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Exercise Prove that, if a and b are non-zero integers with gcd(a,6) d, then the integers & and & are coprime...

Question

Exercise Prove that, if a and b are non-zero integers with gcd(a,6) d, then the integers & and & are coprime

Exercise Prove that, if a and b are non-zero integers with gcd(a,6) d, then the integers & and & are coprime



Answers

Prove that if $a$ and $b$ are integers and $a$ divides $b,$ then $a$ is odd or $b$ is even.

Okay, So in this question, we want to prove that one divides a and a divide zero. So hey, who a divides be even any exists and into just see such that a C is equal to be so In this case, we want to prove that one device a so then way, let's see, is equal to ay a. Then C is into Jack because agent interject, then using the identity of property off multiplication one time CD 21 times Why a which is equal to a so therefore one divide a now to prove that a divide zero See that? See Beezer, then a thought C is equal to a But Sarah now using the zero property of multiplication, this is simply zero So devil, by definition, a divide is here.

Very so here we have that A and B are positive integers I'm just gonna go ahead and soon That is greater than or equal to be. And one of them should be gooder than or equal to the other. We want to prove that the grace common divisor of to a two out of a minus one and two to the power of B minus one is equal to 22 Very So here we have that A and B are positive integers I'm just gonna go ahead and soon that is greater than or equal to be. And one of them should be gooder than or equal to the other. We want to prove that the gray's common divisor that are it's gonna be equal to a might be okay. So that means that the g c d of A and B being able to obesity of being ours also equal to the G c d a b A such that a is equal to be times que plus r okay and by the Oiler uh, division methods. So we we get that the Ukrainian method apologize. The G C D of A and B is equal to the G C d, uh, B and R. Okay. And I also know that our is gonna be equal to might just be right. So it's very ready that there most we can use that in our equation. Right? So for looking at the G c D of two to the power of a minus one and two to the power of B minus one, there's gonna be equal to the common divisor of two to the part of B minus one A my baby. Okay, so that means that the g c d of A and B being able to obesity of being ours, also equal to the G C D a b A might just be right. So it's finally ready that there most. We can use that in our equation. Right? So for looking at the G c D of two to the power of a minus one and two to the power of B minus one, I was gonna be to part of a minus one mod to use the power of B minus one. Okay, so I did. There was used basically this expression to rewrite what we were given. Pulled to the is common advisor of To to kind of be minus one, 22 part of a minus one mod to use the power of B minus one. Okay, so what I did there was used basically this expression to rewrite what we were given. That too, to a minus one mod to be minus one is equal to two to power a mod be well, we know that too. To a minus one mod to be minus one is equal to two to power a mod be, but is equal to two out of B minus one two. However a mod be so I'm gonna use that and put it in our equation above. To get that, the G C D is equal to two out of B minus one two. However a mod be just one. I just won. Okay, well, so then that means that this is equal to the G c d. There's this is equal to if we follow the okay, well, so then that means that this is equal to the G c d. Be the algorithm right where we continuously take the smaller number and okay, the G e g c d ah. Using the module Oh, that repeated calculation we end up with is two to the power of the G C D B minus. So is this is equal to if we follow the algorithm right where we continuously take the smaller number and okay, the G e g c d ah, using the module. Oh, that repeated calculation we end up with is two to the power of the G C D B minus one.

So here we were given A and B are positive integers, and we want to prove that the product of a M. P is equal to the It is common divisor A and B multiplied by the least common multiple. I am V. It's a little let, um, say P one to p k b the primes of the prime factory ization of A and B Thanks. That would give us a in the form of P one to some power times, Pete, to to some power. And that power could be zero, right? If it doesn't exist in there, don't represent be the same way. Of course, their respective powers could be different. So I want to make sure we note that in our notation. Okay, well, so if we have this representation than the greatest common divisor of AM B is gonna be p one times the power of the minimum. Ah, a one and be one then. Then we repeat this process, uh, times the second prime to the power of the minimum A to B to write. And if one of them happened to be zero, then of course, that omen when exists and that's what we would want to see. Okay. And then we also know then that the least common multiple you find similarly right instead is gonna be the maximum. So we have the exact same thing is before I wish I could just copy and paste it, but instead were taking the powers at the maximum. Okay, So do we want to determine the product of the greatest common divisor in these common multiple? So when he used the fat, does Michael know in here? So if the minimum uh, aye, aye. And b I is equal to a I, then he maximum a I am B. I would equal the other one. Right? And in the case where the same, you know, it becomes interchangeable, so it still holds true. And then the converse is true, right? So if minimum is B I, then the maximum it must be a I. So if you want to consider the greatest common divisor of the product and the least common multiple of the product product diversity things then we're gonna get is P one to the minimum. If they won't be one times p two in the minimum all the way. Thio p k to the power of the minimum of a K and B K. It's gonna be that accepts the greatest common adviser and multiply it by in the maximum, which is the least common multiple cancer. Max. It's called the way down to P. K. Okay, well, so we can pull that out. You know, you can rearrange these guys. It's just a bunch of products and gets P one to the power of the minimum, A one B one and also to the power, plus the maximum of the same thing. Okay, we'll do that all the way down to get P Kid to be out of the minimum, plus the maximum. And it's not quite enough room, but who beacon for that. Okay, well, because of the note from before that, if one of them was the middle of the others, the maximum this really just becomes p one a one plus B one times p two a two plus B two all the way to P K A k A plus B k. All right. Well, what is this? Next we can break this up into p one a one times p to a two. Hey multiplied by P. One b one times p to be too early a crossing, and this it's just a and this other sections just be right. So this is really just the product of a Times B. Okay, so you've shown that the greatest common divisor of the product times at least common multiple of the product that's simply equal to the product of A and B together.


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