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10, Prove disprove: Q(i, V2) is a splitting field of f(r) 7 +2r +3 over...

Question

10, Prove disprove: Q(i, V2) is a splitting field of f(r) 7 +2r +3 over

10, Prove disprove: Q(i, V2) is a splitting field of f(r) 7 +2r +3 over



Answers

Prove that the following properties are true for every vector space.
a) 2v = v + v.
b) nv = v + · · · + v, where there are n terms on the right.

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.

I want to prove U minus p is equal to U plus minus fee for factors. You and factors be so we have you might this mean is equal to to be minus and now, using our definition of vector subtraction, we have this equal to hey minus d c he dynasty. We also noticed I sense a c and e t r real numbers he had. This is equal to a plus. Negatives, bus negativity. But then this is just addition. So we have This is equal to you. Okay, Plus negatives. But we also noticed that this negative is nothing but a negative one. So we have a baby negative one times negative one time, Steve. And using our definition escape their multiplication, we can bring them negative one out. So we have a baby, plus maybe one times and bringing this up here we have. This is equal to replacing a B C d. Lift you and B, we have new plus negative one times. And this is equal to for us. Thank you. Be so following a chain of inequalities for the qualities we have. Therefore U minus fee is equal. You did

In the problem we have to prove the properties of vector. So here it is given to be equal to we plus we. So it's a vector. Now here we will prove these vectors so late. Whether we equals a gap plus B J cap plus c kids gap. And then if we would apply the vector with two, therefore this twice of vector V equals two. Twice in two, A cap plus B in the cap plus c K cap. Now this is the left hand side, This is the left hand side. Now when we have the excess of right hand side so here I D. S victory plus victory. So it is a cab plus B, jakob plus C K Camp plus a cap plus B. The cap plus c K cap. Now this equals two, twice off A cap plus twice of B J cap plus twice off. See get gap. Now this is equivalent to twice. That is true. It's taken out and it is a cap plus me, Jacob plus C camp. Therefore it equals two price of feed and this is equal to the right hand side. Sorry, this equals two left hand side. That is allergies. So we have proved that right inside the left hand side are equal. Hence these victories satisfying the property. Now, further we have the other problem that is part B. So in part B we have to verify the end of V. This is written as we plus the blast dot dot dot last v. 10 times. Now this is written as similarly from the solution of that is the solution of jump. That is a problem. A. We have proved that twice of equals B plus B. So here envy equals two. We plus V plus v plus data dot V and times. So we have these are all vectors. This equals two N directory. So we have proved that left hand side because the right hand side, so this is the answer to the problem.


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