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Let M be the abelian group Z/62. Prove that M is Z-module (in other words, given an integer (both positive integers as well as negative integers) and an element m i...

Question

Let M be the abelian group Z/62. Prove that M is Z-module (in other words, given an integer (both positive integers as well as negative integers) and an element m in M, define what element of Mn m should be): You don't need to prove that your definition satisfies the axioms of a module:

Let M be the abelian group Z/62. Prove that M is Z-module (in other words, given an integer (both positive integers as well as negative integers) and an element m in M, define what element of Mn m should be): You don't need to prove that your definition satisfies the axioms of a module:



Answers

Prove or disprove that if A and B are set, then P(A×B)=
P(A) ×
P(B).

Zero is less than B is less than a. Let's consider the function A four X. Is equal to expire in and is it and teacher great and equal to do now there is a theory called mean value to you. This function is a polynomial. It's continuous and differentiable. In an interval beat way. It's continuous enclosing to Albert way and differential. An opening to a gateway. Right? Very function is continuous in a closed into a gateway and differential and opening to a gateway then like green just mean value to them like ranges mentality and says that F a B minus ffb divided by a minus B is equal to a dash of something between be any. A very powerful freedom in calculus. Lagrange is humanitarian. Let's use this. So what is my fook? Fook will be a parent minus B. Baron divided by a minus B. Is equal to a dash of. See what is the derivative of this function? It is an ex power in minus one. This is the radio off function. Experiment by power. So what will be a flash of C. N. C. Bar and minus one? All right now you can see here sees between B. And so that means he's less than me. You see is less than a C. Bar and minus one is less than a bar and minus one. Multiplying and pull tight N. C. Bar and minus one is less than in a bar in minus one. So this quantity is less than in a par and minus one. So that would imply a par n minus B part and divided by a minus B is less than any power and minus one for all and greater than or equal to two for one. What will happen is it will be exactly equal because a pas one minus people. One divided by a minus B. It's compared with one paper, one minus one. This is one and 180 is one. So for one when it is one, this particular inequality is actually an equal. But when you increase one, that is when you take higher than 12345 all natural numbers starting with two. This inequality will hold two and we included by Lagrange is mean value.

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

Question given us when it took problematic particular election that one Q plus two cube plus three Q two and Q is equals to one plus two plus three. Dylan all square. Not approved by mathematical election. We just solid for let's say for and equals to one. And in that case our left hand side is one Q. Which is equal to the right hand side. Since that is one square and that is also one. Now when you look for any calls to K now we assume that it is true for N equals two. K. Assumed it is true and therefore we get one Q plus two cube plus kick you is equals to one plus two plus. Okay hold square. And now we need to prove for any close to K plus one. So when we write this year we get left hand side as once Q two Q bless KQ plus K plus one. Thank you. And our left hand side. Sorry, 19 side is one plus two plus K plus k plus one whole square. Now let's start with the left hand side. So here this entire term can be replaced by this value which is one plus two. Bless less clear square. And we lived with K plus one. Okay. So now we have two dumps and we need to simplifies us that we get this. If both are equal now to do that we can just neglect this step. And what we can use is summation formula. So here this is one plus two plus three T. K plus one cube. And this is some of Uh huh. Where are cube? Sorry? Where are they starting from? One till it's a K plus one. And we know some mention of our cube is if they are going to keep this one. So this escape plus one square. Okay. Plus one. And here we have K plus one plus one since N plus one whole square by four. And on simplifying get K plus one square. Okay. Plus two. Great. 44 So this is our left hand side. Now let's solve the right hand side part. So here we can apply submission formula inside the square down. So this is one plus two plus three plus K. K plus one. So this can be written as some of our our starting from one tree. K plus one whole square summation of our is given by formula N. But here an escape plus one. K plus one. Okay. Plus one plus one by two whole square. And this gives us K plus one square. This becomes K plus two squared over. And that is the same as this and total by induction. Thank you.

Question and this. We need to prove or disprove. The P. He crossed. Be equal to be a grass B. B. We have to prove all this probe. Okay? Now let us zoom. Mhm. E. To be M. And be to the end. So from here, B. A. Will be equal to tourists. To the power. Um And B. B. Would be equal to to raise to the power hand. So right inside would be equal to be a Class B. B. Okay. These are the numbers that's where Modelers E. Gross B. B. Would be equal to tour is to the power M. Into uh to raise with the power in to raise to the power um into to raise to the power. And so to restore power M. Pleasant mm left hand side. Let us get the value of A. N. B. First E. N. B. That is M into end. To be a Crosby would be cool to to restore the power. I am aiming to end. Okay. And from here, from both both of these exhibitions, we can see that to raise to the podium. Plus in equal to tourists. The power aiming to end, which is not possible because M plus N. Cannot be equal to mm two. And so it is not true. It is disproved. Thank you


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