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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 ...


5

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


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

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

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

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

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).


2

No, it’s not true that every closed set in $\Bbb R$ with the lower limit topology is a union of sets of the form $\Bbb R\setminus[a,b)$. For example, $\{0,1\}$ is a closed set, because its complement is the open set $$\bigcup_{x<0}[x,0)\cup\bigcup_{0<x<1}[x,1)\cup\bigcup_{x>1}[x,x+1)\;.$$ Let $\tau$ be the lower limit topology. It’s not hard to ...


1

Hint So $L_n = \phi^n + \Phi^n$ and $L_{n-1} = \phi^{n-1} + \Phi^{n-1}$, and we have $$ L_{n+1} = L_n + L_{n-1} = \phi^n + \Phi^n + \phi^{n-1} + \Phi^{n-1} = \phi^{n-1} (\phi+1) + \Phi^{n-1} (\Phi+1) $$ and now use the fact that $\phi$ and $\Phi$ solve $x^2 = x+1$, so $\phi+1=\phi^2$ and $\Phi+1 = \Phi^2$. Can you finish the proof?


1

The standard way of proving this is to explicitly construct such a number. Here is one way one could do this: If $r = d-c <\sqrt{2}$ then first find (argue for the existence of) a rational number $x$ that is smaller than $r$. Now consider taking steps of length $x$ towards $\sqrt{2}$, i.e. consider the numbers $x,2x,3x,\ldots$. Show that sooner or ...


1

Define the famous $(q_n)$ by: $$q_0 = 1, \text{and} \ \forall \ n \in \mathbb N, q_{n+1} = \frac12 \left(q_n + \frac2{q_n} \right)$$ It can be shown that $\forall \ n$, $q_n \in \mathbb Q^{+}$ and that $q_n \to \sqrt2$. Then, for $\epsilon = d - c>0$, $\exists$ $n_0 \in \mathbb N$, $\forall$ $n \ge n_0$, $|q_n - \sqrt2| < d - c$. Take $q = ...


1

The topology you use on $C $ is the product topology, so basic open sets are products of an open set in $E $ and an open set in $\mathbb R $. We can write $$ C=\bigcup_{t>0}\{x:\ p (x)<t\}\times (-\infty,t). $$


1

As noted in the comments ( and in your add to the question) we have a simple geometric solution of this problem simply noting that the axis of two non aligned consecutive segments always intersect in a point that is the center of the circumference passing through the extremes of the segments . You can find also an analytical solution noting that a ...


1

This is equivalent to Cesaro's theorem holding for $a_n\rightarrow\infty$ in the original form, after relabeling. Suppose $S_n$ diverges to $+\infty$. Then for any $\epsilon>0$, there exists an $N(\epsilon)=N>0$ such that $S_n>\epsilon$ for all $n> N$. Then for $n>N$: $$b_n=\frac{1}{n}(S_1+\cdots +S_N)+\frac{1}{n}\sum_{k=N+1}^nS_k\geq ...


1

I'll try to sketch out a proof similar to the one you provided, and then I will flesh out why this approach does not work (as an aside, you may find this thread on fake induction proofs to be of use in picking up on faulty reasoning sometimes used in induction proofs). Claim: For every non-negative integer $n, 10n=0$. Base case: $10\cdot 0=0$. ...


1

If you're concluding that something is bijective because it's not one-to-one, then you don't know the definition of bijective. "Bijective" simply means both injective and surjective. A function that is not one-to-one, i.e. not injective, is not bijective. As far as determining the target space: that is given. The target space is the space to the right of ...


1

Looks correct to me. Your proof is very detailed and commendably rigorous—I think that's a good thing, given that the result might, at first sight, seem counterintuitive. To highlight the intuitive content of the result, let me present a less formal but perhaps more illuminating argument. Think of $\bigcap \mathcal G$ as the set of all points that satisfy ...



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