# Tag Info

13

The contrapositive of the statement If $\overbrace{\text{$ab$and$a+b$have the same parity}}^{\large P}$, then $\overbrace{\text{$a$is even and$b$is even}}^{\large Q}$. is If $\overbrace{\text{$a$is odd or$b$is odd}}^{\large\lnot Q}$, then $\overbrace{\text{$ab$and$a+b$have different parities}}^{\large\lnot P}$. Note that $Q$ is the ...

9

Your result is an immediate consequence of the following proposition. Proposition. Suppose $X\subseteq Y$. Then $\mathscr P(X)\subseteq\mathscr P(Y)$. Proof. Let $E\in\mathscr P(X)$. Then $E\subseteq X\subseteq Y$ so that $E\subseteq Y$. Hence $E\in\mathscr P(Y)$. This proves $\mathscr P(X)\subseteq\mathscr P(Y)$. $\Box$ Do you see how your problem is now ...

8

\begin{align} |S| & = |S-T| + |S\cap T| \\[8pt] |T| & = |T-S| + |S\cap T| \end{align} If $|S-T|=|T-S|$, then the two right sides are the same, so the two left sides are the same. We can also write a proof explicitly dealing with bijections. You ask why one would "assume" a bijection exists. The bijection $g$ that you write about is not simply ...

7

You can proceed directly as follows: $2x = (x+y) + (x-y)$ which must be irrational as it is the sum of a rational and an irrational. So $x$ is irrational. Similarly $2y = (x+y) - (x-y)$ is irrational.

6

not induction, but maybe useful to note: firstly, since 3 is a prime, we have $n^3 \equiv_3 n$ (Fermat's little theorem) secondly $2n \equiv_3 -n$ (since $3n = 2n + n \equiv_3 0$) adding these two results: $$n^3 + 2n \equiv_3 n-n =0$$

6

By Spectral Theorem, $A$ is orthogonally similar to a diagonal matrix, i.e $$P^{T}AP=\pmatrix{\mu_1 \\ & \ddots \\ && \mu_n}$$ where $\mu_i>0$ is eigenvalue of $A$, and $\space P^{T}P=P^{-1}P=I$. For any $v$, let $v=Pu$. Then $$v^TAv=u^TP^TAPu=\sum_{k=1}^n\mu_ku_k^2>0$$

5

You need to show us some effort in the future. First, to show two sets are equal, we normally pick an element of the first set, show it is contained in the second, then pick an element in the second, and show it is contained in the first. If we suppose $x \in A$, then $x=2k$ for some integer $k$. Since $x = 2k$, $x = 2(k-1)+2$, and since $k-1$ is an ...

5

Take $N\in\Bbb N$ such that $|a_n-L|<1$ for $n\ge N$ and let $$M=\max\{|a_1|,\ldots,|a_N|,L+1\}.$$

5

Let $f:\{1,\ldots,n\}\to X$ be a surjection. Suppose, for the sake of contradiction, that $X$ has at least $n+1$ distinct elements $\{x_1,\ldots,x_{n+1}\}\subseteq X$. Since $f$ is a surjection, there exists, for each $i\in\{1,\ldots,n+1\}$, some $k_i\in\{1,\ldots,n\}$ such that $f(k_i)=x_i$. Since $f$ is a function and the $(x_i)_{i=1}^{n+1}$ are distinct, ...

4

We have that $F_n>F_{n-1}$ then $$F_{n+1}=F_n+F_{n-1}>2F_{n-1}>2\cdot2^{(n-1)/2}=2^{(n+1)/2}$$

4

HINT: For each $x\in X$, let $A_x=\{k\mid f(k)=x\}$. Then each $A_x$ is non-empty. Use that to construct an injection from $X$ into $\{1,\ldots,n\}$.

4

HINT: No, you can’t assume that $A=C$ and show that the inclusions hold: that’s the converse of what you’re supposed to prove, and an implication and its converse are not logically equivalent. Use the fact that $A\subseteq B$ and $B\subseteq C$ to show that $A\subseteq C$. You’re given that $C\subseteq A$, so the rest is straightforward.

4

The given inequality is equivalent to $a^3-a=a(a^2-1)>0$. By multiplying both sides by $a^2+1$, which is always positive, we get $a(a^2-1)(a^2+1)>a^2+1>0$, or $a^5-a>0$.

4

I would write it as: Let $P(n)$ stand for the expression: $$\forall x\leq n(x\not\in A)$$ Then use the assumption that $A$ has no least element to prove that $P(1)$ and $P(n)\implies P(n+1)$. Thus, we've shown that $\forall x:x\not\in A$, which means $A$ is empty. That's essentially the same as your proof, but uses less set notation.

4

Usually, a proof by contradiction of the statement $p \implies q$ is when you assume that the opposite of the desired conclusion is true (i.e., assume the negation of $q$ is true), and follow a few logical implications until you reach a statement that somehow explicitly or implicitly contradicts an initial assumption from the statement $p$. Meanwhile, a ...

4

The Lambert W-function is a function $W(z)$ which solves $z=W(z)e^{W(z)}$. It is a multi-valued function. In this case, you are trying to solve: $$e^{x\pi i/2} = x$$ of: $$\frac{-\pi i}{2}=\frac{-x\pi i}{2}e^{-x\pi i/2}$$ So $$x\frac{-\pi i}{2} = W(-\pi i/2)$$ or $$x =\frac{2i}{\pi} W\left(\frac{-\pi i}{2}\right)$$ I don't think you can do better ...

3

You won't be surprised to learn that in the last seventy-plus years since Tarski's book was first published in English, many other books have been appeared which will perhaps serve better as introductions to modern logic. And if you have downloaded my Teach Yourself Logic, you will have seen my "entry-level" suggestions on formal logic at the beginning of ...

3

You cannot go like this from $k=0$ to $k=1$ (i.e. $k=1$ cannot be expressed in the form $m+l$ as you wrote).

3

$N(t)$ is not equal to $N(t-s)+N(s)$, but $$N(t) = \Big(N(t) - N(s)\Big) + \Big(N(s)\Big)$$ and the two expressions inside the $\Big(\text{big parentheses}\Big)$ are independent of each other (whereas $N(t-s)$ and $N(s)$ are not independent of each other). So \begin{align} \Pr(N(s) = k \mid N(t) = n) & = \frac{\Pr(N(s)=k\ \&\ N(t) =n) ...

3

Yes, your work is all correct. Except a minor issue: the opposite of $|a_n| < \epsilon$ is $|a_n| \ge \epsilon$, not $|a_n| > \epsilon$. Instead of writing $P_n(X)$ as $|a_n| < \epsilon \; \forall n > X$, you may have found it clearer to write it as $\forall n > X \; |a_n| < \epsilon$. Then your entire statement would have been $$... 3 In general, if A\subseteq B, then \mathscr P (A)\subseteq \mathscr P (B) because every subset of A is a subset of B. More formally, if a\in \mathscr P (A), we need to show that a\in \mathscr P (B). But this is trivial, since if x\in a, then x\in B which implies that a\subseteq B which is the same as a\in \mathscr P (B). Now take ... 3 X\subset Y implies every element of X is an element of Y, so subsets of X are subsets of Y, so \mathcal{P}(X)\subset\mathcal{P}(Y). Finally, for Y=\mathcal{P}(X) you have \mathcal{P}(X)\subset\mathcal{P}(\mathcal{P}(X)). 3 Hint:$$\frac{a_{n+1}}{n}=\frac{a_{n+1}}{n+1}\frac{n+1}{n}.$$Or perhaps more to the point,$$\frac{a_{n+1}}{n+1}=\frac{a_{n+1}}{n}\frac{n}{n+1}.$$We've shown a_{n+1}/n\to l, we know n/(n+1)\to1, hence a_{n+1}/(n+1)\to l. And now this implies that a_n/n\to l. Given \epsilon>0 there exists N so |a_{n+1}/(n+1)-l|<\epsilon for all ... 3 If f(a)=c and f(b)=d, then$$\begin{align} \int_a^b f(x) \,\,dx+\int_c^d f^{-1}(y) \,\,dy &=\int_a^b f(x) \,\,dx+\int_a^b f^{-1}(f(x)) f'(x) \,\,dx\\\\ &=\int_a^b f(x) \,\,dx+\int_a^b x f'(x) \,\,dx\\\\ &=\int_a^b \left(f(x)+x f'(x)\right) \,\,dx\\\\ &=\int_a^b (xf(x))' \,\,dx\\\\ &=bf(b)-af(a)\\\\ &=bd-ac \end{align}$$Now, let ... 3 What would the proper negation look like? It turns out that, in this case, there are a number of ways you can go in how you want to prove this claim, not just via direct proof or contrapositive but also how you frame the question logically as well. I'll outline what I think is the clearest and easiest way of going about it. Claim: Let ... 3 You have the contrapositive right. You must negate P and Q separately and prove that the negation of Q implies the negation of P. To expand on this, for "a and b are even" to be false, you only need one of a and b to be odd, so the negation is "a is not even or b is not even". And for the statement "a+b and ab have the same parity" ... 3 Here is a simple proof that K(n) is not only exponential but 'super' exponential in the sense that for all constants C, there is some n_0 such that |K(n)|\geq C^n for all n\gt n_0. Let's rewrite your series as \sum_n\frac{a_n}{a_{n+1}} so that we don't run out of indices; in other words, K(n)=a_n. (For convenience's sake I'm going to take ... 3 y = \frac{3x^2+2y}{x^2+2}, multiplying both sides of the equation by x^2+2 results in an equivalent equation because that term is never 0 (in the reals at least). You end up with yx^2 + 2y = 3x^2+2y subtract 2y from both sides (always legitimate). yx^2=3x^2 Since x\neq 0 we can divide both sides by x^2 and get y=3 3 Proof: We first must note that \pi_j is the unique solution to \pi_j=\sum \limits_{i=0} \pi_i P_{ij} and \sum \limits_{i=0}\pi_i=1. Let's use \pi_i=1. From the double stochastic nature of the matrix, we have$$\pi_j=\sum_{i=0}^M \pi_iP_{ij}=\sum_{i=0}^M P_{ij}=1 Hence, $\pi_i=1$ is a valid solution to the first set of equations, and to make it a ...

3

You need to show independence of increments, i.e. if $0\le a<b<c<d$ then $(N_1(d)+N_2(d)) - (N_1(c)+N_2(c))$ is independent of $(N_1(b)+N_2(b)) - (N_1(a)+N_2(a))$, and similarly for more than two intervals. You can prove that by using independence of increments of each of the two processes separately plus independence of $N_1$ and $N_2$. You also ...

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