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CPSC 121Winter 1, 2020[9 marks] In this question; WC give you three theorems; cach with "proof . Unfortunately, each proof contains mistake Find the mistake in...

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CPSC 121Winter 1, 2020[9 marks] In this question; WC give you three theorems; cach with "proof . Unfortunately, each proof contains mistake Find the mistake in each proof, and explain why it is mistake (you do not have to write a correct proof)_[3 marks] Theorem: If 6n2 24n + 8 2 0 then 24 Proof: when 2 4, 6n > 24, which mcans that 6n22 24n > 0, and S0 6n2 _ 24n + 8 _ 0. QED b. [3 marks] Theorem: For every positive integer a and prime p, the integer aP ~ 0 is divisible by p. Proof: We

CPSC 121 Winter 1, 2020 [9 marks] In this question; WC give you three theorems; cach with "proof . Unfortunately, each proof contains mistake Find the mistake in each proof, and explain why it is mistake (you do not have to write a correct proof)_ [3 marks] Theorem: If 6n2 24n + 8 2 0 then 24 Proof: when 2 4, 6n > 24, which mcans that 6n22 24n > 0, and S0 6n2 _ 24n + 8 _ 0. QED b. [3 marks] Theorem: For every positive integer a and prime p, the integer aP ~ 0 is divisible by p. Proof: We use proof by contradiction. Suppose that there are integers and p for which aP is not divisible by p. This is not true; WC cahl scc by picking 2 and p = 5: aP _ a = 25 _ 2 = 30 which is divisible by 5. Because the negation of the theorem is false, the thcorem must be truc_ QED [3 marks] Theorem: For any three functions f, g and h from N into Rt, if f e O(h) and g € O(h) , then (f + g) € O(h), where f + g is the function from N into R defined by (f + 9)(n) = f(n) + g(n). Recall that f € O(g) if 3c € R+ano € NVn eN,n > no - f(n) < cg(n). Proof: Consider three unspecified functions f, and Suppose that f e O(h). which mcans that there exist values n] . C1 such that for every n 2 n1, f(n) < ch(n)_ Suppose also that g O(h) , which mCans that for every 2 n1, g(n) < ch(n). Pick c = 2C1 and no n1, and consider an unspecified positive integer 2 no. Then (f+9)(n) = f(n)+g(n) < c1h(n) + ch(n) = ch(n), required. QED



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It seems likely that $\sqrt{n}$ is irrational whenever the natural number $n$ is not the square of another natural number. Although the method of Problem 13 may actually be used to treat any particular case, it is not clear in advance that it will always work, and a proof for the general case requires some extra information. A natural number $p$ is called a prime number if it is impossible to write $p=a b$ for natural numbers $a$ and $b$ unless one of these is $p,$ and the other $1 ;$ for convenience we also agree that 1 is not a prime number. The first few prime universe are 2 $3,5,7,11,13,17,19 .$ If $n>1$ is not a prime, then $n=a b,$ with $a$ and $b$ both $<n ;$ if either $a$ or $b$ is not a prime it can be factored similarly; continuing in this way proves that we can write $n$ as a product of primes. For example, $28=4 \cdot 7=2 \cdot 2 \cdot 7$ (a) Turn this argument into a rigorous proof by complete induction. (To be sure, any reasonable mathematician would accept the informal argument, but this is partly because it would be obvious to him how to state it rigorously.) A fundamental theorem about integers, which we will not prove here, states that this factorization is unique, except for the order of the factors. Thus, for example, 28 can never be written as a product of primes one of which is $3,$ nor can it be written in a way that involves 2 only once (now you should appreciate why 1 is not allowed as a prime). (b) Using this fact, prove that $\sqrt{n}$ is irrational unless $n=m^{2}$ for some natural number $m$ (c) Prove more generally that $\sqrt[k]{n}$ is irrational unless $n=m^{k}$ (d) No discussion of prime numbers should fail to allude to Euclid's beautiful proof that there are infinitely many of them. Prove that there cannot be only finitely many prime numbers $p_{1}, p_{2}, p_{3}, \ldots$ $p_{n}$ by considering $p_{1} \cdot p_{2} \cdot \ldots \cdot p_{n}+1$.

Let's consider the equation. Do you? Is equal to M plus two squared minus. So I m plus two squared minus m square. Okay, so what does this equation give us? Well, we have the difference here between the square of a whole number. So some whole number m we have the difference between the square of a whole number and that same hole. Number plus two, right. The difference between a square of a whole number and the square of that whole number plus two. Okay, so, for given a table here, um, where we have a table. If I m what's we have? Ah, m d tables. What? M do you Okay, so we're given while when m is one while ideas eight. Right, Because one plus two is 33 squared is nine on the nine minus. One squared is nine minus one, which is eat. So I m is one is eight. And when it is, too de is 12. When m is the re, we'll give her that. D is 16. Okay? And then complete the table. Saw what em is. Four. Well, what is D? Where All when m is four. We have four plus two, which is six in six squared. So six where it is 36. You have 36 minus ah. Foursquare foursquare to 16th of 36. Minus 16. Is it, uh, 20? Right. Okay, so when m is four d is 20. How about one? M is five. Well, when M is five, we've got five plus two, which is seven and seven square, which is Ah, 49 3 of 49 minus fire square. So 49 minus 25 is, uh, 24. Okay. And then, well, we have one more. Complete someone. M is six. When m is six. What is D when m is 66 plus two, which is 678 and eight squared is ah, 64 minus, uh, 36. Right. You have 64 minus 36 which is a 28 uh, 30 40 50 58 912 weeks ago. Yeah, 28. So when m is six d is 28. Okay, so there's a table. No, that's basically, um, part a of the question here Says to complete table for different values of M and E. Okay, So now Well, what are we kind of noticed about D. We see that we have a constant difference, right? Anywhere. The difference between any two is well, four, right? If we're adding for each time to go from 8 12 12 to 16. 16 to 2020 24 24 28 we're adding as we go up by one. Um, in our amas, we called by four and are these were constant difference here, So make a conjecture about what a simpler equation might look like. Well, we could think that this would probably be a linear equation right now as you look a d. I mean, it doesn't look to be linear, right? With m plus two squared and mine is it almost looks quadratic. I see something squared. But if we were to explain, expect expand this out. Well, we get em. Plus two squared off. That is M plus two times M plus two. So D right is equal to while m plus two squared is m plus two times I m plus two and then minus m squared. Will this equals? Well, I haven't m squared. And then plus to m plus two m and distributing years. I m squared plus two m plus two m's as plus four AM they were plus four. And then I studies minus m squared on here. Well, look, this is Evan M squared minus m squared. So that become zero. And what do I get? De is just equal to four m plus four, right? And now we see it in a linear form. So, yes. Um, the original equation didn't look linear, But what if you could take the table? We see that we have that constant difference. Ah, four. Um, so we assume and a conjecture that this is a video equation, and we have to do is expand it out. And we see that, Yes, we're in the form MX plus B. So, um, there is our little your form, and our equation is linear, all right?

Well for a bunch of things about prime numbers. And we can basically this this uh pie, this prime pie which I will call it. Prime Pie basically says that how many how many prime numbers are there below? A certain number? Okay so a certain energy. And so they asked us calculate the numbers pi to pie of 25 pie of 100. Well we can 5 29. Again they tell us they use the receive of Aristotle tins. Uh huh tough themes. Yeah I remember how, I don't know how to pronounce that exactly. But it basically says just you know find you know divide by to get rid of all the even numbers and divide by three. Get rid of anything that's a multiple three. And then we divide by the next that's five. We get rid of all the multiple five then get rid of all of the multiples of seven and then get rid of all the multiplicity of 11. And I keep doing that until we get up to at least, you know, halfway through. Um And that would be that there would be possibly no more. If we get up to 15 can't there will be no more prime numbers after that. The guy via as once, nothing divided by 50. So for nine we get one for 25 we get 2357 11, 13, 17, 19 and 23. So you get nine, there's nine prime numbers less than is it less than or equal to uh problems that are less than or equal to 25. So 100. Um We need to do a little more work and we can find that in fact we get 10 29 31 37 41 43 47 so on. And counting all these up, we get 25. So there's 25 primes that are less than or equal to 125 crimes that are less than 100 too because the highest, the largest one is 97. Now they say, okay, they tell us that Kaos when he was a teenager said that the um postulated at least that the limit as N. Goes to infinity of part of the N. Divided by N. Over the natural log of an is one. Um And then uh 100 years later, that was actually proven. I'm not sure how you actually proved that. That's a, it's, it would be quite an interesting proof to try to um work through. But anyway, we'll just actually even just see how it was done. So they say, well let's do this for a few values. So we got for 100. We have, so this is 25 and then we have um you know, 100 um times the natural log of minus 100. So and over natural log of um And over natural log event. So we just brought that up under the numerator. Um and that gives us 1.15 Okay so it's it's close to one and then if we go to 1000 or they gave us they gave us this pi to 1000 is 1229 dividing by this factor we get 1.11 point 16 which is a little worse. Um But if we go to um 10,000 we get 1.13 Um If we go keep going um we can figure it out for higher and higher numbers. But then I just made a plot which is I think that they do estimate the crimes um by inspection. Uh Let's see here. So let's see, did they say provide evidence for the truth of this, thereby computing? Okay, so I I just made a plot here. So this is n this is um applauding pie to the 10 to the end, all over 10 to the end, divided by natural law, Gavan tend to the end. So basically the factors of 10 and so you know, when and as one for 10. Well, we just have, you know, we only have four. And so we get a value down here about 0.9 than for 100. We're up here at 1.15 for 1000. We're here for 10,000 were here 100,000, we're here a million were here. Um 10 million. 100 billion. Now you can see that, you know, here's one. So these things do appear to be ascent. Toting to one as we get larger and larger and so then they ask us well to approximate the number of number of primes that are less than or equal to one billion. Well, we're using this this formula here, we can plug that in and we get four point 83 times 10 to the seventh. So that would be our approximation using this, but in fact we can use Mathematica or some, you know, tool to figure out how many primes there are less than a billion. And we get 5.8 times 10 to the seventh. So again, this is a little bit of an underestimate, it looks like this thing is approached as converging from above. Um So yeah, that this would again be a slightly bigger than one. So again be slightly, you know, we're basically this thing seems to give us a lower bound on how many primes there actually are. Well, this this thing here, I should say this 10 to the end are are whatever, you know, whatever this is divided by the natural log of that gives us a seems to give us a lower bound on the number of primes there are.


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