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(2) Prove thatS := {2 € € : lel = 1}is a subgroup of CX_ Hint: You may use the following properties: If 2, W € C, thenIzwl = Izllwl:If 2 = a + ib,...

Question

(2) Prove thatS := {2 € € : lel = 1}is a subgroup of CX_ Hint: You may use the following properties: If 2, W € C, thenIzwl = Izllwl:If 2 = a + ib, 2 = 0, then2-1a _ ib);(2)

(2) Prove that S := {2 € € : lel = 1} is a subgroup of CX_ Hint: You may use the following properties: If 2, W € C, then Izwl = Izllwl: If 2 = a + ib, 2 = 0, then 2-1 a _ ib); (2)



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.

Well we have a question in which we need to proof one plus a place is choir one. Place a place is quite plus AQ plus areas to the power and equal to one minus areas to the power and plus one. And that's when by one minus k. Far uh not equal to one. Otherwise it will become and finite or I'm defined this is the some national geometric provisions geometric series. Okay so let us start with the proof step 11. They will be assuming let is the through four and equal to zero. Oh if we plug in an equal to zero this will be areas to the power zero equal to one minus areas without one by one minus eight. Now it is to power zero. We're playing here. No plumbing here. Okay one equal to one minus a by one minus eight. So remember we have plugged in an equal to zero here for left hand side and and equal to zero. Head for right inside one. So we can say that we can see that left hand side equal to right inside. So this is true. Now you will be assuming that let us assume that this is true for and equal to Okay, let's win. So for an equal to keep this one, we will be just writing sorry an equal to K after zero will be assuming this to be true for anyone to care. So this will be one. Place a place a square. Place a cute place. Plus areas to the power. Okay, equal to it. Is the barricade equal to one minus areas to the father. Shapeless one by one minus a. Okay, so we'll be using this expression expression number one, equation number one. We should save it all. This is it has right inside and left inside. The most important is we have to prove that we are this is true. Far and equal to K plus one. This we have to prove. So let us play in here. One plus A plus a squared plus AQ players. It is the power care plus it is the power K plus one. Equal to one minus areas to the power unit planning. And K plus one in place of K. K plus one plus one, divide by one minus E. No from equation when we have value of this expression as one minus one minus it is to the paperless one divided by one minus A and plus a building. And as it is plus, it is about a pleasant sorry, equal to one minus areas to the power K plus two by one minus a. Okay, if we just would prove that left inside it right inside so we'll be able to prove the expression Okay, let us concentrate on this. This will become one minus areas to the provider. K plus one. So we are writing LHs and here will be one mindset because we're just adding it up. Bless one minus a. Into eight days to the power Okay. Place when mhm Equal to one minus it is to the power K plus one plus areas to the power people is when minus here is to the power kate glass. When minus when? Yeah, I just opened the back it uh It is the power people. Yes, divided by one minus K. These two will get cancelled out and we will be left with one minus Oh sorry this is this indicate person because I am A has our one and this is the same, basically the same. So we have to edit the power of Okay, place to divided by one of my essay which is it will to our right hand side, it is equal to uh right inside. So we can see that that expression is being approved using mathematical induction. Thank you so very much.

Okay, so we're trying to decide for which values of an to to the end is greater than in squared. So there's a couple of ways of going about this, probably the easiest way of going about. It is just trial and error. So we could try playing in a couple of images. Now you don't want to stop it one. Um because if you stop that one, you can actually get the wrong conclusion. So you plug in and is able to one on the left hand side, you get to on the right hand side, we get one and to act two is greater than one. Okay, So that's good, but if you plug in and is equal to two, you get to to the power of two. On the left hand side, on the right hand side, you get to the power of two. Well, these are equal, so this is no longer true. So let's keep trying if we do N is equal to three, we get to to the power of three on the left hand side, on the right hand side, we get three to the power of to back. In this case it's the opposite right. The left hand side is eight, the right hand side is nine. And we do four. We get to to the power of four. We get forward to the power of to in this case they're both 16, so they're equal. So for the first four cases, it's ambiguous as to what values of and make this statement true. If we do five, we get to to the power of five on the left hand side, on the right hand side, we get five to the power of to now this is 32 on the left hand side and 25 on the right. And so this is great of them. And then we can keep going. And what we see is that everything after five, including five, is going to have this relationship. And so our claim is going to be um to to the end is greater than and squared so long as and is greater than equal to five. And we want to prove this by induction. Okay? So to prove this by induction, this is our new claim. We're going to establish our base case. Our base case is going to be show that it's true and it is equal to five. Well, when I was able to five, we have two to the power of five on the left hand side, which is 32 we have five to the power of two on the right hand side, which is 25. That's in fact through the 32 is greater than 25. So we'll say chuck mark, basis case checks out. Okay. And then we'll say our inductive hypothesis is that we're going to say suppose that for some fixed imager K, the statement is true to the power of K is greater than Okay squared. Okay. Now, in order to do our inductive hypothesis, we're gonna actually have to establish an algebra so I'm going to do this in a different color off to the side. Just that we don't confuse it with our actual proof. There's an interesting thing here and you'll see why it's relevant just one second. But it turns out that for integers, if K is greater than or equal to three, then two K squared is greater than K squared plus two K plus one. And I promise this will be relevant to something and you don't have to do it in this order. You would have seen why this would be important later on if you continue with the problem. Now to see this, we can just find where these two things are equal by setting them equal to each other. Okay, rearranging really quick gives us k squared minus two K minus one. Whose roots are if you were to plug in the quadratic formula, the roots would be one plus or minus route to which as decimals are approximately 2.4 something. It doesn't really matter. And the other one is minus point 06 something. Okay, but the point is that any time K is greater than 24? We actually have that two K squared is greater than K square post to keep this one. And if you were to see this, if you were to put both of these things, two k square, it looks something like this. Okay. Whereas K plus one squared is like shifted over but it's less narrow. Okay, I'm not exactly right? Yeah, clean that up a bit. So let's narrow a little bit wider. And so any time we're above this number? Right here. Okay, this one right here, which is the two K squared one is going to be above the K plus one squared one. Okay. Mhm. Now, why is that relevant? Because when we start to go to our inductive are inductive step, we're going to look at two K +12 to the K plus one, which is the same thing as writing two times two to the K. But by our inductive hypothesis, we know that this has to be greater than two times k squared, right? Because our inductive hypothesis is that to the power of K is greater than K squared. So we just insert that for two to the power of K here. But we just showed and read that two K squared is greater than this is strictly greater than K squared plus two. K plus one. Which factored is just K plus one squared. And so in other words we have the two to the K plus one is greater than K plus one squared. Okay. And so we're ready to make our conclusion. Our conclusion is that we have shown that to the power or two to the power of five I suppose, which on the basis case two to the power of five is greater than five power of two. And that if for some Okay, But in particular some K greater than equal to five to the power of K is greater than okay squared. Then it's also true that two to the power of K plus one is greater than cables, one squared. And so it must be true for all in all right. Hopefully this was clear seeing another

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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