Definition of Divisibility (and Why We Need It)

Video Series: Divisibility in the Integers (Step-By-Step Tutorials for Number Theory)

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

00:00
in this video you’ll learn the definition of divisibility and why you need it
did you know that one of the most basic concepts in number theory is that of
divisibility the precise definition of a number is to
be divisible by another number is essential for defining concepts such as
prime numbers so let’s see it hi everyone welcome back i’m dave
this video definition of divisibility and why we need it is part of the series
divisibility in the integers step-by-step tutorials for number theory
so let’s see what we’re going to cover today we’re going to talk about
i’m first going to talk about the why why we need the definition
then i’m going to talk about the actual definition
then we’re going to give examples and then we’re going to talk about some
other ways of using this definition here and then we’re going to talk about how
many divisors you can have and then we’re going to work on some

00:01
exercises together so let’s get started so the y first
why we need uh divisibility the definition so notice that the sum and difference
and product of any two integers is an integer so for example two times three
these are integers this is another integer
or three minus three that’s an integer times four that’s minus twelve which is
an integer so you can take the product or the sum
or the difference of any two integers and you get back an integer
however the same cannot be said about the ratio of two integers so for example
if i take the integers two and three the ratio is not another integer so this is
not an integer and so by the way um the integers
are the numbers zero plus or minus one plus or minus 2 plus or minus 3 and so on

00:02
um so this is the integers here and if you take any two integers and
find the ratio well sometimes you might get an integer
back for example minus 4 over 2 is minus 2 and that is an integer but it’s not
always the case if you take two integers then it’s an integer
so we have the problem of divisibility or we have the problem of taking ratios
is not not closed under taking ratios we simply cannot divide any two integers
and get an integer so the study of the integers is to a
great extent the study of divisibility so we can put it in various other terms
but this is a very crucial part of the study of number theory
okay so let’s look at the actual definition now
so we’re going to say that a and b are integers that’s our starting place

00:03
and we’re going to assume that a is not zero we’re not going to define
what a divides b means when a is zero so this is very beautiful notation in my
opinion a divides b and we’re just going to use a vertical
bar that’s so efficient is so meaningful to me but any case
this notation right here represents these words right here so here’s notation
and here’s the meaning of it there exists an integer c such that
is equal to a times c so b here is equal to a times c now
notice this notation is a vertical bar so that’s very different than

00:04
this kind of bar here or sometimes you’ll see it like this with a slant to it
like you might see two over three or two or three like this
so this is different notation here this is a vertical bar this is not a vertical
bar it’s not a vertical bar this is referring to division
which is not the topic of this video this video is about uh divisibility
so we say a divides b and here’s the notation for it and
here’s the words for it so these three symbols here one two three symbols right
here represents this whole sentence and that’s what i love about these three
symbols here it’s so efficient way of writing this whole sentence out
it’s just so very intuitive to me and i hope that once you get past this
video here that it starts to getting getting that way for you
in any case let’s look at some examples here now so here are some

00:05
examples of divisibility so we’re going to say 3 divides 6. now why is that true
or in fact let’s just ask true or false true or false so 3 divides 6 would
remain i need to replace the a with the three and a b with a six
so does there exist an integer c so let’s just write this out so three
divide six means exists an integer there exists an integer c such that
six equals three times c so can you figure out what the c is
and the answer is yes c is two and so we’ll actually say
much shorter here three divides six is true because 6 equals 3 times 2

00:06
and 2 is an integer so 2 is an integer you can say 2 is an integer
so we could say three divide six is true because six is equal to three times two
and two is an integer this last part is very important here whatever you
multiplying by here is an integer all right let’s look at another example
what about 6 divides 24 true or false so we’ll play the same thing again
do you think it’s true or false so 24 is equal to 6 times what 4
where 4 is an integer so yes 6 does divide 4. what about 8 divides 0
so 8 divides 0 would mean we could say 0 is equal to 8 times something where

00:07
that something is an integer where c is an integer do we know of any c where
c times a to zero the answer is yes it’s zero
where zero is an integer so this is true so c is true
now you cannot ask what is 0 divide something else because we didn’t define that
all right so what about minus 5 divides minus 55
true or false let’s see minus 55 is equal to minus 5 times c
where c is an integer can we find this c
if we can find the c then we can say yes how about c is 11 right so c is 11
where 11 is an integer this is a true statement so this is a true statement

00:08
one more minus 9 divides 909 true or false so let’s see here we have 91909
is equal to minus 9 times c where c is an integer where c is an integer
what would that integer be so we can divide by 9 and get 101 right so 101
where 101 is an integer that’s a true statement so e is true
all right other definitions so let’s look at other definitions
there’s other ways of saying a divides b so let’s remember our definition here

00:09
a divides b remember is shorthand for um there exists an integer c
such that there exists an integer c such that b is so b this the one right here
b is equal to a times c so remember these three symbols right here
represents this whole sentence right here so we can also say we can also use
instead of the word divides we can say a is a divisor of b
so this is when you when you say a is the divisor of b
you also mean these three symbols and you also mean this sentence here so you
can say a divides b or you could say a is a divisor of b
either one of those is fine another way is a is a factor of b
so if you try to factor b you get a in here so a is a factor of b
is another way of saying this one a is a factor of b is another way of saying
this one whenever you say a is a factor of b you have these three symbols and

00:10
you have this sentence here they’re all the same there’s two more ways
b is a multiple of a we can say b is a multiple of a
all of these mean the same thing as same as and we use the same notation
for that one more way b is divisible by a so we say b is
divisible by a and we write these three symbols and we have this sentence here
there exists an integer c such that so here’s four different ways of saying
a divides b so now we actually have five all together
so this is the definition of divisibility and there’s multiple ways
of of of saying it all right so um let’s look at how many how many divisors
can you do so any integer in except zero has just a finite number of divisors

00:11
so why is that true so for example if a divides in where a and n are
positive integers right so then we have n is equal to a times some integer k
in this case i use a k instead of a letter c so
once we know that that that this n can be factored with an a in it
then what we can say is that since k is positive also right a is positive n is
positive that means that k must be positive also we can see that a times k
has to be greater than or equal to a because k could be one
but generally speaking a multiple of a here is going to be larger than a
under the assumptions that a and n are positive now i mean
you’re going to have absolute value with
the 2 here because you’re going to allow for the cases where they’re both
negative so let me give you an example like that so let’s say n is

00:12

  1. and i and i say here that there’s going to be a finite number of divisors
    so we can actually list them all right so one way to write it as is minus 1
    times -28 and one in 28 right so there’s there’s one way and
    then we got we got four and seven and we got minus four and seven
    so each time we work this out with positive integers
    like these two right here or these two right here we we also get the
    double we also get the negatives oops negative seven here so
    that’s why we’re saying two times the absolute value of n
    it’s bounded by that it can’t be that many we’re obviously not going to get 28
    here because uh you know for example three doesn’t
    doesn’t work you cannot factor a three from a 28

00:13
so there’s going to be less than 28 of them but generally speaking you may have
all the factors and so two because you’re gonna have positive factors and
you’re gonna have negative factors here okay so let’s look at some exercises now
this is where it gets fun so here’s up the first um exercise here
we’re gonna show that the sum of two even or two odd integers is even
we’re also going to show that the sum of an odd and an even is odd so actually
number one has three different parts to it so i’m gonna say this is part a here
so it says show that the sum of two even is even right so let m and n be even
be even integers and what operation are we asked to do the sum

00:14
right so m and n are even integers right so what does that mean so there exists
[Music] integers s and t such that so m is even so in other words you can
factor a two out and n is even so it’s two times t
right so to say that m is even it means it’s two times an integer s
and to say that n is even is two times an integer t
so s and t are integers i said that right here s and t are integers so m is
even and n is even now let’s look at the sum we’re trying to do the sum of two
even integers is even so let’s see here then what is m plus n it’s 2s plus 2t
and you can see we can factor out a 2 2 times s plus t

00:15
now notice that s and t are integers so is s plus t where s plus t is an integer
is an integer hence m plus n is even so this would be how we would write out
a proof we’re taking two even integers m and n
and we’re saying that they’re even so we can write them out as 2s and 2t
now when we look at the sum we write out the 2s and the 2t and we factor
and then we notice that this is also an integer so for n plus n to be even it
needs to be 2 times an integer so i say very clearly that this is an integer and
so n plus n is two times an integer so n plus n is even
so we did that part of number one here the sum of two even
is even now let’s do the sum of two odd integers is even so we would need to

00:16
adapt this to say that two odds are even so let’s do that um let’s do it right
here though so we can just kind of compare the two and so because if you’re
just learning how to do this you might want to just compare them so
here let’s say here let m and n be odd integers
okay so what does it mean to be odd so there exists integers s and t such that
so m is odd right so it’s two times an integer plus one and
n is odd also so it’s two times this t plus one okay so
m is odd and n is odd and the way to say the m is odd is you have this s and the
way to say that n is odd is you have this t and they’re both integers so then

00:17
we’ll look at the sum looking at the sum of two odds so they
don’t have to be the same odds so i have two different odds m and n
then m plus n is two s plus one plus two t plus one
which if you work it out is two times s plus t plus one
right because it’s going to be a two s a two t and a two
when i factor out that two i’m gonna get two times one so i get a two where
this is an integer right here s plus t plus 1 is an integer
and that’s exactly what we need hence m plus n is even
hence n plus n is even so the proof follows very much the same as this one
right here the only difference is here we have these plus ones here because

00:18
they’re odds okay and then we have a part c here also
show that the sum of an odd and an even is odd
so that proof will look also very much like a and b but let’s go ahead and get
some space down here so we’ll say that’s part c here part c
i said show that the sum of an odd and an even so i need an odd number and i
need an even number so let m be an odd be an odd integer and n be
an even integer then there exists integers s and t such that so m is

00:19
odd so i’m going to use this s so m is equal to 2s plus 1 because it’s odd
and n is even so i’m just going to say it’s 2 times an integer t so then
m plus n right we’re looking at the sum of an odd and an even so i’m looking
at the sum of an odd and an even so m which is 2s plus 1
plus the n which is two t which is now i can only factor out the
two from the s and the t and i’m left with the one here where s plus t
right here is an integer so we have exactly what we mean by an odd so
where s plus t is an integer hence m plus n is odd

00:20
all right so there we go there’s part c there so that’s um a exercise number one
let’s go on to look at exercise two so show that the product of two odd
integers is odd so again uh this one has multiple parts
just because it can be worded so easily as multiple statements here
so let’s say here this is number two now show that the product of two odd
integers is odd so um let’s look at this part a so we’ll call this here uh 2a
and let’s say so let m so what do we need here the product of
two odds so let m and n be odd integers so there exists integers s and t

00:21
such that m is odd to s plus one and n is odd so two t plus one
so they’re both odds we’re taking the product of two odds is odd so here we go
then m times n will be two s plus one times two t plus one
and if we work this out this will be two s so this will be four s t plus
two s plus two t plus one times one is one and now we can factor out a two we’re
going to get a two s t plus an s plus a t plus one
see what i did there this has a two in it this has a two in it this has a two
in it so i factor out the two from all those i get 2 times an integer plus 1 so
i have to say that this is an integer where 2 s t plus s
plus t is an integer and it is an integer right

00:22
i mean s and t are integers we said that very clearly so s time t is an integer
s is an integer s is an integer so add them all left we get an integer so this
is all an integer so we get two times an integer plus one so hence
m times n the product is odd okay so now we got so we did the product
of two odds is odd and also show that the product of two integers is even
if one of them is even so one of them is even let’s do that now so 2b so

00:23
if either if either or one of them is even so let be an integer and let n be
two times t for some integer t then what is what is the product here m times n
so that’ll be m whatever that is and maybe even or odd we don’t know but n is 2t
and so that would be 2 times mt where m times t is an integer so m is an
integer and t is an integer so m times t is an integer hence
so it’s two times an integer so m times n is even
and it you know we didn’t say anything about the m
so if if if if either or one of them is even so if this right here or m could be

00:24
even two but if just one of them is even the product is even there’s number two
all right so that if a and b are positive integers and a divides b
then a is less than or equal to b so perhaps we can look at that right here so
assume that a is in fact strictly greater than b
what would be so wrong with that assume that a strictly bigger than b
now a and b are positive integers so since a and b are
are positive integers they’re both positive and a divides b we see that um

00:25
k is positive where um b is equal to a times k and this is um
greater than or equal to a but actually let’s think of it like this k is
positive because of this statement here right a
and b are both positive so k has to be positive so if k is positive then we can
now multiply both sides by k here and get a to the k a times k is greater than
b times k so this will be greater than b times k
so this doesn’t make any sense right here b is greater than b times k
so where k is positive integer k is a positive integer okay so hence
in fact a must be less than or equal to b so i think that kind of gives you the
idea there behind that okay so um number four here

00:26
show that the square of ever uh showed that the square of every odd integer
looks like this so here we go number four so the square of an odd right so let
m be an odd integer [Music] so there exists so there exists an integer
s such that [Music] m is odd right so m is equal to two sum
two times some integer plus one and now we’re looking at the square of
an odd so i want to square this odd so um or i’ll say then

00:27
um the square so m squared is two s square two s plus one squared which gives us
four s squared plus four s plus one right and so the square of an odd
so we’re getting 4s squared plus 4s plus 1. now we have possibility here for s
in order to get the 8k we’re going to have to break this down further and
ask is s even or is s odd so let’s look at this case over here so case s is even
then s is equal to two t or t is an integer and so then what will happen to m

00:28
squared so this will be four and then two t squared plus four s which is two t
plus one and so then this will be four how many
how many we have out of this four times four so we’ll have 16 t squared plus
eight t plus one and can we factor an eight out of here and we’re going to get
two t squared plus t plus one so we’re going to get a two t squared
we’re going to get a t left here and then we’re still stuck with this one
left over so we look like eight times an integer plus one
and that’s what we need right here so it looks like this looks like a times
integer k plus one and we’ll just say here that in this case k is 2 t squared
plus t which is an integer so we got 8 k plus 1. now what about the case when

00:29
when when the square here this s here is odd so case s is odd
so that’s the case now so then s is equal to two t plus one or some uh integer t
and so now what is m squared become m squared is
four times this two t plus one squared plus two time uh plus another four
so that’s uh two s times one ten times two so it’s four s
so four and then two t plus one and then finally the plus one here
so i just plugged in two t plus one here and here and i get the m squared looks
like this so let’s see if we can reduce this
so this will be four and then this is four t squared plus four t plus one

00:30
plus an eight t plus a four plus a one eight t plus four plus one
okay so now what do we get here 16 t squared plus 16 t plus four plus eight t
plus four plus one and what is everything that has an eight
so it looks like we’re going to get an eight out of all this and what do we get
here 2t squared sorry let’s go here with this is a 8 times a 2t squared plus a
so this will be 24 so this would be 3t because i’m factoring out an 8 and then
4 and 4 gives us an 8 so i’m going to factor out an 8 so i got a 1 here left
and then i got this one here left here also let me move me up here
you can see that a little better so now this is our k on this problem right here

00:31
so k is two t squared plus three t plus one and that’s an integer there
there we go so there we go this um m squared here looks like an 8k
plus one in the case where s is odd so we got two
um cases here because we looked at this right here we could only factor out a
four and we couldn’t factor out an a but it asks us to factor out an eight here
all right very good next one okay so um this time we’re going to be looking at
a product of every two integers of the form 6k plus 5 is of the form 6k plus 1.
so we got two integers of this form right here so let’s go here

00:32
let’s move me down here and let’s go with number five here so let m end in b
of the form the of the form six k plus five
so m is equal to let’s say six s plus five and n is six t plus five
so i’m saying that m looks like this six k plus five so there’s some integer
where i have six s plus five and n also looks like 6 k plus 5 so
there’s some integer t here so 6 t plus 5 where s and t are integers
okay so i’m showing that the product of these two
looks like six k plus one so let’s look at the product then m times n
which is six s plus five and this is six t plus five

00:33
and so let’s see what are we going to get here 36 st plus 30 s plus 30 t plus 25
and now we need to have a 6 factored out of all this and look at the remainder
there so i can factor six here six here six here
and this 25 here i’m going to think about it as a 24 plus one
so i’m just thinking about that as a 24 plus one there
so we’re going to get here a 6 and here i got a 6 left over so 6 s t
plus here i have a 5 5 s plus a 5 t and now this 25 think about it is 24
plus 1 for the 24 part i’m going to get a plus 4 so that gives me a 24 there
and then i got a 1 left over so there’s my 6 and there’s my k
so k in this case is six s plus five s plus five t

00:34
plus four and that’s an integer so there’s our integer right there so
now we show that m times n also looks like six k
plus one where this is an integer right here
all right very good so there’s number five now number six
so number six is show that any integer of the form is also of this form right
here and then i’ll talk about that second part there okay so number six here
so let m have this form right here six k plus five so let m um have the form
six k plus five right so there exists an integer let’s call it s such that

00:35
m is equal to six s plus five okay very good
so how can i get a 3j plus 2 from here so all right so then
m is so this 6 here i’m going to factor out a 3
and this 5 i’m also going to factor out a three
but um in fact let’s just write it like this here let’s say it’s um 3 times 2s
that gives me my 6s and this 5 let’s write it as 3 plus 2.
so out of these two right here i’m going to factor out a three so m is
three times two s plus one plus two and so now our j here that we need the j
is two s plus one so j is two s plus one and this double check is two s plus one
an integer and the answer is yes two s plus one is an integer because we said s
here is an integer so we know two s plus

00:36
one is an integer now but not conversely
so if you take something that looks like 3j plus 2
how do we know it doesn’t necessarily look like 6k plus 5 well all you have to
do is to show a counter example so the conversely part is
what if i choose j is zero so if j is zero then three j plus two is just two
and three j plus two does not look like six k plus five so two is not
of the form six k plus five now if you need some convincing of that
try to find the k what is 6k plus 5 equal to 2 is there some k
is there some integer k well you move the 5 over you get a minus
three and then you divide by six that’s not going to be an integer

00:37
so we’ll just say this right here two is not of the form
six k plus five so it doesn’t work backwards if you start off with six six
k plus five then you can get three j plus two
but not conversely just because you look like three j plus two like this one
doesn’t mean you’ll look like six k plus five all right very well so
if you have any questions or comments i hope that you use the comment section
below please like this video if you enjoyed it and don’t forget it’s part of
the series divisibility in the integers step-by-step tutorials for number theory
link is below in the description and i want to say thank you for watching and
i’ll see you next time if you like this video please press this
button and subscribe to my channel now i want to turn it over to you math can be
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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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