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Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!

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Loki:
Off with your head.

ankhtahr:
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.

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

Here it is

I'm doomed.

Skewbrow:
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.

--- End quote ---

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:

Loki:
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.
--- End quote ---

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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