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.

The technique of almost disjoint forcing was introduced in MR0289291 (44 #6482). Jensen, R. B.; Solovay, R. M. Some applications of almost disjoint sets. In Mathematical Logic and Foundations of Set Theory (Proc. Internat. Colloq., Jerusalem, 1968), pp. 84–104, North-Holland, Amsterdam, 1970. Fix an almost disjoint family $X=(x_\alpha:\alpha

At the moment most of those decisions come from me, at least for computer science papers (those with a 68 class as primary). The practice of having proceedings and final versions of papers is not exclusive to computer science, but this is where it is most common. I've found more often than not that the journal version is significantly different from the […]

The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here. It is consistent (assuming large cardinals) that there is an extension of Le […]

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Stefan, "low" cardinalities do not change by passing from $L({\mathbb R})$ to $L({\mathbb R})[{\mathcal U}]$, so the answer to the second question is negative. More precisely: Assume determinacy in $L({\mathbb R})$. Then $2^\omega/E_0$ is a successor cardinal to ${\mathfrak c}$ (This doesn't matter, all we need is that it is strictly larger. T […]

The power of a set is its cardinality. (As opposed to its power set, which is something else.) As you noticed in the comments, Kurepa trees are supposed to have countable levels, although just saying that a tree has size and height $\omega_1$ is not enough to conclude this, so the definition you quoted is incomplete as stated. Usually the convention is that […]

The key problem in the absence of the axiom of replacement is that there may be well-ordered sets $S$ that are too large in the sense that they are longer than any ordinal. In that case, the collection of ordinals isomorphic to an initial segment of $S$ would be the class of all ordinals, which is not a set. For example, with $\omega$ denoting as usual the f […]

R. Solovay proved that the provably $\mathbf\Delta^1_2$ sets are Lebesgue measurable (and have the property of Baire). A set $A$ is provably $\mathbf\Delta^1_2$ iff there is a real $a$, a $\Sigma^1_2$ formula $\phi(x,y)$ and a $\Pi^1_2$ formula $\psi(x,y)$ such that $A=\{t\mid \phi(t,a)\}=\{t\mid\psi(t,a)\}$, and $\mathsf{ZFC}$ proves that $\phi$ and $\psi$ […]

Yes, the suggested rearrangement converges to 0. This is a particular case of a result of Martin Ohm: For $p$ and $q$ positive integers rearrange the sequence $$\left(\frac{(−1)^{n-1}} n\right)_{n\ge 1} $$ by taking the ﬁrst $p$ positive terms, then the ﬁrst $q$ negative terms, then the next $p$ positive terms, then the next $q$ negative terms, and so on. Th […]

Yes, by the incompleteness theorem. An easy argument is to enumerate the sentences in the language of arithmetic. Assign to each node $\sigma $ of the tree $2^{

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!