
then how will this change the ordering right here right
so let’s prove d now we can try to adapt the proof perhaps
so we still have this right here and then now we’ll say since c is less
than zero we have minus season p so then now i’m going to use a um minus c
uh everywhere and so let’s just basically start over here so i’m gonna say here

00:58
minus c times um you know minus c times a times b and i’ll just use the dot here
and so this will be minus c times a minus c times b plus or differently
c times b minus c times a and we can you know
write that out right there this is in p right here now why is this in p so this
is in p and this is in p so this product here is in p
and when we write the product right here when we write it out like that
now we know that cb is greater than ca or um in other words

00:59
the other way to write it is ca is less than cb right there
so um and then you know we have communicative law
so we can write it as a times c is less than c times b
however you want to however you want to write that because we have the community
of law holding right now all right so um one more property here
um actually two more if a is in r and it’s not zero then its square must be
positive right here so let’s look at how proof of e would look and so now
we have the trichotomy property so if a is not 0 then a’s in p

01:00
or minus a is in p so we got those two cases right there so
let’s look at if a is in p if a is in p then a squared which is a times a
is in p right if a is in p then remember p is closed under
multiplication so if i pick two things in p its product would be in p and so
a squared is greater than zero well we can write it like that a squared minus a
minus zero is in p um in fact probably i’d like to write it
like that then a squared which is a a squared minus 0 which is a times a
is in p i’ll just write it back again and so [Music] since

01:01
a squared is a squared minus 0 is in p we know
that a squared is greater than zero uh so that’s just by definition of p
uh or by definition of sorry but um by definition of inequality there or our
definition of the ordering all right so nothing too fancy there
just need to make sure and justify every step right
so let’s justify one is greater than zero so proof of f
one is greater than zero um well we already well just say by e
why e is there really anything else more to it than that
we know one is in the real numbers we have an axiom for that

01:02
um axiom what was it m three gave us the existence of one
and actually it said in the axiom that it was not equal to zero
um now by the definition or or part e a squared is greater than one so one
squared which is one times one just by definition of what the squared means
and 1 is and so 1 squared which is 1 times 1 is greater than 0
1 times 1 is just 1 by def by that that would be m3 again um and so yeah
so i guess we should write it out all right so let’s start over here so since

01:03
1 squared is 1 um by e um 1 squared or just by e 1 is greater than 0.
all right so just by using part e here and axiom m3 all right
so um the further you get away from the axioms the more and more theorems you
prove the less you want to actually be referring to an axiom um
you know once you get to a third chapter in some book then you
usually don’t apply back to the axioms but when you’re first starting out like
this it would be good idea to write down all the axioms used
so how do we show that one is greater than zero
so one way would be to say one minus zero is in p
because that would be by the definition of this right here
so how we know one minus zero is in p so we know that one minus zero is one

01:04
which is one squared and so we can use part e there to get that
um you know just some very basic simple facts there to get this proof right here
all right so there’s really not much more to it than that but just to
practice it out um so let me know in the comments and below
if you want to see some more exercises worked out over these basic properties of
real numbers the algebraic and the ordering there is one more thing i want to do
before we go though and this is an important property it kind of gives you a
feeling for this real analysis thing that we’re going to
talk to so we’re going to use the symbol epsilon for the first time here
and when we write this proof out for this right here

01:05
um you know as i mentioned earlier in this episode you want to not only be
able to write proofs out but also make some examples out
that help you understand the proof so basically you want to think about each
and every proof as a draft you want to think about as
revising the draft over and over again but as you’re making your draft you may
want to have some examples that you’re working out that are for your own
personal use you’re not going to like show anyone your examples because you
really just want to show your proof proof and so everything i put over here is
what i want to show people and everything i put over here i don’t want
to show people it’s just working it out making making sure that i understand it
because very seldom does someone actually write down a theorem and then
just write down a proof and wow that was the first time that was ever proven
usually things like this come along with lots of examples so i want to show you
an example of examples of how to make this proof

01:06
uh or how to understand this theorem right here
so let’s look at this theorem it says if you pick any real number
and you know that that real number is greater than or equal to zero
but you also know that it’s less than epsilon for every epsilon greater than
zero then in fact this number that you picked
is actually equal to zero so if you know it’s bounded above or perhaps equal to
zero but you know it’s less than every every little epsilon now by the way
whenever you see this epsilon it’s not really a
arbitrary symbol that was just chosen for no reason
for example i could have written this theorem as if a is in r and
as 0 is less than or equal to a is less than b and then said for every

01:07
for every b greater than 0 then a is equal to 0. in fact if you just
look at our previous statements that we’ve proven
this would seem more natural way because i used nothing but a’s and b’s in all
the examples that we’ve worked before a’s b’s and c’s right
and so this seems like a very natural way to write it but there’s some
kind of built-in intuition when you’re using that epsilon and here’s the secret
whenever you see an epsilon you want to think about epsilon as being something
small something small something very very small
but it doesn’t have to be and so and so that and so it’s just every epsilon any
real number greater than zero but at the same time your intuition
should should let you lead you to look at the cases where
epsilon is small like point zero zero one or point zero zero zero one right
and so that’s what our intuition is is that epsilon is something small so

01:08
i’m going to look at um some examples over here so let’s take a to b3 and
so that’s certainly a real number and let’s look at this uh here 0 is less
than 3 is less than epsilon and epsilon is greater than 0. so
here here’s an example i said as3 and it satisfies
this seemingly satisfies this property right here but it doesn’t really satisfy
this property right here can i choose a is three
so is three less than everything greater than zero no
it’s you know it’s not less than for example three over two
just cut the three in half for example i mean this is
false in other words a cannot be three a cannot be three because

01:09
three over two is greater than zero right three over two is greater than
zero and this says for all epsilon so if
i just pick one then i should get a true
statement but i don’t and so a cannot be three what if i choose say a is five
then i’ll have this right here for all epsilon greater than zero
and is this really true can we make this true
can we make 5 less than anything greater than 0
and of course not 5 is not anything less
5 is not less than anything greater than 0. for example i can choose epsilon is
say five over two right um five is not less than five over two it’s not true so
this is not true i cannot choose a to be equal to five in fact what this theorem
is saying is that the only way you’re going to get true this is false again
um 5 is not less than any number greater than 0 right so this

01:10
is false again in fact what this theorem is saying is that
you can keep running through your cases all you want you’re always going to get
false the only way to get a true theorem is a must be zero
and so i’m going to use this kind of argument over here these examples over
here to actually construct my argument so here we go i’m going to suppose on
the contrary suppose um that a is greater than zero
so let’s just write it like this assume for a contradiction contradiction that
a is greater than zero in other words i want it to be equal to zero so i’m going
to assume that that’s not true i already know it’s a greater than or equal to it
right so i’m assuming what if it’s strictly greater than zero right so then take
epsilon to be equal to i’m going to define the

01:11
epsilon to be whatever half of this a is so let’s say a over 2.
let’s make that a better a a over 2. right so then we have that a over two
is greater than zero a over two is greater than zero and we have
now there’s no contradiction there and zero is greater than equal to a
is less than a over two and you know that’s a contradiction because
whatever a you give me it’s not less than the a over two
you know no matter what a you give me in fact the only way that you can get this
is if a is a zero so this is a contradiction is a contradiction hence

01:12
you know we cannot have a squared zero a e0
so that must happen right there a must be zero so you know if you if you try to
um take this right here for all epsilon greater than zero
well i can counter prove that if you assume it’s greater zero i can counter
prove it and say it’s false it’s a contradiction so in fact this a has to
be zero right here um and so that is a an example of you know looking at a
theorem and trying to understand how to write a proof
just by looking at some examples right here right that’s not part of the proof
right there this would just be the proof right here and so
that’s going to be it for this episode right here i hope you join us on future
episodes and where we get into the real numbers sequences in series
limits derivatives integration everything you know and love about

01:13
calculus but now we’ll do it with proofs there we go
so i hope you have a great day and i’ll see you in the next episode
if you enjoyed this video please like and subscribe to my channel
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About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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