# Tag Info

8

Let $m=\max\{|(x+y)_i|\}$. Then the maximium is achieved for some index, $\nu$ say. Then $$m=|(x+y)_\nu|\stackrel{(1)}=|x_\nu+y_\nu|\stackrel{(2)}\le |x_\nu|+|y_\nu|\stackrel{(3)}\le \max\{|x_j|\}+\max\{|y_k|\}$$ where $(1)$ follows from the definition of vector addition, $(2)$ is the triangle inequality, and $(3)$ follows from the definition of maximum.

5

Here is a formal proof. Fix $1\leq i \leq n$, then it is always the case that $$|x_i|\leq \max\{|x_j|\}.$$ In particular, and using the triangle inequality, we have $$|x_i+y_i|\leq |x_i|+|y_i|\leq \max\{|x_j|\} + \max\{|y_k|\}.$$ Since we pick an arbitrary $i$, the previous inequality is true for all $1\leq i \leq n$, hence also true for the $i$ giving ...

5

If $n\ge 4$, then $n!$ is divisible by $4$, but $x^2+1$ cannot be. For $n=3$, $6$ is not of the form $x^2+1$. $n=0, 1, 2$ were addressed in a comment above.

4

Following my hint in the comments, one of the numbers $\{n,n+2, n+4\}$ must always be a multiple of $3$. However the hypotheses of the problem are that all three are prime. Hence one of them must be the specific prime $3$, as that is the only prime number that is also a multiple of $3$. So there are three cases: $n=3$. Excluded, since $p>3$ forbids ...

3

I'd use less symbols and more words. The aim is to show that, if $x\in D^c$, then $x\in A^c$. So, let $x\in D^c$. By hypothesis, $x\in B^c$, that is, $x\notin B$. Since $A\subseteq B$, we also have $x\notin A$ (otherwise $x\in B$). So $x\in A^c$. Your proof is good, anyway.

3

For the record, there is a very simple proof using the Myhill–Nerode criterion. Let $p_i$ be an enumeration of all primes. Since $0^{p_i} 1^{p_i} \notin L$ while $0^{p_j} 1^{p_i} \in L$ (for $j \neq i$), we see that the words $0^{p_i}$ are pairwise inequivalent with respect to $L$, and so $L$ is not regular. If you really need to use the pumping lemma, let ...

2

You're close. The basis case is ok. The problem is with the final line. It's an expansion of the reversal definition for each character. So while it is correct for any individual string, it implies that the whole proof for all strings is infinite in length, which is disallowed. The problem is fixed by using the inductive hypothesis which the induction ...

2

I think there's a way more simple and intuitive proof. First, as you observed, it is obvious that 2 is an upper bound. Now, to prove it is the supremum. Assume that $M$ is the supremum and $M<2$. Of course, $M>2$ is trivially impossible since $2$ is an upper bound as well and thus any $M$ bigger than $2$ cannot be a supremum. Now, let ...

2

It's quite tempting to use quantifiers and connectives when starting to write proofs. But nothing beats the clarity of plain English. Because you've got the basic idea, but the proof is not very well written. Instead, I'd suggest something of the following nature: Let $x\in D^c$ be an arbitrary element. By the assumption $D^c\subseteq B^c$ we have that ...

2

You need not be discouraged. Rubin often uses proofs that are elegant and concise, rather than easy-to-follow. I understand you desire to "see through" a problem, as you mentioned, but if you can at least follow what Rudin's proof is doing, then you are strengthening your mind for challenging proofs. In time, you will begin to see patterns in all the proofs ...

2

If it's alright, I'll just offer my own personal experience. Like you, I also studied real analysis out of Rudin as a high school junior. I have since learned that Rudin's text -- while very efficient -- is notoriously concise, and that there are far more student-friendly texts out there. Like you, I was often very frustrated with my inability to solve ...

2

The well ordering principle in this case is just the assertion that if you have a collection of integers with a particular number represented at most once then one of them is the smallest. As for s - 1 $\in$ N, it must be, because he just established that s is at minimum 2 and 2 - 1 = 1, which is in N. However, s - 1 $\in$ N is a more general statement and ...

2

The $(\Rightarrow)$ part is good. The $(\Leftarrow)$ part is wrong, unfortunately. You can't say that $abx=bx$. Assume $ab\mid c$; then, for some $x\in R$, $(ab)x=c$ and so $a(bx)=c$. Then $b(ax)=c$ and therefore $b\mid c$. Note that this part doesn't require $a$ is a unit.

2

In decimal basis, your number $n$ can be written as $n = d_0 + d_1 10 + \cdots + d_k 10^k \equiv d_0 + d_1 + \cdots + d_k \pmod3$. I used $10^m \equiv 1 \pmod 3$ for all natural $m$. Moreover, it follows that the sum of the digits of $n$ is the remainder of $n$ divided by 3.

2

Any isometry which is orientation-reversing and not a reflection is a glide reflection. Since any reflection is orientation-reversing, and any rotation is orientation-preserving, their combination is orientation-reversing as well. So essentially all you have to show is that the operation is no reflection. But if you want to follow the approach you already ...

1

The statement is not true. For example, let $S$ be the union of the open interval from $(0,0)$ to $(1,0)$ and the open interval from $(1,0)$ to $(2,0)$. Then $S$ is not convex but its closure, the closed interval from $(0,0)$ to $(2,0)$, is convex. A similar example works in any $\Bbb R^n$, even for $n=1$.

1

base case $n = 1 \Rightarrow 10 \equiv 1\mod 3$ holds. Assume that holds for $n = k$. Then for $n = k+1$, we have: $$10^{k+1} \mod 3 = (10\mod 3)*(10^{k}\mod 3) = (10^k \mod3) \equiv 1\mod 3$$ notice that $10^k$ holds by the induction hypothesis.

1

You're going about this the right way. Yes, pick any $x \in (-\infty,0) \cup [1,\infty)$. If $x \in (-\infty,0)$ then $x$ IS LESS THAN zero! So $[x,0)$ will work. :) For $x \in [1,\infty)$ just use something like $[1,x+1)$. I hope this helps!

1

I think that you should add $\alpha < 1$ in your statement. So, you just have to show that there exists at least one $x\in[0, 2\pi]$ such that this inqequality doesn't hold. Do you have any ideas? Just take the derivative of the $\sin(2014x)$.

1

I assume you want the proof explained in a little more detail. Here's what the proof says in English. Lets assume that conditions 1 and 2 hold. We use a proof by contradiction that it must be true for all n>=1. As with all proofs by contradiction, we assume the statement is false and then show it leads to a contradiction. So we assume there is some s for ...

1

$p$ must be odd and greater than $3$. Thus $$~~~~~p + 2 \equiv 2 \pm 1 \pmod 3$$ $$\Leftrightarrow p + 4 \equiv 1 \pm 1 \pmod 3$$ If the sign is plus, $p+2 \equiv 3 \equiv 0 \pmod 3$. If the sign is minus, $p+4 \equiv 0 \pmod 3$. In either case, one of these numbers is divisible by $3$, and since they can't be $3$ themselves by assumption, they cannot be ...

1

For individual components, you have $|x+y|_i\le |x_i|+|y_i|, 1\le i\le n$, thus, from the definition $|x+y|_i\le \max_j|x_j|+\max_k|y_k|\forall i\implies \max_i|x+y|_i\le \max_j|x_j|+\max_k|y_k|$

1

start with zero, the question is then the same as $$2^0+2^1+2^2+...+2^n=2^{n+1}-1$$ The binary for the LHS is $$11111\dots 1$$ with $n$ ones. add $1$ to get $$10000\dots 0,$$ with $n$ zeros in binary. This is the binary for $2^{n+1}$ and are done.

1

The original proofs in Mathematics came from the field of Geometry, so a good place to start is there. I recommend: Famous Problems in Geometry, and how to solve them by Benjamin Bold.

1

this has a computation-lite intuitive solution. think of the marbles as a squadron of $K$ tiny space-craft. suppose craft belonging to the same pile can communicate by walkie-talkie, but if any two are separated a long-range communications link is set up between them. so when a pile of $m+n$ craft is split into two piles of $m$ and $n$ respectively, this ...

1

You are correct in that they share the same proof method, but they are different in their approach. (1) is a direct proof, while (2) is a proof by contradiction. Let's take an example: proving that $1$ is the unique element of the set $A=\{1,2,4,6\}$ with the property that $x^3=1$. Method (1): Let $a,b\in A$ such that $a^3=1$ and $b^3=1$. Now we note ...

1

A function $f: A\to B$ naturally induces a function $\mathcal P(f):\mathcal P(A)\to\mathcal P(B)$ that will work. It's really easy, so try to think about it for a minute, but if you need more help: Another hint:

1

By assumption there is a bijection $f:\>A\to B$. We are asked to construct a bijective function $H:\>{\cal P}(A)\to{\cal P}(B)$. Your $H$ is coming out of thin air; therefore it's difficult to prove anything about it. For this construction we should use the given data, namely the information contained in $f$. How could we use this $f$ to produce a map ...

1

The decomposition into $xyz$ is dependent on what input string $0^p1^q$ you choose, therefore, $k$ is a function of $p$ and $q$, so you cannot just set $q = k+1$ afterwards. I would choose $0^p1^{(p-1)!}$ for some $p$ prime at least two greater than the pumping length, since then $xy^0z = xz = 0^k1^{(p-1)!}$ for $k<p$. This is possibly not the most ...

1

Substitute $\varepsilon = 1/n$, so we are looking at $$\lim_{n \to \infty} \int_{1/n}^n t^{x-1} e^{-t} \, dt$$ You probably want to show that you have a Cauchy sequence, viz., given $\delta>0$, there is an $M$ so that $$\left\lvert \int_{1/n}^{n} t^{x-1} e^{-t} \, dt - \int_{1/m}^{m} t^{x-1} e^{-t} \, dt \right\rvert < \delta$$ whenever ...

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