Prove that if n is composite then there are integers a and b...

Key Concepts

Factorization

Factorization is the process of decomposing a composite number into its constituent factors. By factorizing a composite number into two non-trivial factors, one immediately obtains a candidate pair (a, b) whose product is exactly n, thus satisfying the condition that n divides the product, while ensuring that neither factor on its own equals n.

Composite Numbers

A composite number is an integer greater than one that can be expressed as a product of two or more positive integers that are not equal to one or the number itself. This concept is crucial because the existence of non-trivial factors implies that the number can be broken down into smaller multiplicative components which become candidates for the integers a and b in the problem.

Divisibility

Divisibility is a basic concept in number theory where an integer n divides another integer m if there exists an integer k such that m = nk. In this problem, divisibility is employed to conclude that while the product of the chosen non-trivial factors yields a multiple of n, neither factor taken individually is large enough (or structured in a way) to be divisible by n.

Transcript

00:01 All right, so this problem asks us to prove that if n is composite, then there are integers a and b such that n divides a b, but n does not divide either a or b.

00:13 All right, so first of all, for problems like this, we need to kind of think about our quires, right? they are saying if n is composite, right? if n is composite, that means that we don't get to choose our own handy composite number.

00:32 We got to show this for all the composite numbers.

00:36 But what we need to show is that there are integers, okay? there are integers a and b.

00:43 So we do get to choose this a and b.

00:45 We don't get to choose the n, though.

00:48 So sometimes, if i really don't know where to start, like i said, we can't do this in the proof, but sometimes i will just see if i can glean any sort of intuition about the problem by choosing what i'm not supposed to choose, all right? so for n, let's just choose a composite number and see, well, if i did just choose one, how would i do this, right? so if i just chose one, let's go with six.

01:16 That's an easy one, right? i immediately know it divides to six.

01:20 One, two, three, and six.

01:22 What i'm looking for are integers a and b, all right? we want n to divide a and b, but we don't want it to divide a or b, okay? so i would think, well, what's something that n does divide? what does my six here? what does it divide? well, the easiest one is it divides itself, right? six is six times one, right? six divides six, okay? can we use that? well, because six is composite, i do know two integers that when i multiply them together, they're six, all right? so if i choose for my a and b, if i choose two and i choose three, all right, a equals two, three is my b.

02:25 All right, now i've got that a, b is six, all right? and if we think about it, our n is dividing our a, b here.

02:36 It's dividing six, right? because what this means is six divides six, all right? but six can't divide.

02:52 It does not divide two.

02:54 And six does not divide three.

02:56 Why? well, a and b are too small right now.

03:00 They're not even up to six.

03:02 We can't even really talk about them dividing them.

03:05 They're above zero, but they're below six, our n.

03:08 So we're going to try to think about kind of turning this into a proof right here for any n, all right? so now we need to be more precise, okay? let n be composite...

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