Problem 1 asks to show that the relation defined as follows is antisymmetric: Given a set the relation is defined on the subsets of by setting for iff where denotes set-theoretic difference of sets.

To show that is antisymmetric, we need to show that whenever are such that and then Suppose satisfy these assumptions. We need to show that they are equal as sets, i.e., that they have the same elements.

By definition, holds iff and holds iff Recall that if are sets, then It follows that iff every element of is also an element of and that iff every element of is also an element of But these two facts together mean precisely that and have the same elements, i.e., that as we needed to show.

Problem 2(a) of quiz 6 asks to consider the relation defined on by setting for iff and to show that is an equivalence relation. This means that is reflexive, symmetric, and transitive. Problem 2(a) of quiz 5 asks to show one of these properties.

To show that is reflexive, we need to show that for any we have that i.e., that But is certainly divisible by 5.

To show that is symmetric, we need to show that for any if it is the case that then it is also the case that Suppose then that This means that i.e., there is an integer such that But then showing that also i.e.,

To show that is transitive, we need to show that if and both and hold, then also holds. But if and then and But then it is certainly the case that Since this proves that, indeed or as we needed to show.

(If one feels the need to be somewhat more strict: That means that there is an integer such that Similarly, means that there is an integer such that But then showing that there is an integer such that namely, we can take )

Problem 2(b) asks to find all natural numbers such that where is as defined for problem 2(a).

That means exactly the same that Since and an integer is a multiple of 5 iff it ends in 5 or 0, we need to find all natural numbers such that ends in or Since and is even for all naturals we actually need to find all natural numbers such that ends in 8. For this, we only need to examine the last digit of the numbers and we find that these last digits form the sequence which is periodic, repeating itself each 4. This means that the numbers we are looking for are precisely i.e., the natural numbers of the form with

43.614000-116.202000

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I was working with one of students(Josue Gomez) in your class.
From problem 2 (b,)
I guess 5 divides 0 since 5*0 = 0. so need to say n is the form 4k + 3 where k is greater than or equal to 0.
I understand 0 is not natural number. I guess k can be 0 also.

Craig: For a while, there was some research on improving bounds on the number of variables or degree of unsolvable Diophantine equations. Unfortunately, I never got around to cataloging the known results in any systematic way, so all I can offer is some pointers to relevant references, but I am not sure of what the current records are. Perhaps the first pape […]

Yes. Consider, for instance, Conway's base 13 function $c$, or any function that is everywhere discontinuous and has range $\mathbb R$ in every interval. Pick continuous bijections $f_n:\mathbb R\to(-1/n,1/n)$ for $n\in\mathbb N^+$. Pick a strictly decreasing sequence $(x_n)_{n\ge1}$ converging to $0$. Define $f$ by setting $f(x)=0$ if $x=0$ or $\pm x_n […]

(1) Patrick Dehornoy gave a nice talk at the Séminaire Bourbaki explaining Hugh Woodin's approach. It omits many technical details, so you may want to look at it before looking again at the Notices papers. I think looking at those slides and then at the Notices articles gives a reasonable picture of what the approach is and what kind of problems remain […]

The description below comes from József Beck. Combinatorial games. Tic-tac-toe theory, Encyclopedia of Mathematics and its Applications, 114. Cambridge University Press, Cambridge, 2008, MR2402857 (2009g:91038). Given a finite set $S$ of points in the plane $\mathbb R^2$, consider the following game between two players Maker and Breaker. The players alternat […]

Yes. This is a consequence of the Davis-Matiyasevich-Putnam-Robinson work on Hilbert's 10th problem, and some standard number theory. A number of papers have details of the $\Pi^0_1$ sentence. To begin with, take a look at the relevant paper in Mathematical developments arising from Hilbert's problems (Proc. Sympos. Pure Math., Northern Illinois Un […]

It is easy to see without choice that if there is a surjection from $A$ onto $B$, then there is an injection from ${\mathcal P}(B)$ into ${\mathcal P}(A)$, and the result follows from Cantor's theorem that $B

Only noticed this question today. Although the selected answer is quite nice and arguably simpler than the argument below, none of the posted answers address what appeared to be the original intent of establishing the inequality using the Arithmetic Mean-Geometric Mean Inequality. For this, simply notice that $$ 1+3+\ldots+(2n-1)=n^2, $$ which can be easily […]

First of all, $f(z)+e^z\ne 0$ by the first inequality. It follows that $e^z/(f(z)+e^z)$ is entire, and bounded above. You should be able to conclude from that.

Yes. The standard way of defining these sequences goes by assigning in an explicit fashion to each limit ordinal $\alpha$, for as long as possible, an increasing sequence $\alpha_n$ that converges to $\alpha$. Once this is done, we can define $f_\alpha$ by diagonalizing, so $f_\alpha(n)=f_{\alpha_n}(n)$ for all $n$. Of course there are many possible choices […]

I disagree with the advice of sending a paper to a journal before searching the relevant literature. It is almost guaranteed that a paper on the fundamental theorem of algebra (a very classical and well-studied topic) will be rejected if you do not include mention on previous proofs, and comparisons, explaining how your proof differs from them, etc. It is no […]

Dr. Caicedo,

I was working with one of students(Josue Gomez) in your class.

From problem 2 (b,)

I guess 5 divides 0 since 5*0 = 0. so need to say n is the form 4k + 3 where k is greater than or equal to 0.

I understand 0 is not natural number. I guess k can be 0 also.

Thanks for your lecture!

Ei

The

natural numbersbegin with 0. That is the definition we are using. The integers that are larger than 0 are, naturally, thepositive integers.Oh thank you!