James' Test 1

Note: On hindsight, I should have covered more of Lagrange's theorem than how the Euclidean property and reflexivity impliews transitivity and symmetry, especially since I only showed how to prove a relation satisfies properties of an equivalence relations and never showed how to prove PROPERTIES from other properties. I think I will give a lecture on quantifiers and proving if an "arbitrary" element has the property, then they all do. I decided to remove one of these questions and let him do the other at a later date.

James' answer to 10 (which also includes the answer to 9): Okay. Lets say that xRy. By reflexivity, xRx, and by Euclidian Relations, because xRy and xRx, yRx. Next, say that yRz. By reflexivity, yRy, and by Euclidian Relations, because yRz and yRy, xRy. By Euclidian Relations, because yRz and yRx, zRx. By reflexivity, zRz, and by Euclidian Relations, because zRx and zRz, xRz. And because we know that xRy, yRz and xRz, we know that a relation that is reflexive and Euclidian is transitive.

GRADE:

Questions 1-8 + EC: 70/80!!!!!

Questions 9-10: 20/20 (done at a later date).

Total: 90/100 (Guess I got to make the test harder).

Comments: Overall, I am thoroughly impressed! He was able to get a counterexample that most of the students of Math 109 missed when I was their homework grader. I

1. Awesome. When I first saw this from dana, I immediately thought of groups. But I forgot to check for associativity. But since T has order 2 (tangling twice brings you back to identity, so RTT = R = TTR), the only real case we got to check is for the pattern you gave, rotation followed by tangle followed by rotate. 10/10

2. Good. 10/10

3. Spot on! 10/10

4. Good. Could have said 1/n for positive integer n, but primes also work since there are infinitely many. 10/10

EC. WHOA!!!! Did you come up with by yourself?! The Dyadic numbers, have you heard of them before? http://en.wikipedia.org/wiki/Dyadic_rational. Did you wikipedia this or did you really derive? Because if you did derive this by yourself...wow. Dude..I purposely gave you this question so you would get this wrong and can add it to the list of questions to ask Ravi. Because in the class I graded almost ALL the students on their 1 WEEK take home final got this wrong. And I am sure none of them could google the answer. If you did google the answer, at least you would have gotten a better score than so many Stanford students. Btw, you forgot to check closure under addition, but by changing denominators, this is still obvious. I am literally in disbelief (in a positive way). (Add 9 out of 10 possible points)

5. Good! I am guessing the first sentence is where you implied closure. 10/10.

6. Whoops, just realized 5 implies 6. 10/10.

7. For the arguement about the inverse, you are supposed to show x^-1 commutes with any element y of G given that x commutes with any element y of G. Your proof only showed that x^-1 commutes with x. The correct arguement would be: we know for any y in G and x in the center, xy = yx. Multiplying both sides on the LEFT by x^-1 gives us y = (x^-1)yx. With this new equation, multiplying both sides on the RIGHT by x^-1 gives us y(x^-1)= (x^-1)y. 8/10.

8. Circular reasoning: We are trying to prove there is a bijection between two finite sets, which is equivalent to saying they have the same size. The correct proof would have been: Since we are given a fixed x, consider the mapping for G to xG, f(y)=x*y. We can see the following: f is

One to One: if f(g) = f(w), by definition of x, xg=xw. Multiplying the left on both sides by the inverse (which exists since x is an element of a group), g = w.

Onto: For any element in xG, it is of the form xg_i for some g_i in G. But f(g_i) maps to the element; since I could have picked any element in xG, every element in xG can be mapped from f and an element of G.

However, you do have some of the right ideas about sizes and mappings here, so I will give you 3/10.

9-10, you were doing incredibly well, and I think you are confused about equivalence classes because of my notation (reflexivity means it is in its own class, so a in R(b) or a in R(c) is not an application of reflexivity. a in R(a) is an application though). I wil do question 9 for you, but I would like you to do question 10 very formally, and that will be graded for 9 and 10. I will teach you better notation than armstrong after this is done, because writing xRy instead of x is in R(y) is so much nicer.

Reflexivity means for any element x, x is in R(x) (so any element is in its own class.) Euclidean means for any elements x, y, and z, if x is in R(y) and x is in R(z), y is in R(z). I want to prove that for any element x and y, if x is in R(y), the y is in R(x).

Ok, assume x is in R(y). By reflexivity, x is in R(x). By Euclidean property, x is in R(y) and x is in R(x), y is in R(x). Since, x, y, and z were arbitrary elements, this holds for all x, y, and z.

9,10 update:

Trivial comment: Euclidian Relation is a single name.

Okay. Lets say that xRy. By reflexivity, xRx, and by Euclidian Relations, because xRy and xRx, yRx. Next, say that yRz. By reflexivity, yRy, -Good.

by Euclidian Relations, because yRz and yRy, xRy -small typo, yRz should be yRx. By the way, since you only assumed xRy, you just reproved reflexivity. Awesomeness.

By Euclidian Relations, because yRz and yRx, zRx. -Good.

By reflexivity, zRz, and by Euclidian Relations, because zRx and zRz, xRz. -good.

And because we know that xRy, yRz and xRz, we know that a relation that is reflexive and Euclidian is transitive. -GREAT

Good job and thanks for redoing this one. Spot on. Let me know if you have questions on homework 3.

P.S, how is precalc going? I just taught my first precalc class this friday, and I am guessing you are WAY ahead of the course.