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Author Topic: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!  (Read 19740 times)

snalin

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Oh holy shit is humans supposed to be able to read that?
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ankhtahr

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holy crap. Thanks a lot! I don't think I would have managed that on my own. I'm sorry for having you type all that out.
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ankhtahr

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Sorry to say, but I'm not really sure what you did at this step:

Apply the identity      to all terms of the sum, except the first and the last:


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Barmymoo

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As far as I can tell, he put some numbers into a visually pleasing pattern! I have so much respect for people who understand this stuff. It's all I can do to understand the biology I have to study, and at least that's expressed in words.
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There's this really handy "other thing" I'm going to write as a footnote to my abstract that I can probably explore these issues in. I think I'll call it my "dissertation."

snalin

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Oh shit, it's supposed to say k+1 in the second binomial there, not k1. Erp.

So, example, you would turn the term      and turn it into  . Do that for every term of the sum except the first and the last (since they're not applicable to the identity)
« Last Edit: 06 Nov 2013, 14:52 by snalin »
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snalin

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Is the images? not working for anyone else? If I open them in a new tab and replace "White" with "Black", it works. Huh.
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ankhtahr

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'cause they're gifs and not pngs. That's why I recommended png.
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snalin

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Fix'd!
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Schmee

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This was in my uni's Maths and Stats society newsletter:
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Several years ago, A/Prof Barry Hughes set a first-year exam question which asked for a proof that there is no largest prime number.The following was sent to him by one of the graduate students who helped with the marking:

Hi Barry, just for educational value, I compiled a list of the most incorrect answers to question B4(b). It is entitled "15 good reasons why Pure Mathematics is not taught to first year students."
  • Proof by example: “Let x be the largest prime. Then x = 91 but 91 + 6 = 97, which is prime. Therefore, 91 cannot be the largest prime number. Therefore there is no largest prime number.”
  • Proof by oddness: “If n is the largest prime number, then n is odd. Then (n + 1)=2 is even. Therefore, (n + 1)=2 + n is odd. But (n + 1)=2 + n is not divisible by any number except itself. As it is bigger than n, the assumption is wrong, by contradiction.”
  • Proof by intuition: “Prime numbers are integers that can be divided by themselves only; prime numbers are odd with the exception of 2. By intuition as n -> infinity, there will always be an odd number that cannot be divided by any other number besides itself.”
  • Proof by sqrt(2): “that p is divisible by q, i.e. p/q = 2r, where r is an even number. Then p = 2qr so p2 = 4q2r2. But r2 does not exist and q! = 1. Therefore, q must exist. Since q exists, p must be divisible. Therefore, by contrary, there is no largest prime number.”
  • Proof by superinduction: “2 is a prime number. Now assume N is the largest prime. But then N + 1 exists and is also prime. Therefore, by induction there is no highest prime number.
  • Proof by the previous question: “Suppose N is the largest prime. Then let N = (n2)/2. Therefore n = sqrt(2N). But from above (1 + 2 + 3 + ... +n) > N. Hence, there is a larger prime number than N.”
  • Proof by tutorial question: “Let m, n be two integers with m > n + 1. If k is even, mk + nk cannot be expressed in terms of (m + n)(polynomial in m and n) and so is prime. Therefore, as m and n can be any numbers, there is obviously no largest prime number.”
  • Proof by having no idea what a prime is: "Say the largest prime possible is x, then 2x is also a prime since the statement is true for all natural numbers."
  • Proof by experimental data: "Suppose n is the highest prime. Then 2n-1 is also prime. But 2n-1>n so there is no highest prime (Check: 2*2-1=3, 2*3-1=5, 2*5-1=11, 2*11-1=23, so true)"
  • Proof by subscript: “If there is a highest prime, we can number all the primes p1, p2, ... , pn. But as there is no highest natural number, there is always an n + 1 so there must be a pn+1. Therefore, there is no highest prime.”
  • Proof by infinity: “Let n be the highest prime number. But infinity is greater than all numbers so infinity > n. If n is the highest prime this would mean infinity has factors. Therefore, we have a contradiction.”
  • Proof by reverse logic: “All prime numbers are odd. Suppose there were a highest prime. Then we have a highest odd number. But if 2k+1 is the highest odd number, then 2k+3 = 2(k+1)+1 is also an odd number. Therefore, we have a contradiction and therefore we have a contradiction."
  • Proof by denial: “Assume there is a largest prime M. We can add 1 to M until we get another prime number N (M + 1 + 1 + 1 + ... + 1 = N). But then N > M. Therefore, M is not the largest prime number, so there is no largest prime number.”
  • Proof by formula: “As prime numbers are derived via the formula, we can assume it works for n = k giving the highest prime number. But then it also works for n = k + 1, so there is no highest prime number.”
  • Proof by continuity: “Let x be the largest prime number. Then x > all other primes. But then (x + n), the next prime number, does not exist. However, numbers are continuous and so (x + n) does exist. Therefore, there is no x.”
These are all verbatim answers from the 150-odd papers I marked, i.e. about one in ten. My favourite is definitely number 10; I think that is quite ingenious.
— George Doukas
... That took a while to copy over, but I think you'll agree it was worth it.
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Barmymoo

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What would be a correct answer to that question?
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There's this really handy "other thing" I'm going to write as a footnote to my abstract that I can probably explore these issues in. I think I'll call it my "dissertation."

Carl-E

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Euclid had a good one.  If there's a largest prime, then you list them all and multiply all of the primes together into one big number. 

Then add 1 to that result.  It's bigger than the biggest prime, so it's not prime.  And none of the primes will divide it, they'll all have a remainder of 1.  That means that it must be prime, since nothing divides it but 1 and itself. 

And that's absurd - it can't be both prime and not prime, so the assumption (that there's a largest prime) is false, there isn't one. 

Ankhtahr, did you get the 52n - 2n yet?  It's (52)n - 2n, which will always factor as

(52 - 2)((52)n-1 + (52)n-22 + ... + 2n-1), and so there's always a factor of 23. 

(sometimes, it just looks like it needs induction...)
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ankhtahr

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Schmee: The first one cracked me up already. Where the hell did that 91 come from?

"Proof by having no idea what a prime is"

Bwahaha!

Carl: Yes, I had to turn in the exercise sheet on Thursday already.

Oh, and I'm so very sorry to say, but Snalin: I found a much, much simpler proof.

Like I had said, I suspected the whole thing could be solved with the binomial theorem (). I noticed that you can just choose a = 1, and apply it on the term.

Which makes it look like this:
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Carl-E

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"Therefore, we have a contradiction, and so therefore, we have a contradiction"

Pure gold! 
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Loki

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Prove: Each convergent series over K \in {\mathds{R}, \mathds{C}} is bounded.

Is there any sensible answer except "well. duh, if it converges to x, it is bounded by x"?
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Carl-E

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Prove: Each convergent series over K \in {\mathds{R}, \mathds{C}} is bounded.

Is there any sensible answer except "well. duh, if it converges to x, it is bounded by x"?

False! 

Just in R, the sequence 1, -1, 1/2, -1/2, 1/3, -1/3, ... converges to 0, but is bounded above by 1 and below by -1.  The limit is rarely a bound unless the sequence is monotone. 

Though maybe I don't understand the question - what exactly is {\mathds{R}, \mathds{C}}?  It looks like the two-dimensional space R X C, which doesn't really change the answer, but I don't think that's what it means? 

(Topology is almost nothing but a collection of counterexamples... I got  million of 'em!)
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(Topology is almost nothing but a collection of counterexamples... I got  million of 'em!)


On the contrary (heh), only concrete topology is a collection of counterexamples; there are almost no counterexamples in synthetic topology!


EDIT:  Also, Barmymoo said I should probably cross-post this from the ENJOY subforum.
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Loki

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Carl: yes, I realized something similar yesterday in bed.

Sorry about the Latex. The exercise means to say that K is an element of the set containing the real number set and the complex number set. That is, K can be R or C. The cartesian product (aka the two-dimensional space) would have been denoted by R x C.
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Carl-E

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Thanks!  I'm an old TeX haXer, but I was stupid-sleepy and couldn't make it out. 
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ankhtahr

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Hey, it's business as usual, I'm sitting at university with some others, working on an exercise sheet, due the day after tomorrow..

We're sitting here for about 3 hours already, and we're still at the first of 7 exercises. We have to find accumulation points of a series. The first five are similar to this one, the last two are different, and probably more difficult.

I had only three hours of sleep last night, as I had to do another exercise sheet.

I'm loving this.
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Carl-E

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That's the delirium talking. 

Keep it up! 
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snalin

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I have never had to spend more than 4ish hours a day at Uni work. I'm not sure if it's super easy here, if I'm more naturally talented than I thought, or if I'm just good at managing my laziness.

Although I suppose that if I had wanted a good grade in the introductory maths courses, I would have had to work way hard. I didn't do that, and got a D. Aaaand that hasn't mattered a single bit. Calculus is dumb anyways.
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So I'm in introduction to Derivatives. I'm scared to get into AP Calc AB next semester...
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Loki

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Ankh, could you PM me links to one of each of the sheets you are doing?


*looks at sentence* *rewords* Could you show me the current worksheets for all of the courses you are taking?
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snalin

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:x

No, but, like, I don't need it, anyways. Who ever needed derivatives? :D
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Skewbrow

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Maths is, right? I mean, maths is written in plural form, but it is still just a single thing, right? Or is it maths are sexy?

Is.  "Maths" is not the plural of "math"; it's a different abbreviation of "mathematics".  Many words end in s without being plural.

Depends. It's "Maths" in the UK, and "Math" on the other side of the pond (or should I say elsewhere?).

But looks like you guys started having fun while I was gone from this subforum. I don't have anything against doing it here, but I also might recommend that we take it to Math.Stackexchange. There is a real-time editable TeX-interpreter (MathJax) there, so we don't need to use graphics. I'm sure Paul might be able to hack us into the cloud running MathJax (or whatever it is), but we won't have hundreds or thousands of qualified people answering here.

Take it from someone who typed the plainTeX-souce to his dissertation 23 years ago. With the VI-editor that came with Unix.
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ankhtahr

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Currently working on this sheet (sorry for using this shitty owncloud-website, I don't have SSH-access to my server, so I can only use that if I don't want to upload it at any strange service.

Don't really think anyone here can help me with it, if I don't give a translation. I don't have the time to give a translation right now.
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Loki

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You are doing weird stuff.

Re 2b)
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Let G be a group and H1, H2 subgroups of G.

Show that H1 union H2 is a subgroup or find a counterexample

Should be true, imho. The neutral element of G is in H1uH2. For every element x of H1, -x in in H1 (H1 is subgroup), and therefore in H1uH2. The same for H2.
Associativity should be, uh, trivial.

Do you know where to start on the others?
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Skewbrow

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You are doing weird stuff.

Re 2b)
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Let G be a group and H1, H2 subgroups of G.

Show that H1 union H2 is a subgroup or find a counterexample


This is false. Let G be the group S_3 of symmetries of an equilateral triangle. Let H1 and H2 be groups generated by a single reflection (or single transposition, if you are doing it in terms of permutations rather than symmetries of a triangle). There union then contains two distinct reflections, but not their composition (which is rotation or a 3-cycle again depending which way you view this group).

In terms of the group axioms it is the zeroth axiom that fails. This set (the union of H1 and H2) does not have an operation.

It is a "standard" exercise to show that the union of two subgroups is a subgroup, if and only if one of the subgroups is contained in the other. The idea is the same. If x is an element of H1 but not an element of H2, and y is an element of H2 but not of H1, then xy cannot be an element of either H1 or H2. Therefore it is not in the union.
« Last Edit: 21 Nov 2013, 09:51 by Skewbrow »
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Skewbrow

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Don't really think anyone here can help me with it, if I don't give a translation. I don't have the time to give a translation right now.
If the topic of the sheet is something like the first course in abstract algebra (judging from Loki's post), I am qualified. I don't fully trust my mathematical German, but I have read a number of German math books (though it was a while ago). The problem is that my browser is convinced that following up your link is a bad idea, risky even.
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Maths is, right? I mean, maths is written in plural form, but it is still just a single thing, right? Or is it maths are sexy?

Is.  "Maths" is not the plural of "math"; it's a different abbreviation of "mathematics".  Many words end in s without being plural.

Depends. It's "Maths" in the UK, and "Math" on the other side of the pond (or should I say elsewhere?).

That's what Paul is saying - unless for some odd reason you're pluralising the American word "math". If you're using the UK word "maths", it is a singular noun.
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There's this really handy "other thing" I'm going to write as a footnote to my abstract that I can probably explore these issues in. I think I'll call it my "dissertation."

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Sorry, May, Paul. I've just heard this maths vs. math discussion too many times, and jumped to a conclusion.
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ankhtahr

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The problem is that my browser is convinced that following up your link is a bad idea, risky even.

Lemme guess, you're getting a certificate error. I'm using a free certificate issued by CACert. Go to cacert.org, click on "Root Certificates" and install the Class 1 and Class 3 Certificate (PEM is usually the best option). Then the message should disappear.

Or you could just ignore the message.
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snalin

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How far are you along, Ankh? Because I seem to remember this being your starting year, and abstract algebra the first semestre seems a bit crazy. Are you taking pure maths?
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Skewbrow

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I'll be teaching freshman abstract algebra this coming Spring, and Ankh's sheet could well come from my course. Which questions are giving you a headache?
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ankhtahr

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This is the math I need to do for Computer Science here. This is linear algebra. I also have advanced mathematics/analysis.

Currenly I'm still struggling with some of the definitions, but 1 c) and 4 are probably the most difficult.

I'm working a little bit ahead right now, This sheet is only due on tuesday, but I'd like to maybe sleep next week.
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Quote from: Ankhtahr's Teacher
Assignment 1: Let G and H be groups with respective neutral elements e G and  e H
Let f be a homomorphism of groups from G to H. Show that:

c) Let p: G -> G/ ker( f ) be the canonical projection mapping the element g to its coset g ker(f) (sorry, I don't know
what "coset" is in German). Then there exists a unique homomorphism f from  G/ ker( f ) to H such that
f = f o p (using underscore instead of the overline bar, sorry). Furthermore, f is injective and its image
is equal to that of f.

Here we have no choice in constructing the homomorphism f. If g ker(f) is an arbitrary element of the quotient group  G/ ker( f ), we have no choice. Because f = f o p, and g ker(f)=p(g), we must define f(g ker(f))=f(g). Can you follow this?

This means that the uniqueness part will be clear, once we get existence. But there are several catches before we are there. First and foremost you should check that f  is well defined (=wohldefiniert?). This is because distinct elements of the group G may give rise to the same coset. So we need to clear the following obstacle. If g ker(f) = g' ker(f) for two different elements g and g' of the group G, then we must be sure that f(g ker(f))= f (g'ker(f)). In other words, we need to check that f(g)=f(g'). Can you do that? Hint: here it is absolutely crucial that we are talking about cosets of ker(f).

Ok, if that is clear, then the rest is relatively easy. Namely you need to check that f is a homomorphism of groups. This is not very difficult, because you only need to recall what the group operation of G/ ker(f) is, and then utilize the fact that f was known to be a homomorphism of groups.

This exercise is designed to make you once more think through the meaning of the relevant definitions.
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Quote from: Ankhtahr's teacher
Assignment 4: Let M be a set and  P(M) its power set. For all elements A, B of P(M) define the
symmetric difference (Can't draw the Delta symbol here. I'm using '+' instead, because this will become
the addition of the ring you are creating. Similarly 'n' denotes intersection when standing alone).
A + B := (A U B) \ (A n B).
Show that (P(M),+,n) is a ring. You are allowed the assume the associativity of the symmetric difference.

Ok. So the first task is to show that (P(M),+) is a commutative group. Before we start let me recap. The set A+B  consists of those elements of M that belong to either A or B but not to both. This will be another subset of M, i.e. an element of P(M), so '+' is an operation on P(M). We were allowed to assume associativity, so let's skip that. A neutral element? Can you think of a subset S of M with the property that no matter which set A is, then S+A=A. Hint: S should not contain any elements of A, because those would be excluded from S+A.

Proving the existence of inverse is dependant on knowing what the neutral element is, so I stop here, and let you sweat a bit. :evil:

Edit: I am trying to guess a helpful level of chatter and hints here. If it is Greek to you, say so, and I will make another attempt. Getting used to the level of abstractness here requires a fair amount of work, and takes a little while. For example, I would guess that may be top 10% of the students in my course would make any serious headway in your Exercise #4. A vast majority of students here are unprepared to fathom that they can do addition and multiplication on sets. German high school may be a bit better in that regard. The exercise is not very difficult, but you have to absorb the rules of the game of definitions and axioms to stand a chance.
« Last Edit: 22 Nov 2013, 08:55 by Skewbrow »
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ankhtahr

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I want to take this opportunity to thank you, Skewbrow, for saving me a little bit of sleep. While it still took me a lot of time, I don't think I would have come to any solution without your help. It's 03:40 am now, but I'm done with the exercise sheet, which is due tomorrow/today. I'll have to get up at 06:45, so I'll leave now, but I wanted to express my gratitude.
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ankhtahr

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And once again I have another linear algebra exercise sheet for Tuesday.

Here it is

I'm doomed.
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Skewbrow

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I'm impressed at the pace your course is going forward (but wouldn't try this at home). Apparently Germans know something about teaching abstract algebra that we don't.

Since you did well last week with very minimal hints (more like gentle probing), let's just roll with it

Quote from: Ankhtahr's teacher
Assignment 1:
A) Show that the set Q(31/2) of numbers of the form a+b31/2, with a and b ranging over Q, is a field with respect to the usual addition and multiplication of real numbers.
B) Show that the mapping that sends a+b31/2 to a-b31/2 is a well-defined automorphism of fields.
Hint: You are allowed to assume without proof that 31/2 is irrational, i.e. not an element of Q.

In part A your are basically asked to show that the set Q(31/2) is closed under addition, subtraction, multiplication and division. If you have covered a result called "subfield criterion", then that is basically what it says. You can forget about checking the field axioms. They hold in the reals, and this a subset, so you get the axioms for free. The only worry is that the basic arithmetic operations won't take you outside the set. Checking that ( a+b31/2 ) +/- (a'+b'31/2 )
with a,a',b,b' rational can be written in the form  c+d31/2  with c and d rational should be easy. Doing the same for  (a+b31/2 )*(a'+b'31/2 ) takes a bit more work. Expand the product and use the fact that 31/2*31/2=3 is rational. Closure under division is the hard part. As the set is closed under multiplication it suffices to check that the inverses, i.e. the numbers 1/( a+b31/2) can be written in that form. Hint: show that
( a+b31/2)( a-b31/2) is a non-zero rational number. This is where the "free" assumption about irrationality of 31/2 will help you out. If you get stuck, describe exactly how far you made it.

In part B the proof of this mapping being an automorphism parallels very much the proof that complex conjugation is an automorphism of fields. In the complex case the key is that i2=-1=(-i)2. Here the equatiions (31/2)2=3=(-31/2)2 play a similar role. Checking the conditions for this being a homomorphism amounts to just using that equation and expanding everything. (Note to others: that's a lot of work, I'm not doing Ankhtahr's homework here - just giving pointers). Before you go there you seem to be required to check that the mapping is well-defiined. Unlike last week, there were no choices made. You do need to check that if  (a+b31/2 ) is equal to (a'+b'31/2 ), then we must have both a=a' and b=b'. This is another spot where you can take advantage of the irrationality of 31/2.

I'm not gonna bother translating Assignment #2. Too many formulas. Part A you can just brute force (my students would use WolframAlpha for this, but it is good for you to do a bit of pencilwork). Part B is fun. The solution set of that equation is a circle in the complex plane (unless I made a mistake) - just write z=x-iy and see what you get. That pair of inequalities: the first requires that z is closer to zero than to 2i. The second requires that the distance to 2i is less than 3. Draw a picture!

I don't feel like reproducing the last one here either. I doubt the sanity of any teacher who expects to cover all the details needed here within a 90 minute problem session. Is s/he a rookie or something? I tried things like this when the ink on my dissertation would still stick to the fingers, but ... Might 's well get on with it. Part A is just identifying the additive and multiplicative neutral elements and checking out all the axioms. If you are meticulous, it will run a few pages. Part B is more of the same. For the part asking about the field of two elements I just tell you to check what happens with the polynomial X2-X.

This should keep you occupied for a while  :evil:
« Last Edit: 29 Nov 2013, 09:01 by Skewbrow »
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Loki

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So, uhm. Ever know that kind of tasks where you know the answer, but you just don't know how to prove it? I have one of those.

Quote
Prove: every polynomial function p: R -> R (with R as reals) of an odd degree has at least one x-intercept.

Below are the thoughts I have got so far:



I suspect that I have to use Bolzano's theorem, showing that there are values a and b in the reals which satisfy p(a)<0 and p(b)>0, and therefore there must exist an x-intercept between a and b, provided the function is continuous (which a polynomial is. I hope I don't have to show that).

It is fairly easy to show that the condition is true for a function p~(x) = c*x2n+1, with c as a real constant and n an arbitrary natural number, but how do I show that it's also true for
p(x) = c2n+1*x2n+1 + (c2n*x2n + c2n-1*x2n-1 + ... + c1*x1 + c0)?
Ie that the bracketed term doesn't "set off" the signum of p(x) by too much?

It would be easier if the polynomial was symmetrical to the origin, but if I had that as a given, I'd already know it has an x-intercept.


Maybe I am approaching this from the wrong angle. Obviously, a function with c0 = 0 has an x-intercept at the origin.
Let c0 be positive without loss of generality. Then we have our b=0 with f(b)=c0>0.
Then, according to highschool analysis, p*(x) := p(x - c0) should have an x-intercept.
p*(x) = c2n+1*(x-c0)2n+1 + (c2n*(x-c0)2n + c2n-1*(x-c0)2n-1 + ... + c1*(x-c0)1 + c0)

I suspect I can then  somehow show that all those "-c0" terms cancel the final "+c0" out, but I am stuck on how. Probably has something to do with the dreaded Pascal triangle.

Help?
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Skewbrow

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Odd degree polynomial over the reals? What can you say about its limits as x goes to +/- infinity?
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Loki

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...
Thanks.  :psyduck:

I don't remember what properties of polynomials we did or did not prove, so I might have to prove that your hint is applicable in a fairly roundabout way but that has definitely set me on the right track.
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ankhtahr

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Even though I feel bad for bothering you guys with so many exercise sheets of mine, I guess I don't have much of a choice. I really need the points now. (University thread for more info)

So yeah. After missing two lectures because of being sick, I'll have to turn in advanced mathematics tomorrow. The sheet is this one here. We only have to do the tasks with a (K), so 25 and 27.
Sorry Skewbrow, advanced mathematics, not linear algebra. But I have no idea how to even approach these tasks. A friend of mine told me that it would be simple calculating, but he's basically always a week ahead in exercise sheets and I don't know how. My other friends don't really know either. Tomorrow evening (it's evening now as well) we'll be sitting in the university again, trying to do these tasks.
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PthariensFlame

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Ok, so I might actually be able to help you here (thank you, Google Translate!).


25 is all about reducing expressions to simplest form; the expressions just happen to contain limits in them, but the limits are incidental to the overall goal.  Most of them require you to invoke l'Hôpital's rule, possibly multiple times:  if f and g are differentiable functions and there exists some k such that f(k) = g(k) = 0 or f(k) = g(k) = infinity, then limx->k f(x)/g(x) = limx->k f'(x)/g'(x).


27 just requires you to inspect the given functions f and g and determine, in each case, what subset of the function's domain it's continuous on.  For example, the absolute value function is continuous over its whole domain, but its derivative is not (it's missing 0).
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Loki

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As a semi-helpful hint for 27a, you should know by now that if two functions f(x) and g(x) are continuous in x0, then their combinations are also continuous in x0, ie
1) f(x0)+g(x0)
2) f(x0)-g(x0)
3) f(x0)*g(x0)
4) f(x0)/g(x0) (provided that g(x0) is not 0 of course)
are continuous.

27b: The rational numbers Q are dense in R (please don't beat me up as I wasn't able to find the correct translation for "liegen dicht in R"). Which means that for any number p which is in R\Q, there is a number q in Q which is infinitely close to it (and vice-versa). What values would the function take between, say, 1 and 1 plus some infinitely small ε? Does this look like a continuous function?
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The future is a weird place and you never know where it will take you.
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