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Let G be Growp Gd Z(G) Le its Center 0). 2(6) is normal subgrup 6) . Jf cyclic then G 6Yzo) T6 abelian...

Question

Let G be Growp Gd Z(G) Le its Center 0). 2(6) is normal subgrup 6) . Jf cyclic then G 6Yzo) T6 abelian

Let G be Growp Gd Z(G) Le its Center 0). 2(6) is normal subgrup 6) . Jf cyclic then G 6Yzo) T6 abelian



Answers

Let f be an ordered field and x,y,z in F.
Prove that if x<0 and y<z, then xy>xz.

One of the key points about electromagnetism is the fact that there are no magnetic multiples and there are a couple ways to write this down. One is that the divergence of B has to equal zero. And so that's why the magnetic fields that you look at. Um either loop around um or they leave one area come out another area. They don't terminate on like a positive negative charge, a positive charge, they don't emanate from. And another way of saying this is that the magnetic flux enclosed inside of an area has to be equal zero. Those are two equivalent statements, but one looks kind of like a derivative thing and the second one is a integral. Um but they basically mean um no magnetic mina polls. So let's use this principle to solve kind of a mystery. Let's suppose that you had a cylinder to laying on its side and you knew that there was a flux going into the left side and what will give some radius to the and cap here. But let's say that you knew there was flux magnetic flux going into that end cap. Um Let's just make up number, let's say it was 27 times 10 to the minus six. Weber's. And let's say that you also knew at the other end there was a magnetic field pointing outwards for that to be along the access be equals 1.8 times 10 to the minus three Tesla. And you want to ask, is this a physical possibility? Um in the universe of things? Um So we're going to call the inward flux negative and it's in bird on the left, outward on the right, we're going to call that flux magnetic positive. And it's b dotted into the area of the end cap. So a reminder that the area vector for a surface points perpendicular to the surface and has magnitude equal two um the surface area. And so if he's pointing along the axis, it's parallel to the a vector, we just have to be pi r squared for that outwards flux, which is about five micro weber's. Okay. And this shows that for the end caps there's a net flux of minus 27 plus five micro Weber's and that should not be right. Well, what's wrong with that picture? What's wrong with that picture is we didn't account for the full closed area. Okay. So what must be going on is some of the flux is leaving the end gaps and the rest of it is coming out through the sides. And how much leaves the sides make sure that there is no flux enclosed in that tube. So what does that say? Plus 22 micro weber's must be leaving the um sides through the curved tube sides. Hopefully, that picture sets a little bit more than what these words are saying to you. Um So we'll just call that fi mr if I um And yes, now now we're good. Now we have Uh huh. The net flux coming from the enclosed tube equal to zero, which it must be no magnetic monte polls.

Hello. Real question. Envisages when that F B and ordered field and X. So I said enough. Okay. It has also given that if X less than zero and why less than that then we need to prove that X. Y greater than access it. So let us get to hear that if access less than zero, this can be written as minus Act should be greater than zero. Okay, now here, if y is less than that so Zach minus Y should be greater than zero. Okay, no, these two have become positive quantities. Some multiplication of two positive quantities should be always positive, should always be positive. So we stretch it as minus X. Which is a positive quantity. Now into that minus Y. Which is again a positive wants to know should be positive. Let us open the bracket minus X. Z bless X. Y should be positive. Let us add except to both the sides will be having X way this is minus exceed all. It is minus except plus X. Y. And we are adding acceptable the sides greater than exit. So these two will become zero. So from here we are getting X. Y greater than X zet. So this is the thing we need to prove. Thank you.

Hello. Real question. Envisages when that F B and ordered field and X. So I said enough. Okay. It has also given that if X less than zero and why less than that then we need to prove that X. Y greater than access it. So let us get to hear that if access less than zero, this can be written as minus Act should be greater than zero. Okay, now here, if y is less than that so Zach minus Y should be greater than zero. Okay, no, these two have become positive quantities. Some multiplication of two positive quantities should be always positive, should always be positive. So we stretch it as minus X. Which is a positive quantity. Now into that minus Y. Which is again a positive wants to know should be positive. Let us open the bracket minus X. Z bless X. Y should be positive. Let us add except to both the sides will be having X way this is minus exceed all. It is minus except plus X. Y. And we are adding acceptable the sides greater than exit. So these two will become zero. So from here we are getting X. Y greater than X zet. So this is the thing we need to prove. Thank you.


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