0:00

Well as I stressed in the lecture, you've got to be careful in distinguishing

Â between these two notions. You have a notation for a relationship,

Â something that's true or false. A is divisible by a or b divides a.

Â And you've got another notation for a number, the result of dividing a by b.

Â This makes sense when you're talking about the integers.

Â This actually is not defined for the integers.

Â I mean, for some integers, you get an answer.

Â But this is a s- -- I mean, division, actual division as an operation, is an

Â operation not on the integers. It's an operation on the rational numbers

Â or the real numbers. (End of transcription.) in, in the

Â integers all you have is addition, subtraction, and multiplication.

Â You don't have division. What you can do is say whether you have

Â divisibility. And divisibility is define, defined in

Â terms of multiplication, okay. So it's all about the distinction between

Â the 2, you've got a property, assuming it's true or false And you've got a

Â notation for a number. This is an actual number.

Â Okay. So the actual answer I'm going to end up

Â with is the thing here at the bottom, to the question.

Â This concisely and accurately as you can the relationship.

Â This is the relationship. B divide into a or a is divisible by b if

Â and only if a divided by b happens to be an integer.

Â And so the, summarizing what I just said above there, that's a notation that

Â denotes a rational number, rational number has a different set of numbers.

Â This denotes the relation, B divides A, i.e, there is an integer Q, such that A

Â equals QB. And whenever you're dealing with

Â divisibility, with this notion and you're working with the integers, you have to

Â reduce that abbreviation. To this.

Â There is an integer switch at that. That's what that means.Okay?

Â That means that. That little thing means that.

Â Okay? And then to get down to here in the case

Â where you do have divisibility, then of course, the q that's here is the quotient.

Â I mean we use q to stand for the word quotient anyway, but notice that this says

Â nothing about division. Division doesn't arise here.

Â It's all about. The results of multiplying two numbers.

Â So this makes perfect sense when you're talking about the integers.

Â Okay? So, we're not doing, sort of, arithmetic

Â in the sense of calculating here. You know, obviously, everything that's

Â involved here about dividing one whole number by another.

Â You know, even[INAUDIBLE] in elementary school, it is just division of whole

Â numbers. The focus here however, is on what you're

Â doing within certain systems of numbers. We have 2 systems of numbers here.

Â We have the integers and we've got the rationals.

Â They're just 2 separate systems of numbers.

Â In the case of the integers, you can add, subtract and multiply.

Â In the case of the rationals, you can add, subtract, multiply and divide.

Â But they're different systems of numbers. And so the focus here is on what you can

Â do with the integers. That's what number theory's about.

Â In then in the very last lecture, lecture ten, we'll be actually looking at the

Â rationals and the reals. But that's a different system[INAUDIBLE].

Â Numbers. You can do different things with it.

Â So we're taking a more sophisticated look at elementary arithmetic.

Â But it still is, after all, elementary arithmetic.

Â Okay, let's look at numbers two and three. But the issue with all problems like this

Â is you have to express the divisibility property in terms of multiplication.

Â Remember, divisibility is a property of pairs integers.

Â You can't divide integers, all you can do with integers is add, subtract or multiply

Â them and divisibility arises when you Take this definition, a divides b if and only

Â if there is a q, an integer, such that b equals q times a.

Â So the general strategy for dealing with divisibility or here, seems to be the only

Â strategy, is you replace issues like this, you replace a statement like that With a

Â statement about multiplication. Because the point is that there is no

Â operation here, there's nothing, that, that's not an operation to do, there's not

Â an arithmetical operation on the integers. And remember this is all about the

Â integers. So you have to express it in the en, in

Â the language of the integers. And the language of the integers allows

Â you to talk about addition. Subtraction and multiplication, but not

Â division. Okay, so how do you show that I mean, how

Â do you answer this one? What's the proof?

Â Well, this one is, is sort of immediate because the, the very definition of

Â divisibility explicitly excludes A not being in 0.

Â A not equal to 0. It excludes A being 0.

Â Okay? So it's false, and that's the reason.

Â 9 divides 0? Well that's definitely true.

Â And to that show it's true, you simply express.

Â The definite, you can express it in terms of the definition.

Â So you would have to show that there is a q, look at the definition, you have to

Â show there is a q, so it's at 0, equals q times 9.

Â Well of course there is, 0 itself is one of those things.

Â Okay so that's false. It's true, this one's is false, for the

Â same reason a was false. You're not allowed to, to have a equal to

Â 0 in, in the notion, that includes that requirement.

Â This one is definitely true, and the proof is just write it in terms of this.

Â This is basically what you end up having to write.

Â In each case, if you look at these, that's what I'm going to end up having to write.

Â I'm going to end up reformulating it in terms of the definition.

Â That's really all it involves, just reformulate the statements.

Â In terms of the definition. Sure that the definition is true.

Â Okay. Well in this case it's true because q

Â equals 1 makes it true. In this case, we know that there's no such

Â q. I mean, you could argue it just by since

Â any possible q would have to be less than what, less than 7, say.

Â You could actually, explicitly. If you wanted to prove to that even more

Â detail. You would just let q be all of the

Â possibilities that have a chance of being that.

Â Q equals 1, 2, 3, 4, 5, 6, 7. And 7 times is already 49.

Â So, you're, you're, you're out of it. So, you actually only need to go to 6, of

Â course. So You could explicitly list all of the

Â possible Qs if you wanted to. But that would be so trivial.

Â I think you could just leave it like that. At this level, if this was, if we were

Â talking to kids in the elementary school, we would ask them to maybe list all of the

Â possible Qs and make sure that none of them give you the answer 44.

Â But at this level you, you can just take that out That's been obvious, alright?

Â This one's certainly true. You exhibit the Q, now it means Q equals

Â minus six, Q equals negative six. Ditto here, you exhibit the Q and again,

Â it's, it's negative seven. Here you exhibit the Q.

Â And the q is 8. Here, you need to show that for all n 1

Â divides n. Well, that's certainly true.

Â And the reason is that for any n in Z, n equals n times 1.

Â Right? [laugh].

Â That's trivial. Right?

Â 1 divides everything. Fallen in Z, fallen in N, N divides 0,

Â that's true, because, again, for any N in Z, 0 equals 0 times N.

Â And this one, this is one we've gotta be careful with, because if we're quantifying

Â over all of the integers, that includes Zero itself.

Â And you're not allowed to have zero dividing anything.

Â Okay, that's excluded from the definition. So this is the one you have to be careful

Â with, because it includes. Zero.

Â Its not a case where it goes wrong, but you only need one counter example to make

Â unifunds, to make universally quantified statement false, and that one counter

Â example is all it takes to get rid of that one.

Â Okay, so that one is false. Okay?

Â Now we've done them all. Notice it was just a same pattern, express

Â divisibility in terms of the definition of divisibility, and then each case it just

Â drops right out/g. Because this is afterall just elementary

Â whole number arithmetic, you know, it's not that there's anything deep going on

Â here. It's just that we're looking at it in a

Â somewhat more sophisticated fashion than you did when you were the elementary

Â school. Everything you need to know to solve this,

Â you learnt in elementary school. It's just that we've now a little bit more

Â of a sophisticated gloss on it. Okay?

Â Well, as with the previous example, all you have to do is reduce each of these to

Â the definition of divisibility. Remember divisibility is a property of

Â pairs of integers, this isn't division. It's obviously related to division but you

Â don't have division In the integers. What you can do with the integers is you

Â can add them, multiply them, and, and subtract them.

Â I mean, subtraction just being you know, the inverse of, of addition.

Â But you can't divide them. Okay?

Â But you've come to have a property of divisibility, but to discuss divisibility

Â within the integers, you have to reduce it To, to a discretion of essentially

Â multiplication. Okay so, how would you show that a divide

Â 0, that, that's you got divisible [INAUDIBLE]by a, well.

Â You observe that actually, because of the properties of 0, 0 is equal to 0 times a,

Â so in particular, 0 satistifes the requirement for divisibility.

Â Okay? There is a q in z, so it says 0 equals qa,

Â namely, q equals 0. So, by definition, a divides 0.

Â Okay, similarly, in the case of a dividing a, Because of the properties of, of 1, a

Â equals 1 times a. So again, the definition of divisibility

Â is satisfied. And it's satisfied in this case, by

Â letting q be equal to 1. So by definition, a divides a.

Â Okay. So that was that one.

Â And the rest are essentially the same idea.

Â A divides 1 only if a equals plus or minus 1.

Â Okay, well we've got two implications to prove.

Â First of all, let's assume that a equals plus or minus 1.

Â Then again, all you have to do is show that there is something so it's that one

Â equals Q times A. Well, if it equals plus or minus one, it

Â certainly is right? Conversely That should be if, little typo

Â there. Conversely, if a divides one, then for

Â some q, one equals q a, by definition of divisibility, but if one equals q times a,

Â then the absolute value of one is the absolute value of q a, which is the

Â absolute value of q times[UNKNOWN] the value of a And if 1 equals that, then the

Â only possibility because these are positive integers now, is that absolute

Â value of q is absolute value of a is 1, that's the only way you can get 1.

Â And so if the if the absolute value of a equals 1, then a has to be plus or minus

Â 1. [inaudible] 1.

Â Let me just do one more and then let you to do all of the rest.

Â If a divides b and c divides d then ac demands bd.

Â Okay? Well we know that there are q and r so

Â it's a b is qa definition of divisibility. D equals r c definition of divisibility

Â hence multiplying the two together you've got b d is q r times r c which is when you

Â rearrange them q r times a c. So by definition a c divides into b d.

Â And the others are essentially the same idea.

Â In each case you just reduce it. To the question of, of multiplication

Â through the definition of divisibility. So you never actually do any dividing, you

Â express division in terms of multiplication.

Â And you can do that because division is the inverse of multiplication.

Â Okay? So it, the whole thing is going to work

Â out. So these proofs are always typically just

Â 1 or 2 lines. They're really just a matter of

Â translating what it is you're having to prove into divisibility.

Â So the 1st line of any of these arguments really is just a matter of re-expressing

Â what it is you're having to prove. In terms of divisible, in terms of

Â multiplication. By the definition of divisibility.

Â Okay. Well, that's it.

Â