Problem 1 asks to determine (with brief justifications) the truth value of the following statements about integers:

1. is False. To show this we provide a counterexample: Specific integers such that For example,

2. is False. To show this we need to exhibit for each integer an integer such that For example, Note that, although is a fixed integer once we know we are not giving a fixed value of that serves as a simultaneous counterexample for all values of

3. is True. To show this we exhibit for each integer a specific integer such that For example: Note that, although is a fixed integer once we know we are not giving a fixed value of that works simultaneously for all

4. is True. To show this, we exhibit specific values of such that For example:

Problem 2 asks to show by contradiction that no integer can be both odd and even. Here is the proof: Suppose otherwise, i.e., there is an integer, let’s call it such that is both odd and even. This means that there are integers such that (since is odd) and (since is even).

Then we have that or But this is impossible, since 1 is not divisible by 2. We have reached a contradiction, and therefore our assumption that there is such an integer ought to be false. This means that no integer can be both odd and even, which is what we wanted to show.

Note that we have not shown that every integer is either odd or even. We will use mathematical induction to do this.

Problem 3 asks for symbolic formulas stating Goldbach’s conjecture and the twin primes conjecture (both are famous open problems in number theory).

Goldbach’s conjecture asserts that every even integer larger than 2 is sum of two primes:

Here, is the formula asserting that is even, namely, and is the formula (given in the quiz) asserting that is prime. Note we had to add existential quantifiers in order to be able to refer to the two prime numbers that add up to

The twin primes conjecture asserts that there are infinitely many primes such that is also prime.

The difficulty here is in saying “there are infinitely many,” since the quantifier only allows us to mention one integer at a time, and writing something of infinite length such as is not allowed.

We follow the suggestion given in the quiz, and represent “there are infinitely many with [some property]” by saying “for all there is a larger with [some property].”

Advertisements

Like this:

LikeLoading...

Related

This entry was posted on Friday, February 26th, 2010 at 1:24 pm and is filed under 187: Discrete mathematics. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

(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 […]

It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable. (This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here (Intern […]

I learned of this problem through Su Gao, who heard of it years ago while a post-doc at Caltech. David Gale introduced this game in the 70s, I believe. I am only aware of two references in print: Richard K. Guy. Unsolved problems in combinatorial games. In Games of No Chance, (R. J. Nowakowski ed.) MSRI Publications 29, Cambridge University Press, 1996, pp. […]

Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see here. What would be an explicit example of a number $r$ with this property? Short of an explicit example, are there any references addressing this question? A natural approach would […]

Suppose $M$ is an inner model (of $\mathsf{ZF}$) with the same reals as $V$, and let $A\subseteq \mathbb R$ be a set of reals in $M$. Suppose further that $A$ is determined in $M$. Under these assumptions, $A$ is also determined in $V$. The point is that since winning strategies are coded by reals, and any possible run of the game for $A$ is coded by a real, […]

The two concepts are different. For example, $\omega$, the first infinite ordinal, is the standard example of an inductive set according to the first definition, but is not inductive in the second sense. In fact, no set can be inductive in both senses (any such putative set would contain all ordinals). In the context of set theory, the usual use of the term […]

I will show that for any positive integers $n,\ell,k$ there is an $M$ so large that for all positive integers $i$, if $i/M\le \ell$, then the difference $$ \left(\frac iM\right)^n-\left(\frac{i-1}M\right)^n $$ is less than $1/k$. Let's prove this first, and then argue that the result follows from it. Note that $$ (i+1)^n-i^n=\sum_{k=0}^{n-1}\binom nk i^ […]

I think it is cleaner to argue without induction. If $n$ is a positive integer and $n\ge 8$, then $7n$ is both less than $n^2$ and a multiple of $n$, so at most $n^2-n$ and therefore $7n+1$ is at most $n^2-n+1

Let PRA be the theory of Primitive recursive arithmetic. This is a subtheory of PA, and it suffices to prove the incompleteness theorem. It is perhaps not the easiest theory to work with, but the point is that a proof of incompleteness can be carried out in a significantly weaker system than the theories to which incompleteness actually applies. It is someti […]